Question

The pH of 0.5 L of 1.0 M NaCl solution after electrolysis for 965 s using 5.0 A current is:

• A. 1.0
• B. 12.7
• C. 1.30
• D. 13.0

Hint:

Electrolysis of NaCl solution produces NaOH, increasing pH significantly.

Answer 1

The Correct Option is D

Step 1: Understanding the Electrolysis of NaCl Solution
Electrolysis of NaCl solution produces NaOH.
The reaction at the cathode generates OH$^-$ ions, increasing pH.
Step 2: Calculating the Number of Moles of NaOH Formed
Total charge passed: \[ Q = It = 5.0 \times 965 = 4825 { C} \] Moles of NaOH produced: \[ \frac{4825}{96500} = 0.05 { mol} \] Step 3: Calculating pH
Molarity of NaOH: \[ \frac{0.05}{0.5} = 0.1 M \] pOH: \[ {pOH} = -\log(0.1) = 1 \] pH: \[ {pH} = 14 - 1 = 13 \] Thus, the correct answer is (D).

Answer 2

Step 1: Understand the problem
We need to find the pH of 0.5 L of 1.0 M NaCl solution after electrolysis.
Given data:
- Volume (V) = 0.5 L
- Concentration of NaCl = 1.0 M
- Current (I) = 5.0 A
- Time (t) = 965 seconds

Step 2: Calculate total charge passed (Q)
Q = I × t = 5.0 A × 965 s = 4825 Coulombs

Step 3: Determine moles of electrons (n)
Faraday constant, F = 96500 C/mol e⁻
Moles of electrons passed, n = Q / F = 4825 / 96500 ≈ 0.05 mol

Step 4: Electrolysis reaction and OH⁻ generation
At the cathode, water is reduced producing OH⁻ ions:
2H₂O + 2e⁻ → H₂ + 2OH⁻
Each mole of electrons produces 1 mole of OH⁻ ions (since 2 electrons produce 2 OH⁻).
Therefore, moles of OH⁻ generated = moles of electrons = 0.05 mol

Step 5: Calculate OH⁻ concentration
Volume = 0.5 L
[OH⁻] = moles / volume = 0.05 mol / 0.5 L = 0.1 M

Step 6: Calculate pOH and then pH
pOH = -log[OH⁻] = -log(0.1) = 1
pH = 14 - pOH = 14 - 1 = 13

Step 7: Final conclusion
The pH of the NaCl solution after electrolysis is 13.0, indicating a strongly basic solution due to OH⁻ formation.