$100 mL$ of aqueous $NaCl$ (concentrated) is electrolysed between platinum electrodes with a current of $0.965 A$ for $200 s$ . Assume all the chlorine escapes as gas without reacting with the solution, the volume stays $100 mL$ and the solution is well mixed. Find the pH of the solution after electrolysis (two decimals). Take $F = 96500 C mo l^{- 1}$ , $lo g 2 = 0.301$ .

Numerical answer type, tolerance 0.02.

Answer: 12.3

Charge

Q = 0.965 \times 200 = 193 C

, which is

193/96500 = 2.00 \times 1 0^{- 3} mol

of electrons. Cathode:

2 H X_{2} O + 2 e X^{-} H X_{2} + 2 OH X^{-}

gives

2.00 \times 1 0^{- 3} mol

OH X^{-}

; the anode gives

Cl X_{2}

, which escapes, so no

H X^{+}

is formed.

[OH X^{-}] = 2.00 \times 1 0^{- 3} /0.100 = 0.0200 M

pOH = 2 - 0.301 = 1.699

, so

pH = 14 - 1.699 = 12.30