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.

Answer

`pH = 7.97`. Degree of hydrolysis `= 2.36 xx 10^(-5)`

Explanation

`"NaNO"_(2)` is the salt of a strong base (NaOH) and a weak acid `"HNO"_(2)`.
`NO_(2)^(-)+H_(2)OleftrightarrowHNO_(2)+OH^(-)` `K_(h) = ([HNO_(2)][OH^(-)])/([NO_(2)^(-)])`
`rArr (K_(w))/(K_(a)) = (10^(-14))/(4.5 xx 10^(-4)) = .22 xx 10^(10)`
Now, If x moles of the salt undergo hydrolysis, then the concentration of various species present in the solution will be:
`[NO_(2)^(-)] = .04-x , 0.04`
`[HNO_(2)] = x`
`[OH^(-)] = x`
`K_(b) = (x^(2))/(0.04) = 0.22 xx 10^(-10)`
`x^(2) = .0088 xx 10^(-10)`
`x = .093 xx 10^(-5)`
`:. [OH^(-)] = 0.093 xx 10^(-5) M`
`[HO^(+)] = (10^(-14))/(.093 xx 10^(-5)) = 10.75 xx 10^(-9) M`
`rArr pH = - "log" (10.75 xx 10^(-19))`
`= 7.96`
Therefore, degree of hydrolysis
`= (x)/(0.04) = (.093 xx 10^(-5))/(.04) = 2.325 xx 10^(-5)`