Question

The ionization constant of nitrous acid is $4.5\times {10}^{-4}$. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.

Answer 1

$NaN{O}_2$ is the salt of a strong base (NaOH) and a weak acid $\left(HN{O}_2\right)$.
$N{O}_2^{-}+H_{2}O\leftrightarrow HN{O}_2+O{H}^{-}{K}_b=\frac{\left[HN{O}_2\right]\left[O{H}^{-}\right]}{\left[N{O}_2^{-}\right]}\Rightarrow \frac{K_w}{K_a}=\frac{{10}^{-14}}{4.5\times {10}^{-4}}=.22\times {10}^{-10}$
Now, If x moles of the salt undergo hydrolysis, then the concentration of various species present in the solution will be:
$\left[N{O}_2^{-}\right]=.04-x;0.4\left[HN{O}_2\right]=x\left[O{H}^{-}\right]=x{K}_b=\frac{x^2}{0.04}=0.22\times {10}^{-10}{x}^2=.0088\times {10}^{-10}x=.093\times {10}^{-5}\therefore \left[O{H}^{-}\right]=0.093\times {10}^{-5}M\left[H_3{O}^{+}\right]=\frac{{10}^{-14}}{.093\times {10}^{-5}}=10.75\times {10}^{-9}M\Rightarrow pH=-log(1.75\times {10}^{-9})=7.96$
Therefore, degree of hydrolysis
$=\frac{x}{0.04}=\frac{.093\times {10}^{-5}}{.04}=2.325\times {10}^{-5}$