Question

Given
$N{O}_2^{-}+H_{2}O\leftrightharpoons HN{O}_2+O{H}^{-}$, $K_b=2.22\times {10}^{-11}$
Thus, the degree of hydrolysis of 0.04 $M$ $NaN{O}_2$ 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

The correct option is B $2.36\times {10}^{-5}$

$NaN{O}_2$ is a salt of a strong base $\left(NaOH\right)$ and a weak acid $\left(HN{O}_2\right)$. So, hydrolysis reaction will be:
$N{O}_2^{-}+H_{2}O\leftrightharpoons HN{O}_2+O{H}^{-}$

We know degree of hydrolysis is given as
$h=\sqrt{\frac{K_h}{C}}$
and $K_h=\frac{K_w}{K_a}$ (For salt of S.B + W.A)
Now, $K_a=\frac{K_w}{K_b}=\frac{{10}^{-14}}{2.22\times {10}^{-11}}=4.5\times {10}^{-4}$
$K_h=\frac{{10}^{-14}}{4.5\times {10}^{-4}}=2.22\times {10}^{-11}$
and $h=\sqrt{\frac{K_h}{C}}=\sqrt{\frac{2.22\times {10}^{-11}}{0.04}}=2.36\times {10}^{-5}$