The ionization constant of nitrous acid is `4.5xx10^(-4)`. Calculate the `pH` of `0.04 M` sodium nitrite solution and also its degree of hydrolysis.

Explanation

`underset(C(1-alpha))(NO_(2)^(-))H_(2)OhArrunderset(CH)(HNO_(2))+underset(CH)(OH^(-))`
where h is degree of hydroloysis `[OH^(-)]=ch`,
Also, `h=sqrt((K_(H))/(C))=sqrt((K_(w))/(K_(a).C))`
`=sqrt((10^(-14))/(4.5xx10^(-4)xx0.04))`
`=2.36xx10^(-5)`
`:. [OH^(-)]=0.04xx2.3xx10^(-5)`
or `pOH=6.025`
`:. pH= 14-pOH=7.975`