```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,

Kh = [Hx ][NH4 OH]/[NH4+ ]=(Cx.Cx)/C(1-x) = (x2 C)/((1-x))     ...... (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-,    Kb = [NH+4][OH-]/[NH4Oh]    .... (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]

pOH = pKb + log[salt]/[base]

Relation between Hydrolysis constant and Degree of hydrolysis

The extent to which hydrolysis proceeds is expressed as the degree of hydrolysisand 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

pH = 1/2pKw - 1/2 log C - 1/2 pKb

= 7 - 1/2 pKb -  1/2log C
```
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