Question

Calculate `pH` and degree of hydrolysis fo `10^(-2)MNH_(4)CN` solution.
Given that `K_(a)` of `HCN=5xx10^(-10)` and `K_(b)` of (aq.`NH_(3))=2xx10^(-5)` at `25^(@)C`.

Answer 1

`pH=(1)/(2)[14+pK_(a)-pK_(b)]=(1)/(2)[14+10-log5-5+log2]=(1)/(2)[18.6]=9.3`
`(h)/(h-1)=sqrt(K_(h))=sqrt((K_(w))/(K_(a)xxK_(b)))=sqrt((10^(-14))/(5xx10^(-16)xx2xx10^(-5)))=1`
`rArr 2h=1`
`rArr h=(1)/(2)=0.5`