Question

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

• A. 8.96
• B. 7.96
• C. 4.72
• D. 5.89

Answer 1

The correct option is B 7.96

Since $NaN{O}_2$ is the salt of a strong base $\left(NaOH\right)$ and a weak acid $\left(HN{O}_2\right)$ and ${NO}_2^{-}+H_{2}O\leftrightarrow HN{O}_2+O{H}^{-}$

So,

$K_b=\frac{\left[HN{O}_2\right]\left[O{H}^{-}\right]}{\left[{NO}_2^{-}\right]}$

$\Rightarrow \frac{K_w}{K_a}=\frac{{10}^{-14}}{4.5\times {10}^{-14}}=.22\times {10}^{-10}$

Now, If $x$ moles of the salt undergo hydrolysis, then the concentration of various species present in the solution will be:

$\left[{NO}_2^{-}\right]=.04-x;0.04$

$\left[HN{O}_2\right]=x$

$\left[O{H}^{-}\right]=x$

$K_b=\frac{x^2}{0.04}=0.22\times {10}^{-10}$

$x^2=.0088\times {10}^{-10}$

$x=0.093\times {10}^{-5}$

$\therefore \left[O{H}^{-}\right]=0.093\times {10}^{-5}$ M

$\left[H_3{O}^{+}\right]=\frac{{10}^{-14}}{0.093\times {10}^{-5}}=10.75\times {10}^{-9}$ M

$\Rightarrow pH=-\log (10.75\times {10}^{-9})$

$=7.96$