Equilibrium

Chemical Equilibrium

It is an experimental fact that most of the process including chemical reactions, when carried out in a closed vessel, do not go to completion. They proceed to some extent leaving considerable amounts of reactants & products. When such stage is reached in a reaction, it is said that the reaction has attained the state of equilibrium. Equilibrium can be defined as a state at which both the opposite processes take place with equal rate. 

Equilibrium represents the state of a process in which the properties like temperature, pressure, and concentration etc of the system do not show any change with passage of time. In all processes which attain equilibrium, two opposing processes are involved. Equilibrium is attained when the rates of the two opposing processes become equal. If the opposing processes involve only physical changes, the equilibrium is called Physical Equilibrium. If the opposing processes are chemical reactions, the equilibrium is called Chemical Equilibrium.

·       

Physical Equilibrium 

Physical equilibrium can be defined as the equilibrium in physical processes. The different types of physical Equilibrium are briefly described below

(a) Solid – liquid Equilibrium

The equilibrium that exist between ice and water is an example of solid – liquid equilibrium. In a close system, at  0oC ice and water attain equilibrium. At that point rate of melting of ice is equal to rate of freezing of water. The equilibrium is represented as

   H2O(s) ⇔ H2O(l) 

(b) Liquid – Gas Equilibrium

Evaporation of water in a closed vessel is an example of liquid – gas equilibrium. Where rate of evaporation is equal to rate of condensation. The equilibrium is represented as

H2O(l) ⇔ H2O(g)

(c) Solid – Solution Equilibrium

If you add more and more salt in water taken in a container of a glass and stirred with a glass rod, after dissolving of some amount. You will find out no further salt is going to the solution and it settles down at the bottom. The solution is now said to be saturated and in a state of equilibrium. At this stage, many molecule of salt from the undissolved salt go into the solution (dissolution) and same amount of dissolved salt are deposited back (Precipitation). Thus, at equilibrium rate of dissolution is equal to rate of precipitation.

Salt(Solid) ⇔ Salt(in solution)

(d) Gas –Solution equilibrium

Dissolution of a gas in a liquid under pressure in a closed vessel established a gas – liquid equilibrium. The best example of this type of equilibrium is cold drink bottles. The equilibrium that exists with in the bottle is

 CO2(g) ⇔ CO2(in solution)
  

Equilibrium in Chemical Process (Reversible and Irreversible Reactions)

A reaction in which not only the reactants react to form the products under certain conditions but also the products react to form reactants under the same conditions is called a reversible reaction.

Examples are

1.    3Fe(s) + 4H2O(g) ⇔ Fe3O4(s) + 4H2(g)

2.    CaCO3(s) ⇔ CaO(s) + CO2(g)

3.    N2(g) + 3H2(g) ⇔ 2NH3(g)

If a reaction cannot take place in the reverse direction, i.e. the products formed do not react to give back the reactants under the same condition it is called an irreversible reaction.

Examples are:

1.    AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(g)

2.    2M(g) + O2(g) → 2MgO(s)

Note: If any of the products will be removed from the system, reversible reaction will become irreversible one.

Generally, a chemical equilibrium is represented as

Where A, B are reactants and C, D are products.

Note:

The double arrow between the left hand part and right hand part shows that changes are taking place in both the directions.

On the basis of extent of reaction, before equilibrium is attained chemical reactions may be classified into three categories.

(a) Those reactions which proceed to almost completion.

(b) Those reactions which proceed to almost only upto little extent.

(c) Those reactions which proceed to such an extent, that the concentrations of reactants and products at equilibrium are comparable.

The equilibrium state is dynamic and not static in nature. A reaction is said to have attained equilibrium when the rate of forward reaction equals that of backward reaction

If all the reactants and products of any reaction under equilibrium are in same physical state, it is called a homogeneous equilibrium?. For example,

  N2(g) +  3H2(g)  2NH3 (g)
Here, all the reactants and products are in same phase

Heterogeneous equilibrium? is that type of equilibrium in which ?the state of one or more of the reacting species may differ i.e. all the reactants and products are not in same physical state.
2NaHCO3(s)  Na2CO3(s) + CO2(g) + H2O(l)?
Here, reactant is solid while in product all the solid, liquid and gaseous matter states are present.

Ionic Equilibrium

Chemical reactions mostly take place in solutions. Solution chemistry plays a very significant role in chemistry. All chemical substances are made up of either polar units (called ions) or non-polar units. The activity of these entities is more evident and pronounced in solution. The behaviour of these substances depends upon their nature and conditions of the medium in which they are added. It is therefore necessary to understand the principles that govern their behaviour in solution.

This type of equilibrium is observed in substances that undergo ionization easily, or in polar substances in which ionization can be induced. Ionic and polar substances are more easily soluble in polar solvents because of the ease of ionization taking place in the solvent medium. With the dissolution of ionic and polar substances in the solvent, these solutions become rich in mobile charge carriers (ions) and thus can conduct electricity. Substances, which are capable of conducting electricity are called as electrolytes while those substances which are non-conducting are called as non-electrolytes.

Characteristics of Equilibrium State

i) It can be attained only if the reversible reaction is carried out in closed vessel.

ii) It can be attained from either side of the reaction.

iii) A catalyst can hasten the approach of equilibrium but does not alter the state of equilibrium.

iv) It is dynamic in nature i.e. reaction does not stop but both forward and backward reactions take place at equal rate.

v) Change of pressure, concentration or temperature favours one of the reactions (forward or backward) resulting in shift of equilibrium point in one direction.

Introduction of Law of Mass Action and Equilibrium Constant

Guldberg and Waage established a relationship between rate of chemical reaction and the concentration of the reactants or, with their partial pressure in the form of law of mass action. According to this law,

“The rate at which a substance reacts is directly proportional to its active mass and rate of a chemical reaction is directly proportional to product of active masses of reactants each raised to a power equal to corresponding stoichiometric coefficient appearing in the balanced chemical equation”.

rate of reaction ∝ [A]a.[B]b

rate of reaction = K[A]a[B]b

where K is rate constant or velocity constant of the reaction at that temperature.

Unit of rate constant (K) = [moles/lit]1–n time–1 (where n is order of reaction.)

Note:   

For unit concentration of reactants rate of the reaction is equal to rate constant or specific reaction rate.

At equilibrium the rate of both forward and backward reactions become equal and after achieving equilibria, the concentration of reactants and products remains constant as shown in figure.

Note:

Active mass is the molar concentration of the reacting substances actually participating in the reaction.

Hence,

Active mass = number of moles/volume in litres

Active mass of solid is taken as unity.

Also, Active mass of reactant (a) = Conc. × activity coefficient

i.e. a = Molarity × f                        for dilute solution f = 1

Applying Law of mass action for general reversible reaction

aA + bB  cC + dD

Rate of forward reaction  [A]a[B]b

or Rf = Kf [A]b [B]b

Similarly for backward reaction               

Rb = Kb[C]c [D]d

At equilibrium Kf[A]a[B]b = Kb[C]c[D]d

The above equation is known as equilibrium equation and Kc is known as equilibrium constant.

The subscript ‘c’ indicates that Kc is expressed in concentrations of mol L–1
At a given temperature, the product of concentrations of the reaction products raised to the respective stoichiometric coefficient in the balanced chemical equation divided by the product of concentrations of the reactants raised to their individual stoichiometric coefficients has a constant value. This is known as the Equilibrium Law or Law of Chemical Equilibrium.

Equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the reaction in the forward direction.

Equilibrium Constant In terms of Partial Pressures (Kp)

For reactions involving gases, it is more convenient to express the equilibrium constant in terms of partial pressure.

The ideal gas equation is written as,

Where c or n/v is concentration expressed in mol/m3 (mol/L3) &  R= 0.0831 bar litre/mol K.

This indicates that at constant temperature, pressure of any gas is directly proportional to its concentration.

For reaction in equilibrium,

We can write,

Using Ideal Gas Equation we get,

PA = CART = [A] RT

PB = CBRT= [B] RT

Pc = CCRT = [C] RT

Pd = CDRT= [D] RT

Substituting the above values we get,

where n = (number of moles of gaseous products) – (number of moles of gaseous reactants) in the balanced chemical equation. The above given equation represents relationship between kc and kp.

Equilibrium Constant of a reaction involving Condensed Phase: The expression for the equilibrium constant of a reaction involving condensed phase (solid or liquid) and gaseous state is obtained by considering the concentrations or partial pressures of only the gaseous species. The concentration of condensed phases being constant are merged with the equilibrium constant.

Some facts about equilibrium constant

a) A very large value of KC or KP signifies that the forward reaction, as written, goes to completion or very nearly so.

b) A very small numerical value of KC or KP signifies that the forward reaction, as written, does not go to any significant extent.

c) A reaction is most likely to reach a state of equilibrium in which both reactants and products are present if numerical value of KC or KP is neither very large nor very small.

d) The equilibrium constant of a forward reaction and that of its backward reaction are reciprocal of each other.

e) If a chemical equation is multiplied by certain factor, its equilibrium constant must be raised to a power equal to that factor in order to obtain the equilibrium constant for the new reaction.

The Reaction Quotient “Q”

Consider the equilibrium

PCl5 (g)  PCl3(g) + Cl2 (g)

At equilibrium 

When the reaction is not at equilibrium this ratio is called ‘QC’ i.e., QC is the general term used for the above given ratio at any instant of time. And at equilibrium QC becomes KC.

Similarly, PCl2 PPCl2 / PPCl2 is called QP and at equilibrium it becomes KP.

If the reaction is at equilibrium, Q = Kc

A net reaction proceeds from left to right (forward direction) if Q < KC.

A net reaction proceeds from right to left (the reverse direction) if Q >Kc

Illustration: For the reaction,

A(g) + B(g)  2C(g) at 25°C, in a 2 litre vessel contains 1, 2, 3 moles of respectively. Predict the direction of the reaction if

(a) K­c for the reaction is 3

(b) Kc for the reaction is 6

(c) Kc for the reaction is 4.5

Solution:      

A(g) + B(g)  2C(g)

 Reaction quotient

Q = 

(a) Q > Kc,therefore backward reaction will be followed

(b) Q > Kc  The forward reaction is followed

(c) Q = Kc The reaction is at equilibrium
  

Significance of the Magnitude of Equilibrium Constant 

·      A very large value of KC or KP signifies that the forward reaction goes to completion or very nearly so.

·      A very small value of KC or KP signifies that the forward reaction does not occur to any significant extent.

·      A reaction is most likely to reach a state of equilibrium in which both reactants and products are present if the numerical value of Kc or KP is neither very large nor very small.

Units of Equilibrium Constant

As we have learned that numerical value of equilibrium constant is a function of stoichiometric coefficient used for any balanced chemical equation.

But, what will happen in the size of units of equilibrium constant when the stoichiometric coefficient of the reaction changes? (If sum of the exponent for product is not equal to reactant side sum)

So for the sake of simplicity in the units of Kp for Kc the relative molarity or pressure of reactants and products are used with respect to standard condition.

(For solution standard state 1 mole/litre and for gas standard pressure = 1 atm) 

Now the resulting equilibrium constant becomes unitless by using relative molarity and pressure.

Illustration . The value of Kp for the reaction 2H2O(g) + 2Cl2(g) 4HCl(g) + O2(g) is  0.035 atm at 400oC, when the partial pressures are expressed in atmosphere. Calculate Kc for the reaction,        

1/2O2(g) + 2HCl(g) —> Cl2(g) + H2O(g)

Solution: 

KP = KC (RT)Δn

Δn = moles of product - moles of reactants = 5 - 4 = 1

R = 0.082 L atm/mol K,

T = 400 + 273 = 673 K

0.035 = KC (0.082 ´ 673)

KC = 6.342 ´ 10-4 mol l-1

KC for the reverse reaction would be 1/KC

KC = 1/6.342×10–4  = 1576.8 (mol l-1)-1

When a reaction is multiplied by any number n (integer or a fraction) the K'C or K'pbecomes (KC)n or (KP)n of the original reaction.

KC for 1/2 O2(g) + 2HCl(g) ——> Cl2(g) + H2O(g)

 is √1576.8 = 39.7 (mol.l-1)-½

Illustration . Kp for the equilibrium, FeO(s) + CO(g) Fe(s) + CO2(g) at 1000oC is 0.4. If CO(g) at a pressure of 1 atm and excess FeO(s) are placed in a container at 1000oC, what are the pressures of CO(g) and CO2(g) when equilibrium is attained?

Solution: Assuming ideal gas behaviour, partial pressures are proportional to the no. of moles present. Since moles of CO2 formed equals moles of CO consumed, the drop in partial pressure of CO will equal the partial pressure of CO2 produced. Let the partial pressure of CO2 at equilibrium be ‘x’ atm. Then, partial pressure of CO will be (1 - x) atm.

Since  K = pCO2 / pco = x / 1 – x = 0.4  =>   x = 0.286

Hence PCO = 1-x = 0.714 atm

Le Chatelier’s Principle

“When an equilibrium is subjected to either a change in concentration, temperature or, in external pressure, the equilibrium will shift in that direction where the effects caused by these changes are nullified”.

Refer to the following Video for Le Chatelier’s Principle

This can be understood by the following example. Overall we can also predict the direction of equilibrium by keeping in mind following theoretical assumption.

PCl5  ——> PCl3 + Cl2

Let us assume that we have this reaction at equilibrium and the moles of Cl2, PCl3 and PCl5 at equilibrium are a, b and c respectively, and the total pressure be PT.

Since

PT  =  (a+b+c) RT / V  

KP = abRT / cV  

Now if d moles of PCl3 is added to the system, the value of Q would be, a(b+d)RT / cV 

We can see that this is more than KP. So the system would move reverse to attain equilibrium.

If we increase the volume of the system, the Q becomes  abRT / cV' where V' > V. Q becomes less, and the system would move forward to attain equilibrium.

If we add a noble gas at constant pressure, it amounts to increasing the volume of the system and therefore the reaction moves forward.

If we add the noble gas at constant volume, the expression of Q remains as Q = abRT / cV and the system continues to be in equilibrium. Nothing happens.

Therefore for using Le-Chatlier’s principle, convert the expression of KP and KC into basic terms and then see the effect of various changes.

Effect of Change of Concentration on Equilibrium

Let us have a general reaction,

 aA + bB  cC + dD

at a given temperature, the equilibrium constant,

Kc = [C]c[D]d / [A]a[B]b 

again if α, β, γ and δ are the number of mole of A, B, C and D are at equilibrium

then, Kc = [γ]c[δ]d / [α]a[β]b 

If any of product will be added, to keep the Kc constant, concentration of reactants will increase i.e. the reaction will move in reverse direction. Similarly if any change or disturbance in reactant side will be done, change in product’s concentration will take place to minimise the effect.
  

Effect of Chenage of Temperature on Equilibrium

The effect of change in temperature on an equilibrium cannot be immediately seen because on changing temperature the equilibrium constant itself changes. So first we must find out as to how the equilibrium constant changes with temperature.

For the forward reaction, according to the Arrhenius equation,

And for the reverse reaction,

R = Reactant, P = Product

It can be seen that

For any reaction, ΔH = Eaf – Ear

From the equation,

logK2/K1 = ΔHo / 2.303R (1/T1 – 1/T2) it is clear that                                      

(a) If ΔHo is +ve (endothermic), an increase in temperature (T2 > T1) will make K2 > K1, i.e., the reaction goes more towards the forward direction and vice-versa.

(b) If ΔHo is -ve (exothermic), an increase in temperature (T2 > T1), will make K2 < K1 i.e., the reaction goes in the reverse direction.

Increase in temperature will shift the reaction towards left in case of exothermic reactions and right in endothermic reactions. 

Increase of pressure (decrease in volume) will shift the reaction to the side having fewer moles of the gas; while decreases of pressure (increase in volume) will shift the reaction to the side having more moles of the gas.

If no gases are involved in the reaction higher pressure favours the reaction to shift towards higher density solid or liquid.

Refer to the following Video for Arrhenius equation

Example . Under what conditions will the following reactions go in the forward direction

(i)  N2(g)+ 3H2(g)  2NH3(g) + 23 k cal.

(ii) 2SO2(g) + O2(g)    2SO3(g) + 45 k cal.

(iii) N2(g) + O2(g)  2NO(g) - 43.2 k cal.

(iv) 2NO(g) + O2(g)   2NO2(g) + 27.8 k cal.

(v) C(s) + H2O(g)   CO2(g) + H2(g) + X k cal.

(vi) PCl5(g)  PCl3(g) + Cl­2(g)- X k cal.

(vii) N2O4(g)  2NO2(g) - 14 k cal.

Solution:     

(i)  Low T, High P, excess of N2 and H2. 

(ii) Low T, High P, excess of SO2 and O2.

(iii) High T, any P, excess of N2 and O2         

(iv) Low T, High P, excess of NO and O2

(v) Low T, Low P, excess of C and H2O  

(vi) High T, Low P, excess of PCl5

(vii) High T, Low P, excess of N2O4.

Example: The equilibrium constant KP for the reaction N2(g) + 3H2(g)  2NH3(g)  is 1.6 × 10-4atm-2 at 400oC. What will be the equilibrium constant at 500oC  if heat of the reaction in this temperature range is-25.14 k cal?

Solution: Equilibrium constants at different temperature and heat of the reaction are related by the equation,  

log KP2 = –25140 / 2.303 × 2 [773 – 673 / 773 × 673]  + log 1.64 10-4

log KP2 = –4.835

KP2 = 1.462 × 10–5 atm–2
  

Effect of Change of Pressure on Equilibrium

The effect of change of pressure on chemical equilibrium can be done by the formula.

Kp = QP × P(Δn)

Case A: When Δn = 0

Kp = Qp

And equilibria is independent of pressure.

Case B: When Δn = – ve

Kp = QP × P–Δn = QP / PΔn

Here with increase in external pressure, will shift the equilibrium towards forward direction to maintain K­p. Similarly, decrease in pressure will shift the equilibrium in the reverse direction.

Case C: When Δn = +ve

 Kp = QP × P–Δn and, effect will be opposite to that of Case ‘B’
  

Effect of Addition of Inert Gases to a Reaction at Equilibrium

1.   Addition at constant pressure

Let us take a general reaction

 aA + bB  cC + dD

We know,

Where,

 nC nD, nA, nB denotes the no. of moles of respective components and PT is the total pressure and n = total no. of moles of reactants and products.

Now, rearranging,

Where n = (c + d) – (a + b)

Now, n can be = 0, < 0 or > 0

Lets take each case separately.

a)   n = 0 : No effect

b)   n = ‘+ve’ : Addition of inert gas increases the  n i.e. is decreased and so is . So products have to increase and reactants have to decrease to maintain constancy of Kp. So the equilibrium moves forward.

c) n = ‘–ve’  In this case  decreases but increases. So products have to decrease and reactants have to increase to maintain constancy of Kp. So the equilibrium moves backward.

2. Addition at Constant Volume : Since at constant volume, the pressure increases with addition of inert gas and at the same time ån also increases, they almost counter balance each other. So  can be safely approximated as constant. Thus addition of inert gas has no effect at constant volume.
  

Effect of Catalyst on Equilibrium

Catalysts are the substances which  alter the rate of a reaction without themselves getting consumed in the reaction.  Since the catalyst is associated with forward as well as reverse direction reaction. So, at equilibrium, rate of forward reaction will be equal to rate of reverse reaction and hence catalyst effect will be same on both forward as well as reverse. Hence catalyst never effect the point of equilibrium but it reduces the time to attain the equilibrium. 

Example: What will be the effect on the equilibrium constant on increasing the temperature. 

Solution: Since the forward reaction is endothermic, so increasing the temperature, the forward reaction is favoured. Thereby the equilibrium constant will increase.

Sample problem Kp for the reaction 2BaO2(s)  2BaO(s) + O2(g) is 1.6 × 10–4 atm, at 400°C. Heat of reaction is – 25.14 kcal. What will be the no. of moles of O2 gas produced at 500°C temperature, if it is carried in 2 litre reaction vessel?

Solution:   We know that

log LP2/KP1 = ΔH / 2.303R [1/T1 – 1/T2] 

log KP2/1.6 × 10–4 = –25.14 / 2.330 × 2 × 10–3 [773 – 673 / 773 × 673] 

=> KP2/1.46 × 10–5 atm 

KP2 = p02 = 1.46 × 10–5 atm 

Since, PV = nRT

1.46 × 10–5 × 2 = n × 0.0821 × 773 

 n = 1.46 × 10–5 / 0.0821 × 773 = = 4.60 × 10-7  

Degree of Dissociation (α)

The degree of dissociation of a substance is defined as the fraction of its molecules dissociating at a given time.

Let us consider the reaction,

2NH3 (g)  N2 (g) + 3H2 (g)

Let the initial moles of NH3(g) be ‘a’. Let  x moles of NH3 dissociate at equilibrium.

Degree of dissociation (a) of NH3 is defined as the number of moles of NH3 dissociated per mole of NH3.

if x moles dissociate from ‘a’ moles of NH3, then, the degree of dissociation of NH3 would be x/a.

We can also look at the reaction in the following manner.

In this way you should calculate the basic equation. So my advice to you is that, while solving problem follow the method given below:

·      Write the balanced chemical reaction (mostly it will be given)

·      Under each component write the initial no. of moles.

·      Do the same for equilibrium condition.

·      Then derive the expression.
  

Dependence of Degree of Dissociation from Density Measurements

?Refer to the following video for Vapour density and degree of dissociation

The following is the method of calculating the degree of dissociation of a gas using vapour densities. This method is valid only for reactions whose KP exist, i.e., reactions having at least one gas and having no solution. 

Since 

pV = nRT 

pV = nRT

Since

P = NRT / V

For a reaction at equilibrium V is a constant and ρ is a constant.   vapour Density α 1/n

 (molecular weight = 2 × V.D)

Here  M = molecular weight initial

m = molecular weight at equilibrium

Let us take a reaction

                        PCl5        PCl3    +   Cl2

Initial moles      C                      0                0

At eqb.            C(1-α)              Cα              Cα

∴  = C(1+α) / C = D/d;     1 + α = D/d = M/m 

Knowing D and d, α can be calculated and so for M and m.
  

Thermodynamics of Chemical Equilibrium

Let ΔG0 be the difference in free energy of the reaction when all the reactants and products are in the standard state and Kc or, Kp be the thermodynamic equilibrium constant of the reaction. Both related to each other at temperature T by the following relation. 

ΔG0 = – 2.303 RT logKc and ΔG0 = – 2.303 RT log Kp(in case of ideal gases) 

Now, we know that thermodynamically,

ΔG0 = ΔH0 – TDS0

 here ΔH0 is standard enthalpy of reaction, and ΔS0 is standard entropy change 

(i) When ΔG0 = 0, then, Kc = 1

(ii) When, ΔG0 > 0, i.e. +ve, then Kc < 1, in this case reverse reaction is feasible showing thereby a less concentration of products at equilibrium rate.

(iii) When ΔG0 < 0, i.e. -ve, then, Kc > 1; In this case, forward reaction is feasible showing thereby a large concentrations of product at equilibrium state.

At equilibrium, 

Example 1.  When PCl5 is heated it dissociates into PCl3 and Cl2. The density of the gas mixture at 200oC and at 250oC is 70.2 and 57.9 respectively. Find the degree of dissociation at 200oC and 250oC.

Solution: PCl5(g) →  PCl3(g) + Cl2(g)

We are given the vapour densities at equilibrium at 200oC and 250oC.

The initial vapour density will be the same at both the temperatures as

it would be  MPCl5 / 2. 

∴ Initial vapour density = (31 + 5 × 35.5) / 2 = 104.25

Vapour density at equilibrium at 200oC = 70.2

∴ Total moles at equilibrium / Total moles initial  = 1 + α = Vapour density initial / Vapour density at equilibrium     = 104.25 / 70.2 = 1.485

 ∴  a = 0.485

At 250oC,   1 + α = 104.25 / 57.9 = 1.8

 ∴  a = 0.8

Sample problem. 

The degree of dissociation of PCl5 is 60%, then find out the observed molar mass of the mixture.

Solution:     PCl5(g)  PCl3(g) + Cl2(g)

                  Initially moles           1                  0     0

                  At eqm.                 1 – α               α         α

 Where α = degree of dissociation = 0.6

Total moles at equilibrium = 1 + a = observed mole

 1 + a = 206.5 / molecular weight observed 

 molecular weight observed = 206.5 / 1 + a = 206.5 / 1.6 = 129.06

Example 2. A system is in equilibrium as  PCl5 + Heat —> PCl3 + Cl2

Why does the temperature of the system decreases, when PCl3 are being removed from the equilibrium mixture at constant volume?

Solution:  When PCl3 are being removed from the system, the reaction moved to the right. This consumed heat and therefore, temperature is decreased.

Example. NO and Br2 at initial partial pressures of 98.4 and 41.3 torr, respectively, were allowed to react at 300K. At equilibrium the total pressure was 110.5 torr. Calculate the value of the equilibrium constant and the standard free energy change at 300K for the reaction 2NO(g) + Br2(g) 2NOBr(g).

Solution: 

2NO(g) + Br2(g) →  2NOBr(g)

 Initial pressure         98.4      41.3                 0         

At eqlbm.                98.4-x       41.3 - x/2      x 

The total pressure at equilibrium is 110.5 torr. 

 98.4-x + 41.3- x/2 + x = 110.5 

 x = 58.4 torr

Now, 1 atm = 760.4 torr,    

x = 7.68 × 10-2 atm. 

pNOBr = 7.68 × 10-2 atm ;     pNO = 98.4-x = 40 torr = 5.26 × 10-2 atm. 

pBr2 = 41.3 – x/2 = 12.1 torr = 1.59 × 10-2 atm. 

ΔGo = -2.303 RT log K = -2.303 (1.99) (300) (log 134)  

 = -2.92 k cal = -12.2 kJ.

[If R is used as 1.99 cal/mol K, then ΔGo will be in cal. If R is used as 8.314 J/mol K, then ΔGo will be in Joules. But KP must always be in (atm)Δn.]

Acids and Bases

The earliest criteria for the characterization of acids and bases were the experimentally observed properties of aqueous solutions. An acid  was defined as a substance whose water solution tastes sour, turns blue litmus red, neutralizes bases and so on. A substance was a base if its aqueous solution tasted bitter, turns red litmus blue, neutralizes acids and so on.

Faraday termed acids, bases and salts as electrolytes and Liebig proposed that acids are compounds containing hydrogen that can be replaced by metals. Different concepts have been put forth by different investigators to characterize acids and bases but the following are the three important modern concepts of acids and bases:
  

·      Arrhenius Concept

 According to Arrhenius concept all substances which give H+ ions when dissolved in water are called acids while those which ionize in water to furnish OH- ions are called bases.

 HA   H+ + A- (Acid)

BOH  B+ + OH- (Base)

Thus, HCl is an acid because it gives H+ ions in water. similarly, NaOH is a base as it yields OH- ions in water.

HCl  H+ + Cl-

NaOH   Na+ + OH-

Some acids and bases ionize completely in solutions and are called string acids and bases. Others are dissociated to a limited extent in solutions and are termed weak acids and bases. HCl, HNO3, H2SO4, HCIO4, etc., are examples of strong acids and NaOH, KOH, (CH3)4NOH are strong bases.

Every hydrogen compound cannot be regarded as an acid, e.g., CH4 is not an acid. Similarly, CH3OH, C2H5OH, etc., have OH groups but they are not bases, they must generate H+ or OH- ions in aqueous solution in order to be defined as acid or base

Actually free H+ ions are highly reaction so they do not exist in water. They combine with water molecules to form hydronium ion (H3O+) molecules, i.e., have strong tendency to get hydrated.

 HX + H2O  H3O+ + X-
                    (Hydronium ion)

The proton in aqueous solution is generally represented as H+ (aq). It is now known that almost all the ion are hydrated to more or less extent and it is customary to put (aq) after each ion.

The oxides of many non-metals react with water to form acids and are called acidic oxides or acid anhydrides.

CO2 + H2O  H2CO3  2H+(aq) +  (aq)

 N2O5 + H2O  2HNO3  2H+(aq) +  (aq)

Many oxides of metals dissolve in water to form hydroxides. Such oxides are termed basic oxides.

Na2O + H2O →  2NaOH    2Na+(aq) + 2OH- (aq)

The substance like NH3 and N2H4 act as bases as they react with water to produce OH- ions.

NH3 + H2O →  NH4OH    NH+4 (aq) + OH- (aq)

The reaction between an acid and a base is termed neutralization. According to Arrhenius concept, the neutralization in aqueous solution involves the reaction between H+ and OH- ions or hydronium and OH-. This can be represented as

H3O+ + OH-   2H2O
  

Limitations of Arrhenius Concept

1.    For the acidic or basic properties, the presence of water is absolutely necessary. Dry HCl shall not act as an acid. HCl is regarded as an acid only when dissolved in water and not in any other solvent.

2.    The concept does not explain acidic and basic character of substances in non-aqueous solvents.

3.    The neutralization process is limited to those reactions which can occur in aqueous solutions only, although reactions involving salt formation do occur in the absence of solvent.

4.    It cannot explain the acidic character of certain salts such as AlCl3 in aqueous solution.

5.    An artificial explanation is required to explain the basic nature of NH3 and metallic oxides and acidic nature of non-metal oxides.
  

·      Bronsted-Lowry Concept - The Proton-Donor-Acceptor Concept

Refer to the following video for Bronsted-Lowry concept 

In 1923, Bronsted and Lowry independently proposed a broader concept of acids and bases. According to Bronsted-Lowry concept an acid is a substance (molecule or ion) that can donate proton, i.e., a hydrogen ion, H+, to some other substance and a base is a substance that can accept a proton from an acid. More simply, an acid is a proton donor (protogenic) and a base is a proton acceptor (protophilic).

Consider the reaction,

 HCl + H2O   H3O + Cl-

In this reaction HCl acts as an acid because it donates a proton to the water molecule. Water, on the other hand, behaves as a base by accepting a proton from the acid.

The dissolution of ammonia in water may be represented as

NH3 + H2O  NH4+ + OH-

In this, reaction, H2O acts as an acid it donated a proton to NH3 molecule and NH3molecule behaves as a base as it accepts a proton.

When an acid loses a proton, the residual part of it has a tendency to regain a proton. Therefore, it behaves as a base.

 Acid  H+ + Base

The acid and base which differ by a proton are known to form a conjugate pair. Consider the following reaction.

 CH3COOH + H2O   H3O+ + CH3COO-

It involves two conjugate pairs. The acid-base pairs are:

Such pairs of substances which can be formed form one another by loss or gain of a proton are known as conjugate acid-base pairs.

If in the above reaction, the acid CH3COOH is labelled acid1 and its conjugate base, CH3COO- as base1. H2O is labelled as base2 and its conjugate acid H3O+ as acid2, the reaction can be written as:

Acid1 + Base2  Base1 + Acid2

Thus, any acid-base reaction involves two conjugate pairs, i.e., when an acid reacts with a base, another acid and base are formed. 

Thus, every acid has its conjugate base and every base has its conjugate acid. It is further observed that strong acids have weak conjugate bases while weak acids have strong conjugate bases.

HCl        +           Cl-                CH3COOH     +   CH3COO-

Strong acid   Weak base      Weak acid    Strong base

There are certain molecules which have dual character of an acid and a base. These are called amphiprotic or atmospheric.

Examples are NH3, H2O, CH3COOH, etc.

The strength of an acid depends upon its tendency to lose its proton and the strength of the base depends upon its tendency to gain the proton.

Acid-Base chart containing some common conjugate acid-base pairs

Acid

Conjugate base

KClO4

(Perchloric acid)

(Perchlorate ion)

H2SO4

(Sulphuric acid)

(Hydrogen sulphate ion)

HCl

(Hydrogen chloride)

Cl-

(Chloride ion)

HNO3

(Nitric acid)

(Nitrate ion)

H3O+

(Hydronium ion)

H2O

(Water)

(Hydrogen sulphate ion)

(Sulphide ion)

H3PO4

(Ortho phosphoric acid)

H2

(Dihydrogen phosphate ion)

CH3COOH

(Acetic acid)

Ch3COO-

(Acetate ion)

H2CO3

(Carbonic acid)

(Hydrogen carbonate ion)

H2S

(Hydrogen sulphide)

(Hydorsulphide ion)

NH3+

(Ammonium ion)

NH3

(Ammonia)

HCN

(Hydrogen cyanide)

CN-

(Cyanide ion)

C6H5OH

(Ethyl alcohol)

C6H5O-

(Phenoxide ion)

NH3

(Ammonia)

OH-

(Hydroxide ion)

CH4

(Methane)

C2H5O-

(Ethoxide ion)

In acid-base strength series, all acids above H3O+ in aqueous solution fall to the strength of H3O+. Similarly the basic strength of bases below OH- fall to the strength of OH- in aqueous solution. This is known as leveling effect.

The strength of an acid also depends upon the solvent. The acids HCIO4, H2S04, HCl and HN03 which have nearly the same strength in water will be in the order of HClO4> H2SO4 > HCl > HNO3 in acetic acid, since the proton accepting ten­dency of acetic acid is much weaker than water. So the real strength of acids can be judged by solvents. On the basis of proton interaction, solvents can be classified into four types: 

·      Protophilic solvents:  Solvents which have greater tendency to accept protons, i.e., water, alcohol, liquid ammonia, etc.

·      Protogenic solvents:  Solvents which have the tendency to produce protons, i.e., water, liquid hydrogen chloride, glacial acetic acid, etc.

·      Amphiprotic solvents:  Solvents which act both as protophilic or protogenic, e.g., water, ammonia, ethyl alcohol, etc.

·      Aprotic solvents:  Solvents which neither donate nor accept protons, e.g., benzene, carbon tetrachloride, carbon disulphide, etc.

HCI acts as acid in H3O+, stronger acid in NH3, weak acid in CH3COOH, neutral in C6H6and a weak base in HF.

HCI   +   HF     →    H2Cl+  +   F-

Base      Acid          Acid             Base
  

·      Lewis Concept of Acids and Bases

Refer to the following video for Lewis concept of acids and bases

This concept was proposed by G.N. Lewis, in 1939. Accord­ing to lewis definition of acids and bases base is defined as a substance which can furnish a pair of electrons to form a coordinate bond whereas an acid is a substance which can accept a pair of electrons. The acid is also known as electron acceptor or electrophile while the base is electron donor or nucleophile.

A simple example of an acid-base is the reaction of a proton with hydroxyl ion.

Acid    Base

Some other examples are:

H3N: + BF3 = H3N → BF3

Base   Acid

H+ +: NH3 = [H ← NH3]+

Acid Base

BF3 + [F]- = [F → BF3]+

Acid  Base

Lewis concept is more general than the Bronsted Lowry concept.

According to Lewis concept, the following species can act as Lewis acids.

1.    Molecules in which the central atom has incomplete octet: All compounds having central atom with less than 8 electrons are Lewis acids, e.g., BF3, BC13, A1C13, MgCl2. BeCL. etc.

2.    Simple cations: All cations are expected to act as Lewis acids since they are deficient in electrons. However, cations such as Na+, K+, Ca2+, etc., have a very little tendency to accept electrons, while the cations like H+, Ag+, etc., have greater tendency to accept electrons and, therefore, act asLewis acids.

3.    Molecules in which the central atom has empty d-orbitals: The central atom of the halides such as SiX4, GeX4, TiCl4, SnX4, PX3, PF5, SF4, SeF4, TeCl4, etc., have vacant d-orbitals. These can, therefore, accept an electron pair and act as Lewis acids.

4.    Molecules having a multiple bond between atoms of dissimilar electronegativity: Typical examples of molecules falling in this class of Lewis acids are C02, S02 and S03. Under the influence of attacking Lewis base, one -electron pair will be shifted towards the more negative atom.
 
  

Lewis Acid &  Lewis Base

The following species can act as Lewis bases.

1.    Neutral species having at least one lone pair of electrons: For example, ammonia, amines, alcohols, etc., act as Lewis bases because they contain a pair of electrons
   

2.    Negatively charged species or anions: For example, chloride, cyanide, hydroxide ions, etc., act as Lewis bases.CN-,   CI-,  OH-

3.    It may be noted that all Bronsted bases are also Lewis bases but all Bronsted acids are not Lewis acids.

Limitations of Lewis Concept of Acid and Base

Since the strength of the Lewis acids and bases is found to depend on the type of reaction, it is not possible to arrange them in any order of their relative strength.

The choice of which definition of acids and bases one wishes to use in a particular instance depends largely on the sort of chemistry that is studied. But Arrhenius concept is perfectly satisfactory and simplest for dealing with reactions in aqueous solutions. It explains satisfactorily the strength of acids and bases in aqueous solutions, neutralisation, salt hydrolysis, etc.
  

·      pH of Acids and Bases 

It is clear from the above discussion that nature of the solution (acidic, alkaline or neutral) can be represented in terms of either hydrogen ion concentration or hydroxyl ion concentration but it is convenient to express acidity or alkalinity of a solution by referring to the concentration of hydrogen ions only. pH measurement is done using pH paper and pH meter

Since H+ ion concentration can vary within a wide range from 1 mol per litre to about 1.0 × 10-14mol per litre, a logarithmic notation has been devised by Sorensen, in 1909, to simplify the expression of these quantities. The notation used is termed as the pH scale.

The hydrogen ion concentrations are expressed in terms of the numerical value of negative power to which 10 must be raised. This numerical value of negative power was termed as pH, i.e.,

pH  = – log aH+ (Where aH+ is the activity of H+ ions).

Activity of H+ ions is the concentration of free H+ ions in a solution. By free, we mean those that are at a large distance from the other ion so as not to experience  its pull. We can infer from this that in dilute solutions, the activity of an ion is same as its concentration since more number of solvent molecules would separate the two ions. For concentrated solutions the activity would be much less than the concentration itself.

Therefore, the earlier given expression of pH can be modified for dilute solutions as,
pH  = – log [H+].

This assumption can only be made when the solution is very much dilute, i.e, [H+]  1M. For higher concentration of H+ ions, one needs to calculate the activity experimentally and then calculate the pH.

pH of a solution is, thus, defined as the negative logarithm of the concentration (in mol per litre) of hydrogen ions which it contains or pH of the solution is the logarithm of the reciprocal of H+ ion concentration.

Just as pH indicates the hydrogen ion concentration, the pOH represents the hydroxyl ion concentration, i.e.,

pOH = -log [OH-] Considering the relationship,

[H+][OH-] = Kw = 1 x 10-14

 Taking log on both sides, we have

log [H+] + log [OH-] = log Kw = log (1 x 10-14)

or -log [H+] - log [OH-] = -log Kw = -log (1 x 10-14)

or pH + pOH = PKw* = 14

i.e., sum of pH and pOH is equal to 14 in any aqueous solution at 25°C. The above discussion can be summarised in the following manner:

[H+]+[OH-]=pH+pOH

pH of Weak Acids and Bases 

Weak acids and bases are not completely ionised; an equilibrium is found to have been established between ions and unionised molecule. Let us consider a weak acid of basicity 'n'.

            AHn    An-  +   nH+

           -C        0     0

teq      C(1-α)   Cα   nCα

[H+] = nCα;  .·. pH = -log10 [nCα]                

For monobasic and,  n=1

 pH = -log10 [Cα]                                       

Dissociation constant of acid Ka may be calculated as

 Ka = [An-][H+]n/[AHn] = [Cα][nCα]n/[C(1-α)]

            =  α [nCα]n/(1-α)       For weak acids, α« 1

.·. (1-α) = 1

            = α[nCα ]n/(1-α)

nCKa = nCα [nCα ]n = [nCα ](n+1) 

= [nCα ] = [nCKa]1/(n+1)

= [H+] = [nCKa]1/(n+1)

.·. pH = -1/(n+1) log10(nCKa)            

For monobasic acid, n = 1

pH = -log√CKα                               

Since Ka = α[nCα]n

ka/α = (nCα)n

 [nCα ] = [Kα/α]1/n = [H+]

pH = -1/n log10(Kα/α)                         

For n = 1       pH = -log10(Kα/α)
  

Limitations of pH Scale

1.    pH values of the solutions do not give us immediate idea of the relative strengths of the solutions. A solution of pH = 1 has a hydrogen ion concentra­tion 100 times that of a solution of pH = 3 (not three times). A 4 x 10-5 N HCI is twice concentrated of a 2 x 10-5 N HCI solution, but the pH values of these solutions are 4.40 and 4.70 (not double).

2.    pH value of zero is obtained in 1 A' solution of strong acid. In case the concentration is 2 N, 3 N, 10 N, etc. The respective pH values will be negative.

3.    A solution of an acid having very low concentration, say 10-8 N, cannot have pH 8, as shown by pH formula, but the actual pH value will be less than 7.

Note: 

1.    Normality of strong acid = [H3O+]
Normality of strong base = [OH-]
.·. pH = -log [N] for strong acids
    pOH = -log [N]    for strong acids

2.    Sometimes pH of acid comes more than 7 and that of base comes less than 7. It shows that the solution is very dilute; in such cases, H+ or OH- contribution from water is also considered, e.g., in 10   N HCI,

      [H+]Total = [10-8]Acid + [10-7]Water

      = 11 × 10-8 M= 1.1 × 10-7 M

 
pH of Mixture

Let one litre of an acidic solution of pH 2 be mixed with two litre of other acidic solution of pH 3. The resultant pH of the mixture can be evaluated in the following way.

Sample 1    

Sample 2

 pH = 2

pH = 3

[H+]=10-2 M

[H+] = 10-3M

V = 1 litre

V= 2 litre

 MlVl+M2V2 = MR(Vi + V2) 10-2 × 1 + 10-3 × 2 = MR (1 +2)

(12 + 10-3)/3  = MR

4 × 10-3 = MR(Here, MR = Resultant molarity)

pH = -log (4 × 10-3)

Total concentration of [H+] or  [H3+O ] in a mixture of weak acid and a strong acid  = (C2 + √(c22 + 4KaC1 ))/2

where C1 is the concentration of weak acid (in mol litre having dissociation constantKa

C2 is the concentration of strong acid

Total [OH-] concentration in a mixture of two weak bases =   √(K1C1+ K2C2)

where K1 and K2 are dissociation constants of two weak bases having C1 and C2 as their mol litre-1 concentration respectively.
  

Periodic Variations of Acidic and Basic Properties
  

(a)  Hydracids of the Elements of the Same Periods

Consider the hydracids of the elements of II period, Viz., CH4, NH3, H20 and HF. These hydrides become increasingly acidic as we move from CH4 to HF. CH4 has negligible acidic properties while HF is a fairly stronger acid. The increase in acidic properties is due to the fact that the stability of their conjugate bases increases in the order

CH-3< NH-2  < OH- < F-

The increase in acidic properties is supported by the succes­sive increase in the dissociation constant.

CH4(=10-58)<NH3 (=10-35)<H20(=10-14)<HF(=10-4)
  

(b)  Hydracids of the Elements of Same Group

1.    Hydrides of V group elements (NH3, PH3, AsH3, SbH3, BiH3) show basic character which decreases due to increase in size and decrease in electronegativity from N to Bi. There is a decrease in electron density in, sp3 -hybrid orbital and thus electron donor capacity decreases.

2.    Hydracids of VI group elements (H20, H2S, H2Se, H2Te) act as weak acids. The strength increases in the order H20 < H2S < H2Se < H2Te. The increasing acidic properties reflects decreasing trend in the electron donor capacity of OH-, HS-, HSe- or HTe- ions.

3.    Hydracids of VII group elements (HF, HCI, HBr, HI) show acidic properties which increase from HF to HI. This is explained by the fact that bond energies decrease. (H-F = 135 kcal/mol, HCI = 103, HBr = 88 and HI = 71 kcal/mol).
  

(c) Oxyacids

The acidic properties of oxyacids of the same element which is in different oxidation states increases with increase in oxidation number.

    + 1   +3     +5         +7

  HCIO  <  HC1O2   <   HC1O3   <  HCIO4

       +4       +6     +3        +5

  H2SO3  <   H2SO4;    HNO2     <  HNO3

But this rule fails in oxyacids of phosphorus.

H3PO2  >  H3PO3  >  H3PO4

The acidic properties of the oxyacids of different elements which are in the same oxidation state decreases as the atomic number increases. This is due to increase in size and decrease in electronegativity.

HC1O4 > HBrO4 > HIO4

 H2SO3  > H2SeO3

But there are a number of acid-base reactions in which no proton transfer takes place, e.g.,

SO2 + SO2  ↔  SO2+ +  S  

Acid1  Base2 Acid2   Base1

Thus, the protonic definition cannot be used to explain the reactions occurring in non-protonic solvents such as COCl2, S02, N2O4, etc.

Example 1:  The hydrogen ion concentration of a solution is 0.001 M. What will be the hydroxyl ion concentration of solution?

Solution: We know that [H+][OH-] 1.0 × 10-14

Given that,    

[H+] = 0.001 M = 10-3 M

So, [OH-] = 1.0 * 10-14/[H+] = (1* 10-14)/10-3 = 10-11M

Example 2:  What is the pH of the following solutions?

(a) 10-3 M HCl

(b) 0.0001 M NaOH

(c)  0.0001 MH2S04

Solution:   HCI is a strong electrolyte and is completely ionised.

HCl  H+ + CI-

So   [H+] = 10-3 M

pH = -log [H+] = -log (10-3) = 3

(b)   NaOH is a strong electrolyte and is completely ionised.

     NaOH ↔ Na+ + OH-
So [OH+] = 0.0001 M = 10-4 M

     pOH = -log(10-4) = 4

        As   pH + pOH = 14

        So   pH + 4 = 14   or     pH=10

Alternative method:   [OH-] = 10-4 M

We know that [H+][OH-] = 1.0 x 10-14

          So  [H+] = (1.0  10-14)/10-14

     pH = -log [H+] = -log(10-10) = 10

(c) H2SO4 is a strong electrolyte and is ionized completely.

     H2SO4 ↔ 2H+ + SO2-4

     One molecule of H2SO4 furnishes ions.

     So [H+] = 2 × 10-4 M

        pH = -log [H+]

     = -log (2×10-4)

     = 3.70

Example 3: Find the pH of a 0.002 N acetic acid solution, if it is 2.3% ionised at a given dilution.

Solution: Degree of dissociation, α = 2.3/100 = 0.023

Concentration of acetic acid, C = 0.002 M

The equilibrium is

CH3COOH  CH3COO- + H+

C(1-α)         Cα           Cα

So  [H+] = Cα  = 0.002 × 0.023 = 4.6 × 10-5 M

pH = -log [H+]= -log (4.6 × 10-5) = 4.3372

Example 4:   Calculate the pH value of a solution obtained by mixing 50 mh of 0.2 N HCl with 50 mL of 0.1 N NaOH.

Solution:    Number of milli-equivalents of the acid

= 50 × 0.2 = 10

Number of milli-equivalents of the base

= 50 × 0.1 =5

Number of milli-equivalents of the acid left after the addition of base

= (10-5) = 5 Total volume of the solution

= 50 + 50 = 100 mL

Thus, 5 milli-equivalents of the acid are present in 100 mL of solution.

or 50 milli-equivalents of the acid are present in one litre of solution.

or 0.05 equivalents of the acid are present in one litre of solution.

The acid is monobasic and completely ionised in solution.

0.05 N HCl = 0.05 M HCl

So     [H+] = 0.05 M

pH = -log [H+] = -log 5 × 10-2 = -[log 5.0 + log 10-2]

= -[0.70-2] = 1.3

Example 5:  What will be the pH of a solution obtained by mixing 800 mL of 0.05 A' sodium hydroxide and 200 mL of OA N HCl, assuming complete ionisation of the acid and the base ?

Solution:   Number of milli-equivalents of NaOH

= 800 × 0.05 = 40

Number of milli-equivalents of HCl

= 200 × 0.1 =20

Number of milli-equivalents of NaOH left after the addition of HCl

= (40 -20) = 20

Total volume = (200 + 800) mL = 1000 mL = 1 litre

20 milli-equivalents or 0.02 equivalents of NaOH are present in one litre, i.e.,

0.02 N NaOH = 0.02 M NaOH (Mono-acidic) and the base is completely ionised.

So     [OH-] = 0.02 M

or     [OH-]= 2 × 10-2 M

pOH = -log(2 × 10-2) = 1.7

We know that,   pH + pOH = 14

So   pH = (14-1.7)= 12.3

Ionic Product of Water
Pure water is a weak electrolyte and it undergo self ionization or auto-protolysis. In this process water molecules splits into  hydrogen ion (H+) and hydroxide ion (OH-).

The equation is shown as:

As we know hydrogen ion is very reactive and it reacts further with water molecules to form hydronium ion(H3O+)

Applying law of mass action at equilibrium, the value of dissociation constant, K comes to

      K = [H+] [OH-]/[H2O]

     or   [H+][OH-] = K[H2O]

Since dissociation takes place to a very small extent, the concentration of undissociated water molecules, [H20], may be regarded as constant. Thus, the product #[H20] gives another constant which is designated as Kw. So,

[H+][OH-] = Kw

The constant, Kw, is termed as ionic product of water.

Refer to the following video for ionic product of water

It is only because of the self ionization of water that it can act as both acid as well as base.

Any substance which increases the concentration of H+ ion would make water acidic.

Similarly, any substance which increases the concentration of OH- ion, would make water basic.

But in pure water, the hydrogen ion (hydroxonium ion) concentration is always equal to the hydroxide ion concentration. For every hydrogen ion formed, there is a hydroxide ion formed as well.

The product of concentrations of H+ and OH- ions in water at a particular temperature is known as ionic product of water. The value of Kw increases with the increase of temperature, i.e., the concentration of H+ and OH- ions increases with increase in temperature.

The value of Kw at 25°C is 1 x 10-14. Since pure water is neutral in nature, H+ ion concentration must be equal to OH- ion concentration.

[H+] = [OH˜] = x

or    [H+][OH-]=x2= 1 x 10-14

or    x = 1 x 10-7 M

or    [H+] = [OH-] = 1 × 10-7 mol litre-1

This shows that at 25°C, in 1 litre only 10-7 mole of water is in ionic form out of a total of approximately 55.5 moles.

When an acid or a base is added to water, the ionic concentration product, [H+][OH-], remains constant, i.e., equal to Kw but concentrations of H+ and OH- ions do not remain equal. The addition of acid increases the hydrogen ion concen­tration while that of hydroxyl ion concentration decreases, i.e.,

[H+] > [OH-]; (Acidic solution)

Similarly, when a base is added, the OH- ion concentration increases while H+ ion concentration decreases,i.e.,

[OH-] > [H+]; (Alkaline or basic solution)

In neutral solution,   

[H+] = [OH-] = 1 x 10-7 M

In acidic solution,

[H+] > [OH-]

or   [H+] > 1 x 10-7 M

and    [OH-] < 1 x 10-7 M

In alkaline solution,      

[OH-] > [H+]

 or   [OH-] > 1 × 10-7 M

and [H+] < 1 x 10-7 M

Thus, if the hydrogen ion concentration is more than 1 x 10-7 M, the solution will be acidic in nature and if less than 1 x 10-7 M, the solution will be alkaline.

[H+]                                                                        Nature of Water

10-0  ,10-1 ,10-2 ,10-3 , 10-4 ,10-5 ,10˜6                          Acidic

10-7                                                                                 Neutral

10-14 ,10-13 ,10-12 ,10-11 ,10-10 ,10-9 ,10-8                    Alkaline

We shall have the following table if OH- ion concentration is taken into account.

  [OH-]                                                                       Nature of Water

0-14  ,10-13 ,10-12 ,10-11 ,10-10 , 10-9 ,10-8                       Acidic

10-7                                                                                   Neutral

10-0 ,10-1 ,10-2 ,10-3 ,10-4 ,10-5 ,10-6                             Alkaline

It is, thus, concluded that every aqueous solution, whether acidic, neutral or alkaline contains both H+ and OH- ions. The product of their concentrations is always constant, equal to 1 × 10-14 at 25°C. If one increases, the other decrease accordingly so that the product remains 1×10-14 at 25o C.

If [H+] = 10-2 M,

then [OH-] = 10-12 M;

the product, [H+][OH-] = 10-2 × 10-12 = 10-14; the solution is acidic.

If [H+] = 10-10 M,

then [OH-] = 10-4 M; the product, [H+][OH-] = 10-10 × 10-4 = 10-14; the solution is alkaline.

Solubility Product 

 If to a given amount of solvent at a particular temperature, a solute is added gradually in increasing amounts, a stage is reached when some of the solute remains undissolved, no matter how long we wait or how vigorously we stir. The solution is then said to be saturated. A solution which remains in contact with undissolved solute is said to be saturated. At saturated stage, the quantity of the solute dissolved is always constant for the given amount of a particular solvent at a definite temperature.

In case the solute is an electrolyte, its ionisation occurs in solution and degree of dissociation depends on the concentra­tion of dissolved electrolyte at a particular temperature.

Refer to the following video for dissolution of salt in water

Thus, in a saturated solution of an electrolyte two equilibria exist and can be represented as:

                        AB  ↔ A+ + B-          

Solid  unionized           ions  (dissolved)

Applying the law of action to the ionic equilibrium,  

Since the solution is saturated, the concentration of unionised molecules of the electrolyte is constant at a particular temperature, i.e., [AB] = K'= constant.

Hence,

[A+] [B-] = K[AB] = KK’ = Ks (constant)

Ks is termed as the solubility product. It is defined as the product of the concentration of ions in a saturated solution of an electrolyte at a given temperature.

Consider, in general, the electrolyte of the type AxBy which is dissociated as:

AxBy  xAy+ + yBx-

Applying law of mass action,

When the solution is saturated,

[AxBy] = K' (constant)

or   Ay+]x[Bx-]y = K [AxBy] = KK' = Ks (constant)

Thus, solubility product is defined as the product of concentrations of the ions raised to a power equal to the number of times the ions occur in the equation representing the dissociation of the electrolyte at a given temperature when the solution is saturated.

Note:  Solubility product is not the ionic product under all conditions but only when the solution is saturated.
  

·      Different Expressions for Solubility Products

Refer to the following video for solubility product

(i) Electrolyte of the type AB:

Its ionisation is represented as:

            AB  ↔ A+ + B-

Thus, Ks = [A+][B-]

AgCl   Ag+ + Cl-;           Ks = [Ag+][Cl-]

BaSO4  Ba2+ + SO-24 ; Ks = [Ba2+][SO-24]

(ii)  Electrolyte of the type AB2:

Its ionisation is represented as:

            AB2↔ A2+ + 2B-

Thus, Ks = [A2+][B-]2

PbCl2  Pb2+ + 2Cl-;      Ks = [Pb2+][Cl-]2

CaF2  Ca2+ + 2F-;        Ks = [Ca2+][F-]2

(iii)  Electrolyte of the type A2B:

Its ionisation is represented as:

            A2B ↔  2A2+ + B2-

Thus, Ks = [A+]2[B2-]

Ag2CrO4  2Ag+ + CrO-24; Ks = [Ag+]2[CrO-24]

H2S   2H+ + S2-;               Ks = [H+]2[S2-]

(iv)   Electrolyte of the type A2B3:

Its ionisation is represented as:

            A2B3 ↔ 2A3+ + 3B2-

Thus, Ks = [A3+]2[B2-]3

As2S3  2As3+ + 3S2-;

Ks = [As3+]2[S2-]3

Sb2S3  2Sb3+ + 3S2-;

Ks = [Sb3+]2[S2-]3

(v)   Electrolyte of the type AB3:

Its ionisation is represented as:

           AB3 ↔ A3+ + 3B-

Thus, Ks = [A3+][B2-]3

Fe(OH)3    Fe3+ + 3OH-; Ks = [Fe3+][OH-]3

AH3    Al3+ + 3l-;      Ks = [Al3+][I-]3
  

·      Solubility Product of a Weak Electrolyte

Let degree of ionization of weak electrolyte AmBn be 'α'.

                        AmBn ↔ mAn+ + nBm-

t = 0                  s          0      0

teq                    s-sα msα  nsα

                        Ksp = [An+]m[Bm-]n

                        = [msα ]m[nsα]n

                        Ksp = mmnn (sα)m+n

Criteria of precipitation of an electrolyte:

A very useful conclusion is derived from the solubility product concept. No precipitation of the electrolyte occurs if the ionic product is less than the solubility product, i.e., the solution has not reached the saturation stage.

Case I:    When Kip<Ksp, then solution is unsaturated in which more solute can be dissolved.

Case II:   When Kip = Ksp, then solution is saturated in which no more solute can be dissolved.

Case III:  When Kip > Ksp, then solution is supersaturated and precipitation takes place.

When the ionic product exceeds the solubility product, the equilibrium shifts towards left hand side, i.e., increasing the concentration of undissociated molecules of the electrolyte. As the solvent can hold a fixed amount of electrolyte at a definite temperature, the excess of the electrolyte is thrown out from the solution as precipitate.

Thus, for the precipitation of an electrolyte, it is neces­sary that the ionic product must exceed its solubility product For example, if equal volumes of 0.02 M AgN03solution and 0.02 M K2Cr04 solution are mixed, the precipita­tion of Ag2Cr04 occurs as the ionic product exceeds the solubility product of Ag2Cr04 which is 2 × 10-12

In the resulting solution,

[Ag+] = 0.02/2 = 0.01 = 1*10-2 M

and  [CrO2-4 ]= 0.02/2 = 0.01 = 1*10-2 M

Ionic product of Ag2CrO4 = [Ag+]2 [CrO2-4]

                                     = (1×10-2)2 (1×10-2)

                                     = 1 × 10-6

1 × 10-6 is higher than 2 × 10-12 and thus precipitation of Ag2CrO4 occurs.
  

·      Common Ion Effect

 Let AB to the weak electrolyte. Considering its dissociation,

AB ↔ A+ + B-

and applying law of mass action we have

 K = [A+][B-]/[AB]

The equilibrium constant, K, has a definite value at any given temperature. If now another electrolyte furnishing the A+ and B- ions be added to the above solution. It will increases the concentration of either A+ ions or B- ions (whichever has been added) and in order that K may remain constant, the concentration of AB must increases, i.e., the equilibrium will shift to the left hand side.

In other words, the degree of dissociation of an electrolyte (weak) is suppressed by the addition of another electrolyte (strong) containing a common ion. This is termed as common ion effect. Acetic acid is a weak electrolyte and its ionization is suppressed in presence of a strong acid (H+ ion as common ion) or a strong salt like sodium acetate (acetate ion as common ion). Similarly, the addition of NH4Cl or NaOH to NH4OH solution will suppress the dissociation of NH4OH due to common ion either NH-4 or OH-.

As a result of common ion effect, the concentration of the ion not in common in two electrolytes, is decreased. The use of this phenomenon is made in qualitative analysis to adjust concentration of S2- ions in second group and OH- ion concentration in third group of analysis.

·      Applications of Solubility Product

(i)  Purification of common salt

Natural common salt consists of many insoluble and soluble impurities. Saturated solution of common salt is prepared and insoluble impurities are filtered off. Hydrogen chloride gas (HCl) is circulated through the saturated solution. HCl and NaCl dissociate into their respective ions as:

NaCl ↔  Na+ + Cl-

HCl  ↔  H+ + Cl-

The concentration of Cl- ions increases considerably in solution due to ionization HCl. Hence, the ionic product [Na+][Cl-] exceeds the solubility product of sodium chloride and, therefore, pure sodium chloride precipitates out from solution.

(ii)  Salting out of soap

Soap is a sodium salt of higher acids. From the solution, soap is precipitated by the addition of concentrated solution of sodium chloride. Soap and sodium chloride are present in the form of ions.

CnH2n+1 COONa ↔  CnH2n+1 COO- + Na+

             Soap

NaCl ↔  Na+ + Cl-

Thus, the concentration of Na+ ions increases considerably on addition of NaCl solution. Hence, the ionic product [CnH2n+1COO-] [Na+] exceeds the solubility product of soap and, therefore, soap precipitates out from the solution.

(iii)  Manufacture of sodium bicarbonate (baking soda):

 In Solvay's soda process. CO2 gas is passed through ammonical brine to precipitate out NaHCO3.

                NH4OH + CO2 → NH4HCO3

            NH4HCO3 + NaCl → NaHCO3 + NH4Cl

    NaHCO3 is precipitated first because of its lower solubility product as compared to those of NH4Cl, NH3HCO3 and NaCl.

    Thus, baking soda (NaHCO3) can be quantifiably estimated.

(iv) Application of solubility product in quantitative analysis

1. Estimation of barium as barium sulphate:

H2SO4 as precipitating agent is added to the aqueous solution of BaCl2.

BaCl2 + H2SO4 → BaSO4 + 2HCl

Precipitation of BaSO4 takes place when its ionic product exceeds solubility product. H2SO4 is added in slight excess to ensure complete precipitation. Large excess of H2SO4 is harmful for complex formation.

2. Estimation of silver as silver chloride:

NaCl solution is added to the silver nitrate solution, slight excess of NaCl is added to ensure complete precipitation.

AgNO3 + NaCl →  AgCl + NaBO3

Again, precipitation of AgCl takes place when ionic product of AgCl exceeds its solubility product.

3. In a similar manner. We estimate lead as lead chromate, calcium as calcium oxalate, etc.

(i)  Precipitation of the sulphides of group II and IV

 Hydrogen sulphide is a weak electrolyte and is used for the precipitation of various sulphides of group II and IV in quantitative analysis.

It ionizes to a small extent in water:

 H2S ↔ 2H+ + S2-

Applying law of mass action,

K = ([H+ ]2 [S2- ])/[H2 S]

The concentrations of S2- ions can be decreased by increasing concentration of H+ions and it can be increased by decreasing concentration of H+ ions. In group II, lower concentration of sulphide ions is required as the solubility products of the sulphides of group II are low while higher concentration of sulphide ions is required in group IV as the solubility products of the suphides of group IV are high.   

The concentration of S2- ions in group II is lowered by maintaining acidic medium. In the presence of HCl, the ionization of H2S is suppressed due to common ion effect. The concentration is so adjusted that only ionic products of the sulphides of group II exceed their solubility products and, therefore, get precipitated. However, CdS has somewhat higher value. For its precipitation, dilution of the solution is done which increases ionization of H2S and thereby increasing concentration of S2- ions.

In group IV, higher concentration of S2- ions is needed. This is done by changing the medium for acidic to alkaline. Ammonium hydroxide is added, the OH- ions furnished by NH4OH remove H+ ions from solution in the form of water molecules as,

                          H+ + OH- ↔  H2O

More of the ionization of H2S occurs and, thus, concentration of S2- ions increases. It becomes so high that ionic products of the sulphides of group IV exceed their solubility products add they get precipitated.

(ii)  Precipitation of III group hydroxides

When NH4OH is added in presence of NH4Cl then precipitation of II group hydroxides takes place, i.e., Al(OH)3, Fe(OH)3 and Cr(OH)3 are precipitated. Solubility product of III group hydroxides is less than those of higher group hydroxides.

 NH4Cl  →  NH+4 + Cl-

NH4OH  ↔  NH+4  + OH-

NH+4  ion furnished by NH4Cl lowers the ionisatin of NH4OH and hence the concentration of hydroxide ion OH-. At low concentration of hydroxide ion only III group hydroxides precipitate.

(v) Fractional Precipitation

It is a technique of separating two or more ions from a solution by adding a reagent that precipitates first one ion and then the second.

Let us suppose 0.1 M Ba2+ and 0.1 M Sr2+ in aqueous solution. K2CrO4 is added as precipitating agent. Ksp BaCrO4 is 1.2 × 10-10 and Ksp SrCrO4 is 3.5 × 10-5.

[CrO2-4]concentration required to precipitate BaCrO4

     = Ksp/[Ba2+] = 1.2 * 10-10/0.1 = 1.2 * 10-9

BaCrO4 will precipitate first because it requires low concentration of CrO2-4  ions. On addition of chromate ions, BaCrO4 starts precipitating when chromate ion concentration reaches 1.2 × 10-9 M. When CrO2-4  ion concentration reaches upto 3.5 × 10-4 M, then SrCrO4 also starts precipitating.

Remaining concentration of Ba2+ when SrCrO4 starts precipitation.

=(Ksp BaCrO4)/[CrO42-] = (1.2×10-10)/(3.5×10-4)= 3.4×10-7 M

% remaining concentration = (3.4 * 10-7)/0.1 * 100

                                              = 0.00034%

(vi) Stability Constant  

Let us consider dissociation of the ion FeBr+.

 FeBr+ ↔   Fe2+ + Br-

Dissociation constant for above equilibria may be given as

Kd = [Fe2+ ][Br- ]/[FeBr+ ]

Reciprocal of dissociation constant is called stability constant.

Ks = [FeBr+ ]/([Fe2+] [Br- ])

Let us consider the formation of complex K2Cd(CN)4, Complex ion is Cd(CN42-)  where oxidation state of central metal Cd2+ is (2+). Complexing process proceeds in four steps as

Here     Ks = K1K2K3K4.
  

·      Significance of Stability Constant

Greater will be the value of stability constant more stable will be the complex.

Note :       

(a) If on addition of a common ion in a salt solution (sparingly soluble), formation of complex ion takes place, then ionization increases, i.e., equilibrium shifts towards right hand direction to maintain the value of Ksp constant. It means, addition of common ion in the case of complex formation increases the solubility of the sparingly soluble salt which is against the concept of common ion effect.

(b) When we add an electrolyte to another electrolyte solution having no common ion, then ionization of the later increases.

(c) For a given electrolyte solubility product is always constant at a particular temperature.
  

·      Solubility of Metal Hydroxides in Acid Medium

 H+ ion furnished by the medium effects the solubility of metal hydroxide, say M(OH)2, because of neutralization of OH- ion by H+ ion.

           M(OH)2  ↔  M2+ + 2OH-

           Ksp of M(OH)2 = [M2+][OH-]2

           [M2+] =  Ksp/[OH-]2

           [H+] [OH-] = Kw = 10-14

           [OH-]2 = 10-28/[H+]2

           From Eqs. (i) and (ii), we have

           [M2+] = Ksp[H+]2/10-28
  

·      Relationship between Solubility and Solubility Product

Salts like Agl, BaS04, PbS04, Pbl2, etc., are ordinarily considered insoluble but they do possess some solubility. These are sparingly soluble electrolytes. A saturated solution of sparingly soluble electrolyte contains a very small amount of the dissolved electrolyte. It is assumed that whole of the dissolved electrolyte is present in the form of ions, i.e., it is completely dissociated.

The equilibrium for a saturated solution of any sparingly soluble salt may be expressed as:

AxBy ↔ xAy+ + yBx-

Thus, solubility product, Ks = [Ay+]x[Bx-]y

Let 's' mole per litre be the solubility of the salt, then

       AxBy ↔  xAy+ + yBx-

                       xs       ys

So                   Ks = [xs]x[ys]y

                           = xx.yy(s)x+y

(i)  1:1 type salts:

Examples: AgCl, Agl, BaSO4, PbSO4, etc.

Binary electrolyte: AB ↔  A+ + B-

                                  s  s

Let solubility of AB be s mol litre-1.

So               Ks = [A+][B-] = s × s = s2

or                s = √Hs

(ii)  1:2 or 2:1 type salts:

Examples: Ag2CO3, Ag2CrO4, PbCl2, CaF2, etc.

Ternary electrolyte:

AB2 ↔ A2+ + 2B-

        s   2s

Let solubility of AB2 be s mol litre-1.

So      Ks = [A2+][B-]2 = s × (2s)2 = 4s3

or       s = 3√Ks/4

              A2B ↔ 2A+ + B2-

                         s  s

   Let s be the solubility of A2B.

                  Ks = [A+]2[B2-]

                      = (2s)2(s) = 4s3

        or      s = 3√Ks/4

(iii)  1: 3 type salts:

    Examples: All3, Fe(OH)3, Cr(OH)3, Al(OH)3, etc.

    Quaternary electrolyte:  AB3 ↔ A3+ + 3B-

         Let s mol litre-1 be the solubility of AB3.

                    Ks = [A3+][B-]3 = s × (3s)3 = 27s4

       or         s  = 4√Ks/27

The presence of common ion affects the solubility of a salt. Let AB be a sparingly soluble salt in solution and A'B be added to it. Let s and s' be the solubilities of the salt AB before and after addition of the electrolyte A'B. Let c be the concentration of A'B.

Before addition of A'B, Ks=s2                      ... (i) 

After addition of A'B, the concentration of A+ and B- ions become s' and (s' + c), respectively.

So Ks = s'(s' + c)                                 .... (ii)

Equating Eqs. (i) and (ii),

s2 = s'(s' +c)

·      Calculation of Remaining Concentration After Precipita­tion

Sometimes an ion remains after precipitation if it is in excess. Remaining concentration can be determined, e.g..

(i)    [A+]left = Ksp [AB]/[B-]

(ii) [Ca2+]left =  Ksp[Ca(OH)2]/[OH-]2

(iii) [An+]mleft = Ksp[AmBn]/[Bm-]n

Percentage precipitation of an ion

= [Initial conc. -Left conc./Initial conc.] × 100
  

·      Simultaneous Solubility

Solubility of two electrolytes having common ion; when they are dissolved in the same solution, is called simultaneous solubility, e.g.,

(i)    Solubility of AgBr and AgSCN, when dissolved together.

(ii)    Solubility of CaF2 and SrF2, when dissolved together.

(iii)   Solubility of MgF2 and CaF2 when dissolved together.

Calculation of simultaneous solubility is divided into two cases.

Case I:   When the two electrolytes are almost equally strong (having closesolubility product), e.g.,

AgBr(Ksp = 5 x 10-13); AgSCN (Ksp = 10-12)

(See Example 23)

Here, charge balancing concept is applied.

Charge of Ag+ = Charge of Br- + Charge of SCN-

[Ag+] = [Br-]    +   [SCN-]

  (a + b) =      a            b

Case II:    When solubility products of two electrolytes are not close, i.e., they are not equally strong, e.g.,

CaF2 (Ksp = 3.4 x 10-11); SrF2 (Ksp = 2.9 x 10-9)

Most of fluoride ions come of stronger electrolyte.
  

·      Ostwald’s Dilution Law

According to Arrhenius theory of electrolyte dissociation, the molecules of an electrolyte in solution are constantly splitting up into ions and the ions are constantly reuniting to form unionized molecules. Therefore, a dynamic equilibrium exists between ions and unionized molecules of the electrolyte in solution. It was pointed out by Ostwald that like chemical equilibrium, law of mass action van be applied to such systems also.

Consider a binary electrolyte AB which dissociates into A+ and B- ions and the equilibrium state is represented by the equation:

                         AB ↔  A+ + B-

Initially t = 0       C    0 0

At equilibrium C(1-α)   Cα    Cα

So, dissociation constant may be given as

K = [A+][B-]/[AB] = (Cα * Cα)/C(1-α) =   Cα2 /(1-α)                                    

For very weak electrolytes,

α <<< 1,  (1 - α ) = 1

 .·.                    K = Cα2

                      α = √K/C 

Concentration of any ion = Cα = √CK .

From equation (ii) it is a clear that degree of ionization increases on dilution.

Thus, degree of dissociation of a weak electrolyte is proportional to the square root of dilution.
  

Limitations of Ostwald's Dilution Law

The law holds good only for weak electrolytes and fails completely in the case of strong electrolytes. The value of 'α' is determined by conductivity measurements by applying the formula Λ/Λ∞. The value of 'α' determined at various dilutions of an electrolyte when substituted in Eq. (i) gives a constant value of K only in the case of weak electrolytes like CH3COOH, NH4OH, etc. the cause of failure of Ostwald's dilution law in the case of strong electrolytes is due to the following factors"

·      The law is based on the fact that only a portion of the electrolyte is dissociated into ions at ordinary dilution and completely at infinite dilution. Strong electrolytes are almost completely ionized at all dilutions and Λ/Λ∞ does not give accurate value of 'α'.

·      When concentration of the ions is very high, the presence of charges on the ions appreciably effects the equilibrium. Hence, law of mass action its simple form cannot be strictly applied in the case of string electrolytes.

Example 1:  The ionization constant of HCN is 4 × 10-10. Calculate the concentration of hydrogen ions in 0.2 M solution of HCN containing 1 mol L-1 of KCN?

Solution:  

The dissociation of HCN is represented as

 HCN ↔ H+ + CN-

Applying law of mass action,

 Ka =   ([H+ ][CN-])/[HCN] or [H+ ] (Ka [HCN])/[CN- ]

In presence of strong electrolyte, the total CN- concentration comes from KCN which undergoes complete dissociation. It is further assumed that dissociation of HCN is very-very small and the concentration of HCN can be taken as the concentration of undissociated HCN.

Thus, [HCN] = 0.2 M and [CN-] = 1M

Putting these values in the expression

[H+] = (Ka [HCN])/([CN- ]) = (4×10-10×0.2)/1 = 8×10-11 mol L-1

[Note: When KCN is not present, the [H+] concentration is equal to √CK i.e., √(0.2*4*10-10) = 8.94 *10-8mol L-1 . This shows that concentration of H+ ions fails considerably when KCN is added to HCN solution.

Example 2:  When 0.100 mole of ammonia, NH3, is dissolved in sufficient water to make 1.0 L solution, the solution is found to have a hydroxide ion concentration of 1.34 × 10-3 M. Calculate Kb for ammonia.

Solution:          

NH3  +   H2O  ↔  NH+4

At equilibrium (0.100 -1.34 ×10-3) M  1.34×10-3 M

                         = 0.09866 M    + OH-

                           1.34 × 10-3 M

 Kb = [NH4+ ][OH- ]/[NK3 ] = (1.34×10-3 × 1.34×10-3)/0.09866

                         =1.8199×10-5

 Example 3:   Ka for HA is 4.9 ×10-8. After making the necessary approximation, calculate for its decimolar solution

(a)   % dissociation

(b)   H+ ion concentration.

Solution:  For a weak electrolyte.

α = √K/C = √((4.9×10-8)/0.1)

            = 7 × 10-4

% dissociation = 100 × α = 100 × 7 × 10-4

            = 7 ×10-2

HA       ↔     H+     +  A-

C(-1-α)      Cα                Cα

[H+] = C × α = 0.1 ×7 × 10-4 = 7 × 10-5 mol L-1

Example 4:  Nicotinic acid (Ka = 1.4 ×10-5) is represented by the formula HNiC. Calculate its per cent dissociation in a solution which contains 0.10 mole of nicotine acid per 2 litre of solution.

Solution:  Initial concentration of the nicotinic acid = 0.10/2 = 0.05molL-1

                           HNiC    ↔  H+ + NiC-

Equilibrium conc. (0.05-x)  x  x

As x is very small, (0.05 - x), can be taken as 0.05

Ka = [H+ ][NiC- ]/[HNiC] = (x × x)/0.05

or    x2 = (0.05) × (1.4 × 10-5)

or    x = 0.83 × 10-3 mol L-1

% dissociation = (0.83×10-3)/0.05×100 = 1.66

Alternative method:   Let α be the degree of dissociation

                             NiHC            ↔          H+     + NiC-

At equilibrium 0.05 (1 -α)   0.05α    0.05α

Ka =  (0.05α×0.05α)/0.05(1-α)

As α is very small, (1-α) →  1.

So,   1.4 × 10-5 = 0.05 α2

or    α√((1.4×10-5)/0.05=1.67×10-2 )

Per cent dissociation = 100 × α = 100 × 1.67 ×10-2  = 1.67

Example 5:  At 30o C the degree of dissociation of 0.066 M HA is 0.0145. What would be the degree of dissociation of 0.02 M solution of the acid at the same temperature?

Solution:  Let the ionization constant of the acid be Ka.

Degree of dissociation at 0.66 M concentration = 0.0145.

Applying α = √Kα/C

0.0145 =  √Kα/0.066                       ..... (i)

Let the degree of dissociation of the acid at 0.02 M concentration be α1.

α1 = √Kα/0.02                               ...... (ii)

Dividing Eq. (ii) by Eq. (i)

α1/0.0145=√(0.066/0.02)=1.8166

or   α1 = 0.0145 × 1.8166 = 0.0263

Example 6:  A solution contains 0.1 M H2S and 0.3 M HCl. Calculate the concentration of S2- and HS- ions in solution. Given  and  for H2S are 10-7 and 1.3 × 10-13 respectively.

Solution:  H2S ↔  H+ + HS-

Kα1=[H+ ][HS- ]/[H2 S]        ....... (i)

Further HS- = H+ + S2-

Kα2=[H+ ][HS2- ]/[H2 S]         ....... (ii)

Multiplying both the equations

Kα1×Kα2=([H+ ]2 [HS2- ])/[H2S]

 Due to common ion, the ionization of H2S is suppressed and the [H+] in solution is due to the presence of 0.3 M HCl.

[S2-]=(Kα1× Kα2[H2S])/[H+]2=(1.0×10-7×1.3×10-13×(0.1))/(0.3)2

Putting the value of [S2-] in Eq. (ii)

1.3×10-13 = (0.3×1.44×10-20)/([HS-])

or [HS-] (0.3×1.44×10-20)/(1.3×10-13)= 3.3×10-8 M

Salt Hydrolysis

Pure water is a weak electrolyte. Weak because it does not completely disassociates  into hydrogen and hydroxide but exist in equilibrium with these two ions. It is  neutral in nature, i.e., H+ ion concentration is exactly equal to OH- ion concentration

[H+] = [OH-]

When this condition is disturbed by decreasing the concentration of either of the two ions, the neutral nature changes into acidic or basic. When [H+] > [OH-], the water becomes acidic and when [H+] < [OH-], the water acquires basic nature. This is exactly the change which occurs during the phenomenon known as salt hydrolysis. It is defined as a reaction in which the cation or anion or both of a salt react with water to produce acidity or alkalinity.

Salts are strong electrolytes. When dissolved in water, they dissociate almost completely into ions. In some salts, cations are more reactive in comparison to anions and these react with water to produce H+ ions. Thus, the solution acquires acidic nature.

M+ +   H2O  ↔  MOH    +  H+

                     Weak base

In other salts, anions may be more reactive in comparison to cations and these react with water to produce OH- ions. Thus, the solution becomes basic.

  A- +   H2O  ↔   HA   +  OH-

                      Weak acid

The process of salt hydrolysis is actually the reverse of neutralization.

Salt + Water  ↔ Acid + Base

If acid is stronger than base, the solution is acidic and in case base is stronger than acid, the solution is alkaline. When both the acid and the base are either strong or weak, the solution is generally neutral in nature.

As the nature of the cation or the anion of the salt determines whether its solution will be acidic or basic, it is proper to divide the salts into four categories.

·      Salt of a strong acid and a weak base.
Examples:  FeCl3, CuCl2, AlCl3, NH4Cl, CuSO4, etc.

·      Salt of a strong base and a weak acid.
Examples: CH3COONa, NaCN, NaHCO3, Na2CO3, etc.

·      Salt of a weak acid and a weak base.
Examples: CH3COONH4, (NH4)2CO3, NH4HCO3, etc.

·      Salt of a strong acid and a strong base.
Examples: NaCl, K2SO4, NaNO3, NaBr, etc.

Refer to the following video for salt hydrolysis

·      Salt of a Strong Acid and a Weak Base

The solution of such a salt is acidic in nature. The cation of the salt which has come from weak base is reactive. It reacts with water to form a weak base and H+ ions.

B+  +   H2O  ↔  BOH  +  H+

                        Weak base

Consider, for example, NH4Cl. It ionises in water completely into NH4 and CF ions. ions react with water to form a weak base (NH4OH) and H+ ions.

NH+4  +  H2O   ↔   NH4OH  +  H+

    C(1-x)                   Cx Cx

Thus, hydrogen ion concentration increases and the solution becomes acidic.

Applying law of mass action,

    ...... (i)

where C is the concentration of salt and x the degree of hydrolysis.

Other equilibria which exist in solution are

NH4OH ↔ NH+4 + OH-,   

   .... (ii)

H2O ↔  H+ + OH-,

Kb = [H+][H-]         ..... (iii)

From eqs. (II) and (iii)

Kw/Kb =[H+ ][NH4 OH]/[NH4+ ] = Kh  .... (iv)

[H+] = [H+ ][NH4+]/[NH4OH] = Kw/Kb ×[NH4+ ]/[NH4 OH]

log [H+] = log Kw - log Kb + log[salt]/[base]

-pH = -pKw + pKb + log[salt]/[base]

pKw - pH = pKb + log[salt]/[base]

Relation Between Hydrolysis Constant and Degree of Hydrolysis

The extent to which hydrolysis proceeds is expressed as the degree of hydrolysis and is defined as the fraction of one mole of the salt that is hydrolysed when the equilibrium has been attained. It is generally expressed as h or x.

h = (Amount of salt hydrolysed)/(Total salt taken)

Considering again eq. (i),

Kh = x2C/(1-x)   or Kh = h2C/(1-h)

When h is very small (1-h) → 1,

H2 = Kh × 1/c

 or   h = √(Kh/C)  = √(Kw/Kb  C)

[H+] = h × C = √(C  Kh)/Kb

log [H+] =  1/2 log Kw + 1  1/2log C - 1/2log Kb

Salt of a Weak Acid and a Strong Base 

The solution of such a salt is basic in nature. The anion of the salt is reactive. It reacts with water to form a weak acid and OH- ions.

A- + H2O;   ↔   HA + OH-

                        Weak acid

Consider, for example, the salt CH3COONa. It ionises in water completely to give CH3COO- and Na+ ions. CH3COO- ions react with water to form a weak acid, CH3COOH and OH- ions.

CH3COO- + H2O  ↔   CH3COOH + OH-

C(1-x)                         Cx   Cx

Thus, OH- ion concentration increases, the solution be­comes alkaline.

Applying law of mass action,

Kh = [CH3COOH][OH-]/[CH3CO-] = (Cx×Cx)/C(1-x) = (Cx2)/(1-x) )  ...... (i)

Other equations present in the solution are:

CH3COOH  CH3COO- + H+,       

Ka = [CH3COO-][H+]/[CH3COOH] ...... (ii)

H2O   H+ + OH-,                

Kw = [H+][OH-]    ....... (iii)

From eqs. (ii) and (iii),

log [OH-] = log Kw - log Ka + log[salt]/[acid]

-pOH = -pKw + pKa + log[salt]/[acid]

pKw - pOH = pKa + log[salt]/[acid]

Considering eq. (i) again,

Kh = cx2/(1-x)      or    Kh = Ch2/(1-h)

When h is very small, (1-h) → 1

or h2 = Kh/C

or h =  √Kh/C

[OH-] = h × C = √(CKh) = √(C  Kw/Ka)

[H+] = Kw/[OH-]

= Kw/√(C*Kw/Ka) = √(Ka Kw)/Kc

-log [H+] = -1/2log Kw - 1/2log Ka + 1/2log C

pH =  1/2pKw +  1/2pKa +  1/2log C

= 7 + 1/2pKa +  1/2log C.

Salt of a Weak Acid and a Weak Base

Maximum hydrolysis occurs in the case of such a salt as both the cation and anion are reactive and react with water to produce H+ and OFT ions. The solution is generally neutral but it can be either slightly acidic or slightly alkaline if both the reactions take place with slightly different rates. Consider, for example, the salt CH3COONH4. It gives CH3COO- and  ions in solution. Both react with water.

Other equilibria which exist in solution are:

CH3COOH ↔  CH3COO- + H+,

Ka =  [CH3COO-][H+]/[CH3COOH]    ..... (i)

NH4OH  NH+4 + OH-,

Kb = [NH+4] [OH-]/[NH4OH]           ..... (ii)

H2O ↔ H+ + OH-

Kw = [H+][OH-]                                    ..... (iii)

From Eqs. (i), (ii) and (iii),

Kh =  Kw/Ka.Kb = [CH3COOH][NH4OH]/[CH3COO-][NH+4]                .... (iv)

Let C be the concentration and h be the degree of hydrolysis

Kh = h2/(1-h)2

When h is small, (1-h) → 1.

Kh = h2

h = √Kh = √Kw/Ka*Kb

[H+] Ka × h

= Ka × √Kw/Ka*Kb

= √Kw  Ka/Kb

-log [H+] = -1/2log Ka - 1/2log Kw + 1/2log Kb

pH = 1/2pKa +  1/2pKw - 1/2pKb

= 7 +  1/2pKa - 1/2pKb

When pKa = pKb, pH = 7, i.e., solution will be neutral in nature.

When pKa > pKb. The solution will be alkaline as the acid will be slightly weaker than base and pH value will be more than 7.

In case pKa < pKb, the solution will be acidic as the acid is relatively stronger than base and pH will be less than 7.

Salt of a Strong Acid and a Strong Base

Such a salt, say NaCl, does not undergo hydrolysis as both the ions are not reactive. The solution is thus, neutral in nature.
  

·      Hydrolysis of Amphiprotic Anion

Let us consider hydrolysis of amphiprotic anion only, i.e., when counter cation is not hydrolysed, example of some salts of this category are NaHCO3, NaHS, Na2HPO4, NaH2PO4. 

Here, H2PO-4 and HPO2-4  are amphiprotic anions. pH after their hydrolysis can be calculated as,

pH of  H2PO-4  in aqueous medium = (pka1 + pka2)/2

pH of H2PO2-4  in aqueous medium = (pka2 + pka3)/2

Here, H2PO2-4  is conjugate base of H2PO-4 and H3PO4 is conjugate acid of H2PO-4.

Similarly, PO3-4 is conjugate base of HPO-24  and HPO-4  is conjugate acid of PO3-4 .

(iv)  Let us consider amphiprotic bicarbonate anion.     

pH HCO-3of  ion after hydrolysis in aqueous medium  = (pka1 + pka2)/2

(v)  Let us consider the hydrolysis of amphiprotic anion along with cation, e.g., NH4HCO3, NH4HS.

In above examples both cations and anions are derived from weak base and weak acids respectively hence, both will undergo hydrolysis in aqueous medium.

When these salts are dissolved in water, [H3O+] concentra­tion can be determined as,

[H3O+] = √ka1[kw/kb + ka2]

pH = -log = √ka1[kw/kb + ka2]
  

Hydrolysis at a Glance

Salt                                              Nature                                    Degree                                 Hydrolysis Constant                                       pH

1. NaCl                           (Strong acid + Strong Base)            No Hydrolysis   

                                             Neutral

2. Ch3COONa                (Weak acid + Strong base)               h = √kw/Cka                             Kh = kw/ka                                        pH=1/2[pkw + pka + logC]

                                                    Base

3. NH4Cl                         (Strong acid + Weak base)              h = √kw/Ckb                               Kh = kw/Ckb                                     pH=1/2[pkw- pkb - logC]

                                                   Acidic

4. CH3COONH4            (Weak     acid + Weak base)             h = √kw/(ka + kb)                       Kh = kw/(ka + kb)                          pH=1/2[pkw + pka - pkb] 

In the case of salt of weak acid and weak base, nature of medium after hydrolysis is decided in the following manner:

(i) If Ka = Kb, the medium will be neutral.

(ii)  If Ka > Kb, the medium will be acidic.

(iii)  If Ka < Kb, the medium will be basic.

The degree of hydrolysis of salts of weak acids and weak bases is unaffected by dilution because there is no concentration term in the expression of degree ofhydrolysis.

Note : Degree of hydrolysis always increases with increase in temperature because at elevated temperature increase in Kw is greater as compared to Ka and Kb.

Buffer Solution

For several purposes, we need solutions which should have constant pH. Many reactions, particularly the biochemical reactions, are to be carried out at a constant pH. But it is observed that solutions and even pure water (pH = 7) cannot retain the constant pH for long. If the solution comes in contact with air, it will absorb CO2 and becomes more acidic. If the solution is stored in a glass bottle, alkaline impurities dissolve from glass and the solution becomes alkaline.

A solution whose pH is not altered to any great extent by the addition of small quantities of either an acid (H+ ions) or a base (OH- ions) is called the buffer solution. It can also be defined as a solution of reserve acidity or alkalinity which resists change of pH upon the addition of small amount of acid or alkali.

Refer to the following video for buffer solutions

So, a buffer solution can be defined as a solution which resists a change in its pH when such a change is caused by the addition of a small amount of acid or base.

This does not mean that the pH of the buffer solution does not change (we make this assumption while doing numerical problems). It only means that the change in pH would be less than the pH that would have changed for a solution that is not a buffer.

There are three types of buffer solutions:

·      weak acid–salt buffer

·      weak base–salt buffer and

·      salt buffer 

General characteristics of a buffer solution

·      It has a definite pH, i.e., it has reserve acidity or alkalinity.

·      Its pH does not change on standing for long.

·      Its pH does not change on dilution.

·      Its pH is slightly changed by the addition of small quantity of an acid or a base.

Buffer solutions can be obtained:

·      by mixing a weak acid with its salt with a strong base, eg; Acetic Acid + Soidum Acetate,  Boric acid + Borax, Phthalic acid + Potassium acid phthalate.

·      by mixing a weak base with its salt with a strong acid,  e.g; (a)  PNH4OH + NH4Cl , (b)  Glycine + Glycine hydrochloride

·      by a solution of ampholyte. The ampholytes or amphoteric electrolytes are the substances which show properties of  both an acid and a base. Proteins and amino acids are the examples of such electrolytes.

·      by a mixture of an acid salt and a normal salt of a polybasic acid, e.g., Na2HPO4 + Na3PO4, or a salt of weak acid and a weak base, such as CH3COONH4.

The first and second type are also called acidic and basic buffers respectively.

Explanation of buffer action

Acidic buffer: 

Consider the case of the solution of acetic acid containing sodium acetate. Acetic acid is feebly ionised while sodium acetate is almost completely ionised.

The mixture thus contains CH3COOH molecules, CH3COO- ions, Na+ ions, H+ ions and OH- ions. Thus, we have the following equilibria in solution:

Feebly ionised                                                                      Completely ionised                                                              Very feebly ionised

CH3COOH   ↔   H+ + CH3COO-                                     CH3COONa ↔  Na+ + CH3COC-                                                       H2O  ↔   H+ + OH-                 

When a drop of strong acid, say HCl, is added, the H+ ions furnished by HCl combine with CH3COO- ions to form feebly ionised CH3COOH whose ionisation is further suppressed due to common ion effect.

Thus, there will be a very slight effect in the overall H+ ion concentration or pH value.

When a drop of NaOH is added, it will react with free acid to form undissociated water molecules.

CH3COOH + OH-    CH3COO- + H2O

Thus, OH- ions furnished by a base are removed and pH of the solution is practically unaltered.

·      Buffer capacity

The property of buffer solution to resist alteration in its pH value is known as buffer capacity. It has been found that if the ratio [Salt]/[Acid] or [Salt]/[Base]  is unity, the pH of a particular buffer does not change at all. Buffer capacity is defined quantitatively as number of moles of acid or base added in one litre of solution as to change the pH by unity,

or   

where δb → number of moles of acid or base added to 1 litre solution and δ(pH) → change in pH.

Buffer capacity is maximum:

1.    When [Salt] = [Acid], i.e., pH = pKa for acid buffer

2.    When [Salt] = [Base], i.e., pOH = pKb for base buffer under above conditions, the buffer is called efficient.

Utility of buffer solutions in analytical chemistry

Buffers are used:

·      To determine the pH with the help of indicators.

·      For the removal of phosphate ion in the qualitative inorganic analysis after second group using CH3COOH + CH3COONa buffer.

·      For the precipitation of lead chromate quantitatively in gravimetric analysis, the buffer, CH3COOH + CH3COONa, is used.

·      For precipitation of hydroxides of third group of qualitative analysis, a buffer, NH4Cl + NH4OH, is used.

·      A buffer solution of NH4Cl, NH4OH, and (NH4)2CO3 is used for precipitation of carbonates of fifth group in qualitative inorganic analysis.

·      The pH of intracellular fluid, blood is naturally maintained. This maintenance of pH is essential to sustain life because, enzyme catalysis is pH sensitive process. The normal pH of blood plasma is 7.4.

·      Following two buffers in the blood help to maintain pH (7.4).

       (a) Buffer of carbonic acid (H2CO3 and NaHCO3)

       (b) Buffer of phosphoric acid (H2P04, HPO2-)

·  Buffers are used in industrial processes such as manufacture of paper, dyes, inks, paints, drugs, etc. 

·  Buffers are also em­ployed in agriculture, dairy products and preservation of vari­ous types of foods and fruits. 

·  Buffer solutions are necessary to keep the correct pH for enzymes in many organisms to work. 

·  A buffer of carbonic acids (H2CO3) and bicarbonate (HCO3−) is present in blood plasma to maintain a pH between 7.35 and 7.45.

·  Buffer solutions are used to increase shelf life of drugs

Henderson's Equation or henderson-Hasselbalch equation
(pH of buffer solution):

Refer to the following video for Henderson's Equation

(i) Acidic buffer:        

It consists of a mixture of weak acid and its salt (strong electrolyte). The ionisation of the weak acid, HA, can be shown by the equation

HA ↔  H+ + A-

 Applying law of mass action,

Ka =   H+A- /[HA] 

It can be assumed that concentration of A- ions from complete ionisation of the salt BA is too large to be compared with concentration of A- ions from the acid HA.

BA   B+ + A-

Thus, [HA] = Initial concentration of the acid as it is feebly ionised in presence of common ion

and [A-] = Initial concentration of the salt as it is completely ionised.

pH = 1 + pKa

and when [Salt]/[Acid] = 0.1 then

pH = pKa -1

So weak acid may be used for preparing buffer solutions having pH values lying within the ranges pKa + 1 and pKa -1. The acetic acid gas a pKa of about 4.8; it may, therefore, be used for making buffer solutions with pH values lying roughly within the ranges 3.8 to 5.8.

(ii)  Basic Buffer:

It consists of a weak base and its salt with strong acid. Ionization of a weak base, BOH, can be represented by the equation.

BOH   B+ + OH-

Applying law of mass action,

Kb =  [B+][OH-]/[BOH]                      

or  [OH-] = Kb [BOH]/[B+]                         

As the salt is completely ionized, it can be assumed that whole of B+ ion concentration comes from the salt and contribution of weak base to B+ ions can be ignored.

BA  B+ + A-              (Completely ionised)

So   [OH-]= Kb[Base]/[Salt]                         

or    pOH = log[Salt]/[Base]  log Kb

or    pOH = pKb + log[Salt]/[Base]                 

Knowing pOH, pH can be calculated by the application of the formula.

pH + pOH = 14

(iii) Salt Buffer

The fundamental principle behind a buffer action is the fact that on adding an acid the system consumes the H+ ion added to produce a weak acid and on adding a  base, it consume the OH– ion added to produce a weak base. This ensures that H+ or OH– ion  added is consumed and the weak acid or the weak base produced gives less H+ or OH– ion as they are weak.

Based on this principle, a solution of a salt of a weak acid and weak base is also called as a buffer. Let us take the example of CH3COONH4. It dissociates as,
CH3COONH4 →  CH3COO– + NH4+.

When H+ ion is added, CH3COO– ion  consumes it to give CH3COOH. When OH– ion is added, CH3COO- ion consumes it to give NH4OH. Hence it acts as a buffer.

Test Your Knowledge

Q.1 Which of the following pairs represent an acidic buffer?

a. NH4OH +NH4Cl  b. CH3COOH + NH4OH   c. CH3COOH +CH3COONa 

Q.2 . Buffer solutions are not used for..

a. qualitative analysis of mixture

b. qualitative analysis of mixtures

c. digestion of food

d. purification of water

Q.3 . Which of the following equations represents the correct form of Henderson's Equation?

a. 

b. 

c. 

d. 

Q.4 . Buffer solution can not  be formed by 

a. mixing a weak acid with its salt with a strong base. 

b. by mixing a weak base with its salt with a strong acid.

c. by a solution of protic solvents.

d. by a solution of ampholyte. 

Q.5. Which of the following buffer is present in our blood plasma?

a. Acetic acid + Sodium Acetate

b.  Carbonic acid + bicarbonate

c. Boric acid + Borax, 

d. Phthalic acid + Potassium acid phthalate

Answer key:

Q.1 

Q.2 

Q.3 

Q.4 

Q.5 

c

d

a

d

b

Solved Examples on Equilibrium

Question 1:  

Calculate  the pH of the solution when 0.1 M CH3 COOH (50 ml) and 01. M NaOH (50 ml) are mixed, [Ka (CH3COOH)=10-5]

Solution:

CH3 COOH  ↔ CH3 COO_ + H+ …(I)

NaOH →  Na+ + OH-

H+ + OH_ H2O …(II)

(I) + (II)

CH3COOH + OH-   CH3COO- + H2O . (III)

0.05-X         0.05-x     x

Keq of eq. (III) = Ka/Kw

conc. of H2O remain constant

109 = x/(0.05-x)2

because value of eq. Const.is very high

here for x» 0.05

let 0.05-x=a

109=0.05/a2

a = 7.0710-6

pOH= 6-log 7.07

pOH= 6 – 0.85

pH= 14-6+0.85 = 8.85

________________________________________________________________________________________

Question 2: 

Calculate the pH at the equivalence point of the titration between 0.1M CH3­COOH ( 25 ml) with 0.05 M NaOH. Ka (CH3COOH) = 1.8  10–5.

Solution: 

We have already seen that even though when CH3COOH is titrated with NaOH the reaction does not go to completion but instead  reaches equilibrium. We can assume that the reaction is complete and then salt gets hydrolysed because, this assumption will help us to do the problem easily and it does not effect our answer.

First of all we would calculate the concentration of the salt, CH3COONa.  For reaching equivalence point,

 N1V1 = N2V2

0.1 ´ 25 = 0.05 ´ V2

 V2 = 50 ml 

Therefore [CH3COONa] = (0.125)/75 =0.1/3

[H+] = 

[H+] = 2.32 ´ 10–5

 pH = – log 2.32 ´ 10–5 = 8.63

___________________________________________________________________________

Question 3: 

Given the solubility product of Pb3 (PO4)2 is 1.5 x 10-32.Determine the solubility in gms/litre.

Solution:  

Solubility product of Pb3 (PO4)2 = 1.5  10–32

Pb3 (PO4)2  3Pb2+ + 2PO43-

If x is the solubility of Pb3 (PO4)2

Then Ksp = (3x)3 (2x)2 = 108 x5

x = 1.692  10–7 moles/lit

Molecular mass of Pb3(PO4)2 = 811

 x = 1.692 ´ 10–7 ´ 811 g/lit = 1.37  10–4 g/lit

 Solubility product is

 Ksp(SrC2O4) = [Sr2+] [C2O42–] = (5.4  10–4)2-    = 2.92  10–7