Explain the term Hydrolysis of salts with examples. Discuss the pH of the following types of salt solutions.
(i) Salts of weak acid and strong base.
(ii) Salts of strong acid and weak base.

Explanation

Hydrolysis of salts in the reverse preocess of neutralisation. In this process a salt react with water to form an acid and a base.
Salt `+` Water `hArr` Acid `+` Base.
The fraction of the total salt that is hydrolysed at equilibrium is called degree or extent of hydrolysis. It is denoted by hy.
(i) Salts of weak acid and strong base: This type of salts produce alkaline solutions on hydrolysis.
`MX+H_(2)OhArr underset("Strong Base")(MOH)+underset("Weak acid")(HX)`
In solution MX and the strong base MOH remains almost undissociated.
`M^(+)+X^(-)+2H_(2)hArrMOH+H_(3)O^(+)+X^(-)`
`M^(+)+2H_(2)OhArrMOH+H_(3)O^(+)`
Applying law of mass action
`K_(h)=([MOH][H_(3)O^(+)])/([M^(+)])`
For the dissociation of weak base
`MOHhArrM^(+)+OH^(-)`
`K_(b)=([M^(+)][OH^(-)])/([MOH])`
Multiply with `K_(h)`
`K_(b)xxK_(h)=([M^(+)][OH^(-)])/([MOH])=([MOH][H_(3)O^(+)])/([M^(+)])`
`=[H_(3)O^(+)][OH^(-)]=K_(w),K_(h)=(K_(w))/(K_(b))`
Degree of hydrolysis:
`underset(C(1-h))(M^(+)+2H_(2)O)hArrunderset(Ch)(MOH)+underset(Ch)(H_(3)O^(+))`
`K_(h)=([MOH][H_(3)O^(+)])/([M^(+)])=((Ch)(Ch))/(C(1-h))=Ch^(2)`
`( :' 1-h=1)`
`h=(K_(w))/C` or `sqrt((K_(w))/(K_(b)xxC))`
Hydronium ion concentration and pH
`[H_(3)O^(+)]=Ch=Csqt((K_(w))/(K_(b)xxC))=sqrt((K_(w)xxC)/(K_(b)))`
`-log[H_(3)O^(+)]=1/2 log K_(w)+1/2logK_(b)-1/2logC`
`pH=1/2pK_(w)-1/2pK_(b)-1/2logC`