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

(i). Calculation of degree of hydrolysis (h) ,
Sodium nitrite `(NaNO_(2))` is a salt of strong base (NaOH) and weak acid `(HNO_(2))` . Its degree of hydrolysis may be caluculated as follows :
`K_(h) =(K_(w))/(K_(a)) =(1.0 xx 10^-14)/(4.5 xx 10^(-4)) =2.22 xx 10^(-11)`
Degree of hydrolysis (h) `=sqrt((K_(h))/(C)) =sqrt((2.22 xx 10^(-11))/(0.04)) = (5.55 xx 10^(-10))^(1//2) =2.36 xx 10^(-5)`
(II) . Calculation of pH of hte solution
The hydrolysis of `NaNO_(2)` in aqueous solution may be represented as follows:
`underset(C (1-h))(NO_(2)^(-)) + H_(2)O hArr underset(Ch)(HNO_(2)) + underset(Ch)(OH^(-))`
`[OH^(-)] = C xx h =0.04 xx 2.36 xx 10^(-5) =9.44 xx 10^(-7)`
`[H^(+)] =(K_(w))/(9.44xx 10^(-7)) =(1.0 xx 10^(-14))/(9.44 xx 10^(-7)) =1.06 xx 10^(-8) M`
`pH =- log [H^(+)] =- log [1.06 xx10^(-8)]`
`=[8 - log 1.06] =8 -0.025 =7.975`