Question

The ionization constant of nitrous acid is 4.5 × 10^–4. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.

Answer 1

NaNO₂ is the salt of a strong base (NaOH) and a weak acid (HNO₂).

\[\ce{NO_2^- + H_2O ↔ HNO_2 + OH-}\]

`"K"_"h" = (["HNO"_2]["OH"^(-)])/(["NO"_2^-])`

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"_"h" = x^2/0.04 = 0.22 xx 10^(-10)`

`x^2 = .0088 xx 10^(-10)`

`x = .093 xx 10^(-5)`

`therefore ["OH"^(-)] = 0.093 xx 10^(-5) "M"`

`["H"_3"O"^+] = 10^(-14)/(.093 xx 10^(-5)) = 10.75 xx 10^(-9)` M

`=> "pH" = - log (10.75 xx 10^(-9))`

= 7.96

Therefore, degree of hydrolysis

`x/0.04 = (.093 xx 10^(-5))/.04 = 2.325 xx 10^(-5)`