Question

Given,
$N{O}_2^{-}+H_{2}O\rightleftharpoons HN{O}_2+O{H}^{-},K_b=2.22\times 10^{-11}$
Thus, degree of hydrolysis of $0.04MNaN{O}_2$ 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)

$NaN{O}_2$ is a salt of weak acid $\left(HN{O}_2\right)$ and strong base. Its aqueous solution is basic due to hydrolysis of $N{O}_2^{-}$ ion



$x=$ degree of hydrolysis

$K_h=\frac{\left[HN{O}_2\right]\left[O{H}^{-}\right]}{\left[N{O}_2^{-}\right]}$

$HN{O}_2\rightleftharpoons H^{+}+N{O}_2^{-}$

$K_a=\frac{\left[H^{+}\right]\left[N{O}_2^{-}\right]}{\left[HN{O}_2\right]}$

Thus, $K_h{K}_a=\left[H^{+}\right]\left[O{H}^{-}\right]=K_w$

$\therefore K_h=\frac{K_w}{K_a}$

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

$\therefore x=\sqrt{\frac{K_h}{C}}=\sqrt{\frac{K_w}{K_{a}C}}$

For $N{O}_2^{-}$ (conjugate base), $K_b=2.22\times 10^{-11}$

$\therefore K_a\left[HN{O}_2\right]=\frac{K_w}{K_b}=\frac{1.0\times 10^{-14}}{2.2\times 10^{-11}}=4.5\times 10^{-4}$

$x=\sqrt{\frac{K_w}{K_a\cdot C}}$

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