Question

Given,
$NO _{2}^{-}+ H _{2} O \rightleftharpoons HNO _{2}+ OH^- , K_{b}=2.22 \times 10^{-11}$
Thus, degree of hydrolysis of $0.04 \,M\, NaNO _{2}$ solution is

• A. $1.14 \times 10^{-2}$
• B. $2.36 \times 10^{-5}$
• C. $0.11$
• D. $4.86 \times 10^{-3}$

Answer 1

Answer: (B)

$NaNO _{2}$ is a salt of weak acid $\left( HNO _{2}\right)$ and strong base. Its aqueous solution is basic due to hydrolysis of $NO _{2}^{-}$ ion



$x =$ degree of hydrolysis

$K_{h}=\frac{\left[ HNO _{2}\right]\left[ OH ^{-}\right]}{\left[ NO _{2}^{-}\right]}$

$HNO _{2} \rightleftharpoons H ^{+}+ NO _{2}^{-}$

$K_{a}=\frac{\left[ H ^{+}\right]\left[ NO _{2}^{-}\right]}{\left[ HNO _{2}\right]}$

Thus, $K_{ h } K_{ a }=\left[ H ^{+}\right]\left[ OH ^{-}\right]=K_{ w }$

$\therefore K_{h}=\frac{K_{w}}{K_{a}}$

Also, $ K_{h}=\frac{C x \cdot C x}{C(1-x)}=\frac{C x^{2}}{1-x}=C x^{2}$

$\therefore x=\sqrt{\frac{K_{h}}{C}}=\sqrt{\frac{K_{w}}{K_{a} C}}$

For $NO _{2}^{-}$ (conjugate base), $K_{ b }=2.22 \times 10^{-11}$

$\therefore K _{ a }\left[ HNO _{2}\right]=\frac{K_{w}}{K_{b}}=\frac{1.0 \times 10^{-14}}{2.2 \times 10^{-11}}=4.5 \times 10^{-4}$

$x=\sqrt{\frac{K_{w}}{K_{a} \cdot C}}$

$\therefore x=\sqrt{\frac{1 \times 10^{-14}}{4.5 \times 10^{-4} \times 0.04}}=2.36 \times 10^{-5}$