Calculate
(i) pH of the solution,
(ii) degree of hydrolysis of NaCN,
(iii) hydrolysis constant for a 0.1M NaCN solution at `25^@C. [Given: K_a(HCN) = 4.0xx10^(-10)`

Explanation

The hydrolysis reaction is
`NH_4^(+) + CN^(-)+H_2O hArr NH_4OH +HCN`
Hydrolysis constant,
(a)`K_h=K_w/(K_axxK_b)`
`K_a=4.0xx10^(-10), K_b=1.8xx10^(-5)`
`=(1.0xx10^(-14))/((4.0xx10^(-10))xx(1.8xx10^(-5)))`
=1.39
(b)Degree of hydrolysis , h
`h=sqrtK_h`
`=sqrt1.39` =1.18
(c ) pH of the solution is
`pH=1/2pK_w+1/2 pK_a-1/2pK_b`
`pK_w`=14
`pK_a=-log (4xx10^(-10))=9.40`
`pKb=-log (1.8xx10^(-5))=4.74`
`therefore pH=1/2(14)+1/2(9.40)-1/2(4.74)`=9.33