Relationship between pH and pOH : 

The ionic product of water, Kw is given by,

Kw = [H₃O⁺] x [OH⁻]

At 298 K, Kw =1 x 10⁻¹⁴.

pK_w = —log₁₀K_w= — log_1o1 x 10⁻¹⁴.= 14

[H₃O⁺] x [OH⁻]  =1 x 10⁻¹⁴

Taking logarithm to base 10 of both sides,

log₁₀[H₃O⁺] + log_1o[OH⁻] = log_1o1 x 10⁻¹⁴

Multiplying both the sides by — 1,

—log₁₀[H₃O⁺] — log_1o[OH⁻] = —log_1o1 x 10⁻¹⁴

‘.’ pH = —log₁₀[H₃O⁺],  pOH = — log_1o[OH⁻]

pK_w = —log₁₀K_w

∴ pH + pOH = pKw

Or pH + pOH = 14

(i) Salts derived from strong acid and strong base :

• Example : NaCl, Na₂SO₄, NaNO₃, KCl, KNO₃.

NaCl :

NaOH     +     HCl      ⇌     NaCl + H₂O

strong base    strong acid            salt

(ii) Salts derived from strong acids and weak bases :

• Example : NH₄Cl, CuSO₄, NH₄NO₃, CuCl₂.

NH₄Cl :

NH₄OH      +    HCI         ⇌    NH₄Cl+ H₂O

weak base   strong acid         salt

(iii) Salts derived from weak acids and strong bases :

• Example : CH₃COONa, KCN, Na₂CO₃.

CH₃COONa :

CH₃COOH   +    NaOH     ⇌    CH₃COONa + H₂O

weak acid        strong base          Salt

(iv) Salts derived from weak acids and weak bases :

• Example : CH₃COONH₄, NH₄CN.

CH₃COONH₄ :

CH₃COOH + NH₄OH       ⇌     CH₃COONH₄ + H₂O

weak acid     weak base              salt

Hydrolysis of the salt of strong acid and weak base ;

Salt CuSO₄ is obtained from strong acid H₂SO₄ and weak base Cu(OH)₂.

When it is dissolved in water it undergoes hydrolysis as follows :

CuSO₄(aq)  +   2H₂O(l)   ⇌     Cu(OH)₂   +   H₂SO₄

salt                                          weak base      strong acid

Cu²⁺(aq) + SO₄^2—(aq) +2H₂O ⇌ Cu(OH)₂(aq) +2H⁺(aq) + SO₄^2—(aq)

Cu²⁺(aq) + 2H₂O (l) ⇌ Cu(OH)₂(aq) + 2H⁺(aq)

OR

Cu²⁺(aq) + 4H₂O (l) ⇌ Cu(OH)₂(aq) + 2H₃O⁺(aq)

Since, [H₃O⁺] > [OH^—], the solution is acidic in nature.

Hydrolysis of a salt of weak acid and strong base : .

Consider a salt of weak acid and strong base like CH₃COONa. In aqueous solution it undergoes hydrolysis as follows :

CH₃COONa(aq)   +    H₂O(l)  ⇌    CH₃COOH(aq)  + NaOH(aq)

Salt                                                    weak acid          strong base

CH₃COO(aq)  + Na⁺(aq)  + H₂O(l) ⇌ CH₃COOH(aq)  + Na⁺ + OH^—(aq)

Since the base NaOH is strong, it dissociates completely while acid CH₃COOH being weak dissociates partially.

Hence [OH^—] > [H₃O⁺] in the solution and the solution reacts basic.

Hydrolysis of the salt of weak acid and weak base :

Consider a salt BA of weak acid (HA) and weak base (BOH).

In aqueous solution it undergoes hydrolysis as follows :

BA(aq)  + H₂O(l)     ⇌    BOH(aq)   +  HA(aq)

salt                                weak base       weak acid

B⁺(aq)  + A^— (aq)  + H₂O(l)  ⇌ BOH(aq) + HA(aq)

The nature of the solution will depend upon relative strength of weak acid and weak base, hence will depend upon their dissociation constants K_a and K_b.

(i) A salt of weak acid and weak base for which K_a > K_b :

Consider hydrolysis of NH₄F.

NH4⁺(aq) + F^—(aq) + H₂O(l)  ⇌ NH₄OH  +   HF

weak base     weak acid

Since K_a (7.2 x 10^—4) for HF is greater than K_b (1.8 x 10^—5) for NH₄OH, the acid dissociates partially more than the base, hence,

[H₃O⁺] > [OH^—] and the solution reacts acidic after hydrolysis.

(ii) A salt of weak acid and weak base for which K_a < K_b:

Consider hydrolysis of NH₄CN.

NH₄⁺(aq)    +     CN^—(aq) + H₂O(l)  ⇌    NH₄OH(aq) + HCN(aq)

weak base         weak acid

Since K_a (4 x 10^—10) for HF is greater than K_b (1.8 x 10^—5) for NH₄OH, the base dissociates more than the acid, hence,

[H₃O⁺] < [OH^—] and the solution reacts basic after hydrolysis.

(ii) A salt of weak acid and weak base for which K_a = K_b :

Consider hydrolysis of CH₃COONH₄.

CH₃COO^— + NH₄⁺ H₂O   ⇌  CH₃COOH + NH₄OH

Salt                                           weak acid       weak base

Since K_a = K_b, the weak acid CH₃COOH and weak base NH₄OH dissociate to the same extent, hence, [H₃O⁺] = [OH^—] and the solution reacts neutral after hydrolysis.

Mechanism of action of an acidic buffer :

(i) An acidic buffer is a mixture of a weak acid and its salt with a strong base. The weak acid dissociates feebly, but the salt dissociates almost completely.

Moreover, due to the common ions, largely supplied by the salt, dissociation of the weak acid is further suppressed.

(CH₃COOH + CH₃COONa) :

CH₃COONa(aa) ——> CH3COO(aq) + Na⁺(aq) (Complete)

(ii) When a small quantity of strong acid (H⁺) is added to this mixture, hydrogen ions combine with acetate ions to form undissociated acetic acid. Thus, addition of an acid does not change the pH of the buffer.

H⁺(aq) + CH₃COO(aq) ⇌ CH₃ + COOH(aq)

This removal of added H⁺ is called reserved basicity.

(iii) When a small quantity of a strong base (OH^—) is added, the hydroxide ions react with the acid producing the corresponding anions and water.

Thus, the concentrations of H⁺ and OH^— in the solution do not change and the pH remains constant.

CH₃COOH(aq) + OH^—(aq) ⇌ CH₃COO^—(aq) + H₂O(l)

This removal of added OH^— is called reserved acidity.

Mechanism of action of a basic buffer :

A basic buffer solution is a solution containing a weak base and its salt with a strong acid. The weak base dissociates feebly, but the salt dissociates completely. Moreover, due to the presence of the common ion, largely supplied by the salt, the dissociation of the base is further suppressed.

(NH₄OH + NH₄Cl) :

NH₄Cl(aq)  ——> NH₄⁺(aq) + Cl^—(aq)    (Complete)

When a small quantity of a strong acid is added to the solution, the hydrogen ions combine with the base producing corresponding cations and water.

Thus, the addition of an acid does not change the pH of the buffer.

H⁺(aq) + NH₄OH(aq) ⇌  NH₄⁺(aq) + H₂O(l)

This removal of added H⁺ is called reserved basicity.

When a small quantity of a strong base is added, the hydroxide ions combine with NH₄⁺ ions to form undissociated NH₄OH. As a result, the hydrogen or hydroxyl ion concentration does not change. Thus, the pH of the solution does not change.

OH^— (aq) + NH₄⁺ ⇌ NH₄OH(aq)

This removal of added OH^—  is called reserved acidity.

Applications of buffer solution

Buffer solution finds extensive applications in a variety of fields. Some of its applications are given.

(i) In biochemical system : pH of blood in our body is maintained at 7.36 - 7.42 due to (HCO₃ + H₂CO₃) buffer. A mere change of 0.2 pH units can cause death. The saline solution used for intravenous injection must contain buffer system to maintain the proper pH of the blood.

(ii) Agriculture : The soils get buffered due to presence of salts such as carbonate, bicarbonate, phosphates and organic acids. Depending on pH the fertilizers are selected.

(iii) Industry : Buffers play an important role in paper, dye, ink, paint and drug industries.

(iv) Medicine : Many medicines particularly in the liquid state have a good stability and optimum activity at a definite pH, for which buffer solutions are used.

• For example penciline preparations are carried out in the presence of a buffer of sodium citrate.
• A buffer solution of magnesium citrate is prepared by adding citric acid to Mg(OH)₂.

(v) Analytical chemistry : In a qualitative analysis, the precipitation of groups, the chemical tests for detection of ions, etc. are carried out at a definite pH.

• For example, precipitation of cations of IIIA are carried in the presence of a basic buffer of pH 8 — 10 is maintained with the use of (NH₄OH + NH₄Cl) buffer.

Relationship between solubility and solubility product :

Consider a saturated solution of a spraingly soluble; electrolyte (or salt) A_xB_y at a given constant temperature.

Let S mol dm^-3 be the solubility of A_xB_y.

A following heterogeneous ionic equilibrium exists.

A_xB_y(s)  ⇌  A_xB_y(dissolved) ⇌  xA^y+   +    yB^x—

(solid)              S                     xS           yS mol dm^—3

Hence final equilibrium will be,

A_xB_y(s) ⇌ xA^y+ + yB^x—

                  xS         yS

By a law of mass action, the equilibrium constant K will be represented as,

K = $\frac{[A^{y+}]×[B^{x-}]}{[A_xB_{y(s)}]}$

∴  K x [A_xB_y(s)] = [A^y+]^x x [B^x—]^y

Since the active mass (concentration) of pure solid

A_xB_y(s) is treated as constant, [A_xB_y(s)] = K’

 K x [A_xB_y(s)] = K x K’ = K_(sp)

Therefore, K_sp = [A^y+]^x x [B^x—]^y

where K_sp is called solubility product of A_xB_y.

At equilibrium the concentrations are,

[A^y+] = xS mol dm^-3

[B^x—] = yS mol dm^-3

 Ksp = [A^y+]^x x [B^x—]^y

= (xS)^x x(yS)^y

K_sp = x^x.y^y.(S)^x+y  …….(1)

Hence solubility S is given by, -

S =$(\frac{K_{sp}}{x^xy^y})^{\frac{1}{x+y}}$  mol dm^-3  …..(2)

The above equations, (1) and (2) give the relationship between solubility and solubility product.

Here x and y represent number of cations and anions respectively from the electrolyte.

Explanation :

A weak electrolyte dissociates partially in aqueous solution to produce cations and anions.

Equilibrium exists between ions thus formed and the undissociated molecules.

BA ⇌ B⁺ + A^-

For such an equilibrium, the dissociation constant

K is defined as

K = $\frac{[B^+]×[A^-]}{[BA]}$

K is constant for the weak electrolyte at a given temperature.

Now, if another electrolyte BC or DA is added to the solution BA, having a common ion either B⁺ or A^-, then the concentration of either B⁺ or A^-, is increased. However, as K is always constant, the increase in the concentration of any one of the ions shifts the equilibrium to left. In other words, the dissociation of BA is suppressed. This is called common ion effect.

For example, the dissociation of a weak acid CH₃COOH is suppressed by adding

CH₃COONa having common ion CH₃COO^- .

CH₃COONa —> CH₃COO^- + Na⁺;

CH₃COOH   ⇌  CH₃COO^-  + H⁺

Common ion effect on dissociation of a weak acid :

Consider the dissociation or ionisation of a weak acid, CH₃COOH in its solution.

CH₃COOH(aq)   ⇌  CH₃COO^-(aq)  + H⁺(aq)

The dissociation constant K_a for CH₃COOH will be,

K_a = $\frac{[CH_3COO^-]×[H^+]}{[CH_3COOH]}$

K_a is constant for CH₃COOH at constant temperature.

If a strong electrolyte like salt CH₃COONa is added to the solution of CH₃COOH, then on dissociation it gives a common ion CH3COO^-

CH₃COONa —> CH₃COO^- + Na⁺;

Due to common ion CH₃COO^-, overall concentration of CH₃COO^- in the solution is increased, which increases the ratio, $\frac{[CH_3COO^-]×[H^+]}{[CH_3COOH]}$

In order to keep this ratio constant, the concentration of H⁺ is decreased, by shifting the equilibrium to the left hand side according to Le Chatelier’s principle.

Thus the ionisation of a weak acid is suppressed by a common ion.

Common ion effect on dissociation of a weak base :

Consider the dissociation or ionisation of a weak base, NH₄OH in its dilute solution.

NH₄OH(aq) ⇌ NH₄⁺(aq)  + OH^-(aq)

The dissociation constant K_b for NH₄OH will be,

K_b = $\frac{[NH_4^+]×[OH^-]}{[NH_4OH]}$

If a strong electrolyte like salt NH₄Cl is added to the solution of NH₄OH, then it gives common ion NH₄⁺

NH₄Cl ---> NH₄⁺ + Cl^-

Due to common ion NH₄⁺ overall concentration of NH₄⁺ is increased, which increases the ratio

[NH₄⁺] x [OH^-]/[NH₄OH]

In order to keep this ratio constant, the equilibrium is shifted to the left hand side which satisfies Le Chatelier’s principle.

Thus the ionisation of a weak base is suppressed by a common ion.