[Single-digit Integer] The pH of 100 L of concentrated solution after the electrolysis for 10 s usin — Electrochemistry Chemistry Question

💡 Solution & Explanation

\textbf{Step 1: Calculate charge and moles of electrons.} \[ Q = I \times t = 9.65 \times 10 = 96.5\ \text{C} \] \[ n_e = \frac{Q}{F} = \frac{96.5}{96500} = 1.0 \times 10^{-3}\ \text{mol} \] \textbf{Step 2: Cathode reaction of concentrated KCl.} In concentrated KCl electrolysis, water is reduced at the cathode: \[ 2\text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\text{OH}^- \] \[ n(\text{OH}^-) = n_e = 1.0 \times 10^{-3}\ \text{mol} \] \textbf{Step 3: Calculate [OH⁻] and pH.} Volume = 100 L: \[ [\text{OH}^-] = \frac{1.0 \times 10^{-3}}{100} = 1.0 \times 10^{-5}\ \text{M} \] \[ \text{pOH} = -\log(10^{-5}) = 5 \] \[ \text{pH} = 14 - 5 = \boxed{9} \]