Question

If $K_a=1.8\times {10}^{-5}$, calculate the degree of hydrolysis of 0$.1M$ solution of sodium acetate at ${25}^{\circ }C$:

• A. $7.45\times {10}^{-5}$
• B. $7.45\times {10}^{-3}$
• C. $7.45\times {10}^{-1}$
• D. $5.5\times {10}^{-2}$

Answer 1

The correct option is A $7.45\times {10}^{-5}$

Sodium acetate is a salt of strong base and weak acid.
$C{H}_{3}CO{O}^{-}\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons C{H}_{3}COOH\left(aq\right)+O{H}^{-}\left(aq\right)$
$C00$
$C-ChChCh$
In this reaction the $\left(K_h\right)$ hydrolysis constant is:
$K_h=\frac{\left[C{H}_{3}COOH\right]\left[O{H}^{-}\right]}{\left[C{H}_{3}CO{O}^{-}\right]}=\frac{(Ch{)}^2}{C(1-h)}$
$C{H}_{3}COOH\left(aq\right)\rightleftharpoons C{H}_{3}CO{O}^{-}\left(aq\right)+O{H}^{-}\left(aq\right)$
$K_a=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H^{+}\right]}{\left[C{H}_{3}COOH\right]}$
$H_{2}O\left(l\right)\rightleftharpoons H^{+}\left(aq\right)+O{H}^{-}\left(aq\right)$
In this reaction the $K_w$ constant is:
$K_w=\left[H^{+}\right]\left[O{H}^{-}\right]$
$\frac{K_w}{K_a}=\frac{\left[H^{+}\right]\left[O{H}^{-}\right]}{\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H^{+}\right]}{\left[C{H}_{3}COOH\right]}}$
$\frac{K_w}{K_a}=\frac{\left[C{H}_{3}COOH\right]\left[O{H}^{-}\right]}{\left[C{H}_{3}CO{O}^{-}\right]}$
$K_h=\frac{K_w}{K_a}=C{h}^{2}h=\sqrt{\frac{K_w}{K_a\times C}}h=\sqrt{\frac{{10}^{-14}}{1.8\times {10}^{-5}\times 0.1}}h=7.45\times {10}^{-5}$