[Single-digit Integer] Percentage of aniline hydrochloride hydrolysed in its M/40 solution at 25°C i — Electrochemistry Chemistry Question

💡 Solution & Explanation

\textbf{Step 1: Use the given cell potential to find pH.} The aniline–HCl hydrogen electrode half-cell gives: \[ E = 0.00 - \frac{0.06}{1}\,\text{pH} \] Given $E = -0.18$ V: \[ -0.18 = -0.06 \times \text{pH} \implies \text{pH} = 3 \implies [\text{H}^+] = 10^{-3}\ \text{M} \] \textbf{Step 2: Apply the hydrolysis equilibrium.} Aniline hydrochloride ($\text{C}_6\text{H}_5\text{NH}_3^+\text{Cl}^-$) undergoes cationic hydrolysis: \[ \text{C}_6\text{H}_5\text{NH}_3^+ + \text{H}_2\text{O} \rightleftharpoons \text{C}_6\text{H}_5\text{NH}_2 + \text{H}_3\text{O}^+ \] Initial concentration $C_0 = \dfrac{M}{40} = 0.025\ \text{M}$ \textbf{Step 3: Calculate degree of hydrolysis.} $[\text{H}^+]$ produced $= h \times C_0$, where $h$ is the degree of hydrolysis: \[ h = \frac{[\text{H}^+]}{C_0} = \frac{10^{-3}}{0.025} = 0.04 \] \textbf{Step 4: Convert to percentage.} \[ \% \text{ hydrolysis} = h \times 100 = 0.04 \times 100 = \boxed{4\%} \]