Question

The pH of an aqueous solution of $C{H_3}COONa$ of concentration C (M) is given by,
A.$7 - \dfrac{1}{2}p{K_a} - \dfrac{1}{2}\log C$
B.$\dfrac{1}{2}p{K_w} + \dfrac{1}{2}p{K_b} + \dfrac{1}{2}\log C$
C.$\dfrac{1}{2}p{K_w} - \dfrac{1}{2}p{K_b} - \dfrac{1}{2}\log C$
D.$\dfrac{1}{2}p{K_w} + \dfrac{1}{2}p{K_a} + \dfrac{1}{2}\log C$

Answer 1

Hint: $C{H_3}COONa$ is salt of a strong base ($NaOH$) and a weak acid ($C{H_3}COOH$).The reaction of a salt with water is called salt hydrolysis. Hydrolysis of salt of strong base and weak acid results in an alkaline solution.

Complete step by step answer:
Sodium acetate ($C{H_3}COONa$) when dissolved in water dissociate completely to acetate ion and sodium ion as shown by the equation,
$C{H_3}COONa \to C{H_3}CO{O^ - } + N{a^ + }$
The acetate ion then take the hydrogen ion furnished by slightly dissociated water to form acetic acid.
$C{H_3}CO{O^ - } + {H^ + } \to C{H_3}COOH$
Hence the concentration of hydrogen ions decreases in solution. Then to maintain a constant value of ${K_w}$, the un-dissociated water further undergo dissociation.
${H_2}O \rightleftharpoons {H^ + } + O{H^ - }$
Then again the hydrogen ions are taken up by acetate ion. This results in an increase in concentration of hydroxyl ions and decrease in concentration of hydrogen ions. Hence the solution becomes basic.
Let C be the initial concentration of sodium acetate and x be the degree of hydrolysis. The hydrolysis reaction can be written as,
$C{H_3}CO{O^ - } + {H_2}O \to C{H_3}COOH + O{H^ - }$
 $C(1 - x)$ $Cx$ $Cx$
Applying the law of chemical equilibrium and taking the concentration of water as constant, the hydrolysis constant, ${K_h}$ can be written as,
${K_h} = \dfrac{{\left[ {C{H_3}COOH} \right]\left[ {O{H^ - }} \right]}}{{\left[ {C{H_3}CO{O^ - }} \right]}}$
${K_h} = \dfrac{{Cx.Cx}}{{C(1 - x)}} = \dfrac{{C{x^2}}}{{1 - x}}$
x is a small value compared to one. Therefore,
${K_h} = c{x^2}$.
Or, $x = \sqrt {\dfrac{{{K_h}}}{C}} $
Now pH can be calculated by,
$pH = - \log \left[ {{H^ + }} \right] = 14 - pOH$
Since the solution is alkaline, we can find out the value of pOH. Then we can subtract this value from $14$ to get pH.
We have $\left[ {O{H^ - }} \right] = Cx$
For the hydrolysis of weak acid and strong base, ${K_h}$ can be written as,
${K_h} = \dfrac{{{K_w}}}{{{K_a}}}$
Hence we can write,
$\left[ {O{H^ - }} \right] = Cx = C\sqrt {\dfrac{{{K_h}}}{C}} = C\sqrt {\dfrac{{{K_w}}}{{{K_a}C}}} = {\left( {\dfrac{{{K_w}C}}{{{K_a}}}} \right)^{\dfrac{1}{2}}}$
Taking negative log on both sides,
$
   - \log \left[ {O{H^ - }} \right] = - \dfrac{1}{2}\log {K_w} - \dfrac{1}{2}\log C + \dfrac{1}{2}\log {K_a} \\
  pOH = \dfrac{1}{2}p{K_w} - \dfrac{1}{2}\log C - \dfrac{1}{2}p{K_a} \\
  pH = 14 - pOH \\
  pH = 14 - \dfrac{1}{2}p{K_w} + \dfrac{1}{2}\log C + \dfrac{1}{2}p{K_a} \\
 $
But $p{K_w} = 14$. Hence we can write,
$
  pH = p{K_w} - \dfrac{1}{2}p{K_w} + \dfrac{1}{2}\log C + \dfrac{1}{2}p{K_a} \\
  pH = \dfrac{1}{2}p{K_w} + \dfrac{1}{2}\log C + \dfrac{1}{2}p{K_a} \\
 $
Hence pH of an aqueous solution of $C{H_3}COONa$ of concentration C (M) is given by $\dfrac{1}{2}p{K_w} + \dfrac{1}{2}p{K_a} + \dfrac{1}{2}\log C$ .
Hence the correct option is D.

Note:
This equation holds only for the hydrolysis of weak acid and strong base. For hydrolysis of strong acid and weak base or weak acid and weak base, the equation to find out pH is different.