Question

For sodium acetate solution in water, the given equilibrium reaction occur:
$C{H}_{3}CO{O}^{-}\left(aq\right)+H_{2}O\left(l\right)\stackrel{hydrolysis}{\rightleftharpoons }C{H}_{3}COOH\left(aq\right)+O{H}^{-}\left(aq\right)$
Which of the following describes the correct relation ?
(where $h$ is the degree of hydrolysis of salt)

• A. $h=\sqrt{\frac{K_w\cdot C}{K_a}}$
• B. $h=\sqrt{\frac{K_w}{K_a\cdot C}}$
• C. $h=\sqrt{\frac{K_a\cdot C}{K_w}}$
• D. $h=\sqrt{\frac{K_a}{K_w\cdot C}}$

Answer 1

The correct option is B $h=\sqrt{\frac{K_w}{K_a\cdot C}}$

Salt of weak acid and strong base. If C is the concentration and h is the degree of hydrolysis, then hydrolysis of anion takes place as following :
$\underset{C(1-h)}{C{H}_{3}CO{O}^{-}}\left(aq\right)+H_{2}O\left(l\right)\stackrel{hydrolysis}{\rightleftharpoons }\underset{Ch}{C{H}_{3}COOH}\left(aq\right)+\underset{Ch}{O{H}^{-}}\left(aq\right)$

$K_h=\frac{\left[C{H}_{3}COOH\right]\left[O{H}^{-}\right]}{\left[C{H}_{3}CO{O}^{-}\right]}=\frac{Ch\times Ch}{C(1-h)}=\frac{C{h}^2}{1-h}$
and,
$C{H}_{3}COOH\rightleftharpoons C{H}_{3}CO{O}^{-}+H^{+}{K}_a=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H^{+}\right]}{\left[C{H}_{3}COOH\right]}{H}_{2}O\rightleftharpoons H^{+}+O{H}^{-}{K}_w=\left[H^{+}\right]\left[O{H}^{-}\right]$
We see that
$K_h=\frac{K_w}{K_a}=\frac{C{h}^2}{1-h}Forh<0.1Then,1-h\approx 1\frac{K_w}{K_a}=C{h}^{2}h=\sqrt{\frac{K_w}{K_a\cdot C}}$