# Question 1cf69

Nov 3, 2016

Anionic hydrolysis of strong conjugate base $C {H}_{3} C O {O}^{-}$ with its ICE-table

$C {H}_{3} C O {O}^{\text{-}} + {H}_{2} O \stackrel{{K}_{h}}{r i g h t \le f t h a r p \infty n s} C {H}_{3} C O O H + O {H}^{-}$

$I \text{ "" "c" "" "" "" "" "" "" " 0" "" "" "" "" } 0$

$C \text{ "-ch" "" "" "" "" "" " ch" "" "" "" } c h$

$E \text{ "c-ch" "" "" "" "" "" " ch" "" "" "" } c h$

Where c is concentration of $C {H}_{3} C O {O}^{-}$ or sodium acetate before hydrolysis and h is the degree of hydrolysis.

So ${K}_{h} = \frac{\left[C {H}_{3} C O O H\right] \left[O H\right]}{\left[C {H}_{3} C O {O}^{-}\right]} \ldots \ldots \textcolor{red}{\left(2\right)}$

$\implies {K}_{h} = \frac{\left(c h\right) \left(c h\right)}{c \left(1 - h\right)} \approx c {h}^{2}$

$\implies h = \sqrt{{K}_{h} / c} \ldots \ldots \textcolor{red}{\left(3\right)}$

Considering ionic equilibrium of dissociation of $C {H}_{3} C O O H$ in aqueous solution.

$C {H}_{3} C O O H + {H}_{2} O \stackrel{{K}_{a}}{r} i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + C {H}_{3} C O {O}^{-}$

K_a = ([H_3O^(+) ][CH_3COO^-])/([CH_3COOH]) ......color(red)((4))#
Considering ionic equilibrium of dissociation of ${H}_{2} O$

${H}_{2} O + {H}_{2} O \stackrel{{K}_{w}}{r} i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + O {H}^{-}$

The ionic product of water
${K}_{w} = \left[{H}_{3} {O}^{+}\right] \left[O {H}^{-}\right] = {10}^{-} 14 \text{ } \ldots . . \textcolor{red}{\left(5\right)}$
Taking log

$- \log {K}_{w} = - \log \left({10}^{-} 14\right) = 14$

$\implies p {K}_{w} = - \log \left(\left[{H}_{3} {O}^{+}\right] \left[O {H}^{-}\right]\right) = \log \left({10}^{-} 14\right) = 14$

$\implies p {K}_{w} = p H + p O H = - \log \left({10}^{-} 14\right) = 14 \ldots \ldots \textcolor{red}{\left(6\right)}$

Now by (4) and (5)

${K}_{w} / {K}_{a} = \frac{\left[C {H}_{3} C O O H\right] \left[O H\right]}{\left[C {H}_{3} C O {O}^{-}\right]} = {K}_{h}$

Again

$\left[O {H}^{-}\right] = c h = c \times \sqrt{{K}_{h} / c} = \sqrt{{K}_{h} c}$

$p O H = - \log \left[O {H}^{-}\right] = - \log \sqrt{{K}_{h} c}$

$\implies p O H = - \log \sqrt{{K}_{h} c}$

$= - \frac{1}{2} \log {K}_{h} - \frac{1}{2} \log c$

$\implies p O H = \frac{1}{2} \left(- \log \left({K}_{w} / {K}_{a}\right) - \log c\right)$

$\implies p O H = \frac{1}{2} \left(- \log {K}_{w} + \log {K}_{a} - \log c\right)$

$\implies p O H = \frac{1}{2} \left(p {K}_{w} - p {K}_{a} - \log c\right)$

$\implies 14 - p H = \frac{1}{2} \left(14 - p {K}_{a} - \log c\right)$

$\implies 14 - p H = 7 - \frac{1}{2} \left(p {K}_{a} + \log c\right)$

$\implies p H = 7 + \frac{1}{2} \left(p {K}_{a} + \log c\right) \ldots . \left(7\right)$

Here $p H = 9.48$
${K}_{a} = 1.8 \times {10}^{-} 5$
$\implies p {K}_{a} = - \log \left(1.8 \times {10}^{-} 5\right)$

$= 5 - \log 1.8 = 4.74$

Inserting the values in (7)

$9.48 = 7 + \frac{1}{2} \left(4.74 + \log c\right)$

$\implies \frac{1}{2} \log c = 9.48 - 7 - 2.37 = 0.11$

$\implies \log c = 0.22$

$c = 1.66 M$