Question

Calculate the hydrolysis constant of a salt containing $N{O}_2^{-}$ ion. If $K_a$ for $HN{O}_2=4.5\times {10}^{-10}$.

• A. $2.8\times {10}^{-6}$
• B. $20.5\times {10}^{-5}$
• C. $2.2\times {10}^{-5}$
• D. $1.1\times {10}^{-3}$

Answer 1

The correct option is C $2.2\times {10}^{-5}$

$N{O}_2^{-}$ ion hydrolyse as
$N{O}_2^{-}+H_{2}O\rightleftharpoons HN{O}_2+O{H}^{-}$
$K_H=\frac{\left[HN{O}_2\right]\left[O{H}^{-}\right]}{\left[N{O}_2^{-}\right]}\dots \dots \left(1\right)$
$K_w=\left[H^{+}\right]\left[O{H}^{-}\right]\dots \dots \left(2\right)$
$HN{O}_2$ ionizes as follows:
$HN{O}_2\rightleftharpoons H^{+}+N{O}_2^{-}$
$K_a=\frac{\left[H^{+}\right]\left[N{O}_2^{-}\right]}{\left[HN{O}_2\right]}\dots \dots \left(3\right)$
From equations (1) and (3)
$K_H=\frac{K_w}{K_a}=\frac{1\times {10}^{-14}}{4.5\times {10}^{-10}}$
$=2.22\times {10}^{-5}$