Question

Calculate the drgee of hydrolysis and `pH` of `0.02M` ammonium cyanide `(NH_(4)CN)` at `298K`. `(K_(a)` of `HCN = 4.99 xx 10^(-9), K_(b)` for `NH_(4)OH = 1.77 xx 10^(-5))`

Answer 1

`K_(h) = (10^(-14))/(4.99 xx 10^(-9) xx 1.77 xx 10^(-5)) = 1.132`
If can be seen that hydrolysis constant `(K_(h))` is not small, and for calculaitng `h`, the equaiton used is:
`h = sqrt(K_(h))/(1+sqrt(K_(h)))=(sqrt(1.132))/(1+sqrt(1.132)) = (1.06)/(1+1.06) = (1.06)/(2.06) = 0.51`
Using the formula
`pH = pK_(a) - log (h) + log (1 - h)`
`=- log (4.99 xx 10^(-10)) - log (0.51) + log (1- 0.51)`
`= 9.30`