The ionization constant of nitrous acid is `4.5 xx 10^(-4)`. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.

Explanation

`K_(a)=4.5xx10^(-4)`
`K_(h)=K_(w)/K_(a)=(1xx10^(-14))/(4.5xx10^(-4))=1/4.5xx10^(-10)`
Let the degree of hydrolysis=h
`K_(h)=(ch^(2))/(1-h)`
`10^(10)/4.5=0.04xxh^(2)`
`:. 0.180 h^(2)=10^(-10)`
`h^(2)=10^(-10)/0.180`
`h=sqrt(5.5)xx10^(-5)=2.36xx10^(-5)`
`NaNO_(2)` is salt of `W_(A)//W_(B)`
`pK_(a)=-log(4.5xx10^(-4)=3.346~~3.35`
`log C=log 0.04=-1.39~~-1.40`
`=1/2 (pK_(w)+pK_(a)+ logC)`
`=1/2(14+3.35-1.40)=7.975`