Practicing Success
A current of 0.193 amp is passed through 100 mL of 0.2M NaCl for an hour. Calculate pH of solution after electrolysis. Assume no volume change. |
11.9 12.82 12.0 11.5 |
12.82 |
The correct answer is option 2. 12.82. We know from Faraday's law, \(Q = it\) Given, \(i = 0.193 A, t = 1hr = 60 × 60 s\) ∴\(Q = 0.193 × 3600\) or, \(Q = 694.8 C\) or, \(Q = \frac{694.8}{96500}F\) or, \(Q = 7.2 × 10^{−3}F\) or, \(Q = 7.2 × 10^{−3} \text{mole of Na}^+\) or, \(Q = 7.2 × 10^{−3} \text{mole of NaOH}\) ∴ \([OH^−] = \frac{7.2 × 10^{−3}}{0.1}\) or, \([OH^−] = 7.2 × 10^{−2}M\) ∴ \(pOH = 2 − log 7.2\) or, \(pH = 14 − pOH\) or \(pH = 14 − 2 −log 7.2\) or, \(pH = 12 − log 7.2\) ∴\(pH = 12.82\) |