Question

The dissociation constant of a substituted benzoic acid at ${25^\circ }{\text{C}}$ is $1.0 \times {10^{ - 4}}$. Find the pH of a $0.01{\text{ M}}$ solution of its sodium salt.
A.$4$
B.$6$
C.$8$
D.$10$

Answer 1

First, we will identify the acid base pair of the salt. Here, the salt gives benzoic acid and sodium hydroxide on hydrolysis so it is a salt of strong base and weak acid. To find the pH of this salt, we will use the formula-
$\Rightarrow {\text{pH = 7 + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{p}}{{\text{K}}_{\text{a}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{logC}}$ --- (i)
Where ${\text{p}}{{\text{K}}_{\text{a}}} = - {\log _{10}}\left( {{{\text{K}}_{\text{a}}}} \right)$ and C is concentration of the salt. Here ${{\text{K}}_{\text{a}}}$is a dissociation constant of acid.
Put the given values in the formula and solve it to get the answer.

Complete step-by-step answer: Given, the dissociation constant of a substituted benzoic acid=$1.0 \times {10^{ - 4}}$
The concentration of its sodium salt=$0.01{\text{ M}}$
We have to find pH of the sodium salt of benzoic acid.
Sodium salt of benzoic acid ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COONa}}$ is prepared from strong base (sodium hydroxide)${\text{NaOH}}$and weak acid (benzoic acid)${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH}}$ which can be written as-
$\Rightarrow$ Salt$\rightleftharpoons$ Weak acid Strong base
The formula of pH of salt of weak acid and strong base is given as-
$\Rightarrow {\text{pH = 7 + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{p}}{{\text{K}}_{\text{a}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{logC}}$ --- (i)
Where ${\text{p}}{{\text{K}}_{\text{a}}} = - {\log _{10}}\left( {{{\text{K}}_{\text{a}}}} \right)$ and C is concentration of the salt
We are given that the dissociation constant ${{\text{K}}_{\text{a}}} = {10^{ - 4}}$ so we get-
$\Rightarrow$ ${\text{p}}{{\text{K}}_{\text{a}}} = - {\log _{10}}\left( {{{10}^{ - 4}}} \right)$
We know that that$\log {m^n} = n\log m$, so on applying this, we get-
$\Rightarrow$ ${\text{p}}{{\text{K}}_{\text{a}}} = - \left( { - 4} \right){\log _{10}}\left( {10} \right)$
Now, we know that ${\log _{10}}10 = 1$ so we get-
$\Rightarrow {\text{p}}{{\text{K}}_{\text{a}}} = 4$--- (ii)
Also given, Concentration =$0.01 = {10^{ - 2}}$
So $\log {\text{ C = lo}}{{\text{g}}_{10}}{\text{1}}{{\text{0}}^{ - 2}}$
And on applying the rule$\log {m^n} = n\log m$, we get-
$\Rightarrow \log {\text{ C = }} - 2{\text{lo}}{{\text{g}}_{10}}{\text{10}}$
We know that ${\log _{10}}10 = 1$ so we get-
$\Rightarrow \log {\text{ C = }} - 2$-- (iii)
On putting the values from eq. (ii) and eq. (iii) in eq. (i), we get-
$\Rightarrow {\text{pH = 7 + }}\left( {\dfrac{{\text{1}}}{{\text{2}}} \times {\text{4}}} \right){\text{ + }}\left[ {\dfrac{{\text{1}}}{{\text{2}}} \times \left( { - 2} \right)} \right]$
On simplifying, we get-
$\Rightarrow {\text{pH = 7 + }}2 - 1$
On solving, we get-
$\Rightarrow {\text{pH = }}8$

The correct answer is option C.

Note: In this type of questions, we have to remember the formula of pH of the different types of salt. They are as follows-
The formula of pH of salt of strong acid and weak base is given as-
$\Rightarrow {\text{pH = 7}} - \dfrac{{\text{1}}}{{\text{2}}}{\text{p}}{{\text{K}}_{\text{b}}} - \dfrac{{\text{1}}}{{\text{2}}}{\text{logC}}$ where ${\text{p}}{{\text{K}}_{\text{b}}} = - {\log _{10}}\left( {{{\text{K}}_{\text{b}}}} \right)$ and C is concentration of the salt
The formula of pH of salt of weak acid and weak base is given as-
$\Rightarrow {\text{pH = 7 + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{p}}{{\text{K}}_{\text{a}}} - \dfrac{{\text{1}}}{{\text{2}}}{\text{p}}{{\text{K}}_{\text{b}}}$ where ${\text{p}}{{\text{K}}_{\text{a}}} = - {\log _{10}}\left( {{{\text{K}}_{\text{a}}}} \right)$ and ${\text{p}}{{\text{K}}_{\text{b}}} = - {\log _{10}}\left( {{{\text{K}}_{\text{b}}}} \right)$
Also, remember that salt of strong acid and strong base do not hydrolyze so its pH will remain $7$.