Question

1. A solution of NaCl is neutral, with an expected PH of 7. If electrolysis was carried out on 500 mL of an NaCl solution with a current of 0.500 A, how many seconds would it take for the PH of the solution to rise to a value of 9? (PH is related to H+ concentration by PH = -log [H+]).

Answer 1

Solution.

$pH + pOH = 14$

$pOH(1) = 14 - 7 = 7$

$pOH(2) = 14 - 9 = 5$

$\Delta [OH^-] = 10^{-5} - 10^{-7} = 0.0000099 \ M$

$n(OH^-) = 0.0000099 \times 0.5 = 0.00000495 \ mol$

$n(OH^-) = \frac{I \times t}{F}$

$t = \frac{n(OH^-) \times F}{I}$

t = 0.96 s

Answer:

t = 0.96 s