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).
NO2- + H2O $\leftrightarrow$ HNO2 + OH-
Kh =$\frac{ [HNO_2][OH^-]}{NO_2^-}$
⇒ $\frac{K_w}{K_\alpha}$ = $\frac{10^{-14}}{4.5 × 10^{-4}}$ = .22 × 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
[HNO2] = x
[OH-] = x
Kh = $\frac{x^2}{0.04}$= 0.22 × 10-10 = x2 = .0088 × 10-10
x = .093 × 10-5
∴[OH-] = 0.093 × 10-5M
[H3O+] = $\frac{10-^{14}}{.093 × 10^{-5}}$ = 10.75 × 10-9M ⇒ pH = -log(10.75 × 10-9) = 7.96
Therefore, degree of hydrolysis = $\frac{x}{0.04}$ = .093 × $\frac{10^{-5}}{.04}$= 2.235 × 10-5