Question

The pH of 100 L of concentrated KCl solution after the electrolysis for 10 s using 9.65 A at 298 K is

Answer 1

9

Answer 2

1. Electrode Reactions: In the electrolysis of concentrated KCl solution with inert electrodes, water is reduced at the cathode, and chloride ions are oxidized at the anode.

   • Cathode (Reduction): $2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq)$
   • Anode (Oxidation): $2Cl^-(aq) \rightarrow Cl_2(g) + 2e^-$ The net reaction produces hydroxide ions ($OH^-$), increasing the pH.

2. Calculate Charge Passed (Q): The total charge passed is given by the product of current ($I$) and time ($t$). $Q = I \times t$ $Q = 9.65 \text{ A} \times 10 \text{ s}$ $Q = 96.5 \text{ C}$

3. Calculate Moles of Electrons ($n_e$): Using Faraday's constant ($F = 96500 \text{ C/mol}$), the moles of electrons transferred are: $n_e = \frac{Q}{F}$ $n_e = \frac{96.5 \text{ C}}{96500 \text{ C/mol}}$ $n_e = 0.001 \text{ mol}$

4. Calculate Moles of $OH^-$ Produced: From the cathode reaction, 2 moles of electrons produce 2 moles of $OH^-$ ions. Therefore, the moles of $OH^-$ produced are equal to the moles of electrons passed. Moles of $OH^- = n_e = 0.001 \text{ mol}$

5. Calculate Concentration of $OH^-$: The volume of the solution is 100 L. The concentration of $OH^-$ is: $[OH^-] = \frac{\text{Moles of } OH^-}{\text{Volume of solution}}$ $[OH^-] = \frac{0.001 \text{ mol}}{100 \text{ L}}$ $[OH^-] = 0.00001 \text{ M} = 1 \times 10^{-5} \text{ M}$

6. Calculate pOH: pOH is calculated using the concentration of $OH^-$ ions: pOH = -log$[OH^-]$ pOH = -log($1 \times 10^{-5}$) pOH = 5

7. Calculate pH: At 298 K, the relationship between pH and pOH is pH + pOH = 14. pH = 14 - pOH pH = 14 - 5 pH = 9