Question

For 0.50 M aqueous solution of sodium cyanide, $\left({\text{pK}}_{\text{b}}{\text{of CN}}^{-}\text{is}4.70\right)$, calculate
(i) Hydrolysis constant
(ii) Degree of hydrolysis
iii) pH
log( 3.15 )=0.498
$\text{Antilog (-9.3)}=5.01\times {10}^{-10}$

• A. $2\times {10}^5,6.3\times {10}^{-3},11.5$
• B. $2\times {10}^3,6.3\times {10}^{-3},11.5$
• C. $4\times {10}^3,6.3\times {10}^{-3},9.5$
• D. $2\times {10}^3,4.2\times {10}^{-3},11.5$

Answer 1

The correct option is A $2\times {10}^5,6.3\times {10}^{-3},11.5$

Salt hydrolysis of NaCN is
$NaCN+H_{2}O\to HCN+N{a}^{+}+O{H}^{-}$
Given,
$p{K}_b=4.70$

For conjugate acid-base pair,
$p{K}_a+p{K}_b=14p{K}_a=14-4.70p{K}_a=9.3p{K}_a=-log{K}_{a}9.3=-log{K}_a{K}_a=5.01\times {10}^{-10}{K}_h=\frac{K_w}{K_a}{K}_h=\frac{1\times {10}^{-14}}{5.01\times {10}^{-10}}{K}_h=2\times {10}^{-5}$

(ii) Degree of hydrolysis,
$\text{h}=\sqrt{\left({\text{K}}_{\text{h}}/\text{c}\right)}=\sqrt{(2\times {10}^{-5}/0.5)}$

$=6.3\times {10}^{-3}$

(iii) $\left[{\text{OH}}^{-}\right]=\text{ch}=0.5\times 6.3\times {10}^{-3}$

$\left[{\text{OH}}^{-}\right]=3.15\times {10}^{-3}$
$\text{pOH}=-log\left[O{H}^{-}\right]$
$pOH=-log(3.15\times {10}^{-3})$
$pOH=3-log\left(3.15\right)$
$pOH=3-0.498=2.501$

$pH+pOH=14$
$\text{pH}=14-2.5=11.5$