$250 mL$ of $0.50 M$ $CuSO X_{4}$ is electrolysed between platinum electrodes with a current of $1.50 A$ for $3860 s$ . Assume complete dissociation of the acid formed, no change in volume and that $Cu X^{2 +}$ is still present at the end. Find the pH of the solution after electrolysis (two decimals). Take $F = 96500 C mo l^{- 1}$ , $lo g 2 = 0.301$ , $lo g 3 = 0.477$ .

Numerical answer type, tolerance 0.02.

Answer: 0.62

Charge

Q = 1.50 \times 3860 = 5790 C = 0.0600 mol

of electrons. Cathode:

Cu X^{2 +} + 2 e X^{-} Cu

removes

0.0300 mol

Cu X^{2 +}

(out of

0.125 mol

present, so copper deposition continues throughout). Anode:

2 H X_{2} O O X_{2} + 4 H X^{+} + 4 e X^{-}

produces one

H X^{+}

per electron, i.e.

0.0600 mol

H X^{+}

[H X^{+}] = 0.0600/0.250 = 0.240 M

, so

pH = - lo g (0.24) = - (lo g 24 - 2) = 2 - (3 \times 0.301 + 0.477) = 0.62