Question

: The hydrolysis constant of aniline hydrochloride in M/32 solution of salt at 298K is:
$Pt|{{H}_{2}}(1\text{ atm) }\\!\\!|\\!\\!\text{ }{{\text{H}}^{+}}(1M)||\dfrac{M}{32}{{C}_{6}}{{H}_{6}}N{{H}_{3}}Cl|{{H}_{2}}(1\text{ atm)}|Pt;\text{ }{{\text{E}}_{cell}}=-0.188V$
(a)- $4.2\text{ x 1}{{\text{0}}^{-2}}$
(b)- $2.1\text{ x 1}{{\text{0}}^{-2}}$
(c)- $4.1\text{ x 1}{{\text{0}}^{-2}}$
(d)- $2.6\text{ x 1}{{\text{0}}^{-3}}$

Answer 1

We can use two formulas to solve this problem as: (i)${{E}_{cell}}=-0.0591\log \dfrac{{{[{{H}^{+}}]}_{Anode}}}{{{[{{H}^{+}}]}_{Cathode}}}$ which will be used to find the emf of the cell and $pH=7-\dfrac{1}{2}p{{K}_{b}}-\dfrac{1}{2}\log C$ which will be used to find the pH.

Complete step-by-step answer: We know that to calculate the emf of the cell we can use the formula:
${{E}_{cell}}=E_{cell}^{\circ }-\dfrac{0.0591}{n}\log \dfrac{{{[{{H}^{+}}]}_{Anode}}}{{{[{{H}^{+}}]}_{Cathode}}}$
Where $E_{cell}^{\circ }$ is the standard potential of the cell. In the question given above, the value of standard emf will be zero. So, the formula will be:
${{E}_{cell}}=-\dfrac{0.0591}{n}\log \dfrac{{{[{{H}^{+}}]}_{Anode}}}{{{[{{H}^{+}}]}_{Cathode}}}$
And the value of n will be one.
${{E}_{cell}}=-0.0591\log \dfrac{{{[{{H}^{+}}]}_{Anode}}}{{{[{{H}^{+}}]}_{Cathode}}}$
Given the concentration of the hydrogen ions in the anode side is 1 and the value of the emf of the cell is -0.188 V. putting these values in the formula, we get:
$-0.188=-0.0591\log \dfrac{1}{{{[{{H}^{+}}]}_{Cathode}}}$
On solving this, we get:
$-\log {{[{{H}^{+}}]}_{Cathode}}=3.18$
We know that the value of pH of the solution is equal to the negative logarithm of the concentration of the hydrogen ions. So, we can call this value as the pH.
pH = 3.18
For aniline hydrochloride, the formula for pH of the solution will be:
$pH=7-\dfrac{1}{2}p{{K}_{b}}-\dfrac{1}{2}\log C$
The given concentration is $\dfrac{1}{32}$ and pH is 3.18, we can calculate the ${{K}_{b}}$. We can write:
$3.18=7-\dfrac{1}{2}p{{K}_{b}}-\dfrac{1}{2}\log \left( \dfrac{1}{32} \right)$
${{K}_{b}}=7.15\text{ x 1}{{\text{0}}^{-10}}$
Now, we can calculate the value of ${{K}_{h}}$ by taking the formula:
${{K}_{h}}=\dfrac{{{K}_{w}}}{{{K}_{b}}}$
The value of ${{K}_{w}}$ is ${{10}^{-14}}$
Now putting the value in the formula, we get:
${{K}_{h}}=\dfrac{{{K}_{w}}}{{{K}_{b}}}=\dfrac{{{10}^{-14}}}{7.15\text{ x 1}{{\text{0}}^{-10}}}=1.39\text{ x 1}{{\text{0}}^{-5}}$
Now, we can calculate the hydrolysis constant as:
$h=\sqrt{\dfrac{{{K}_{h}}}{C}}=\sqrt{\dfrac{1.39\text{ x 1}{{\text{0}}^{-5}}}{1/32}}=2.1\text{ x 1}{{\text{0}}^{-2}}$
So, the hydrolysis constant is $2.1\text{ x 1}{{\text{0}}^{-2}}$

Therefore, the correct answer is option (b).

Note: We have used to value of pH of the aniline hydrochloride as $pH=7-\dfrac{1}{2}p{{K}_{b}}-\dfrac{1}{2}\log C$because the aniline hydrochloride is a weak acid. For weak bases we can use $pH=7+\dfrac{1}{2}p{{K}_{a}}+\dfrac{1}{2}\log C$.