Question
Two litre solution of a bu er mixture containing 1.0 M – NaH2PO4 and 1.0 M – Na2HPO4 is placed in two compartments (one litre in each) of an electrolytic cell. The platinum electrodes are inserted in each compartment and 1.25 A current is passed for 965 min. Assuming electrolysis of only water at each compartment, what will be pH in each compartment after passage of above charge?
(pKa for H2PO4− = 2.15, log 7 = 0.85)
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Solution
Anode: $2H_{2}O(l) \rightarrow O_{2}(g) + 4H^{+}(aq) + 4e^{-}$
Cathode: $2H_{2}O + 2e^{-} \rightarrow H_{2}(g) + 2OH^{-}(aq)$
neq H+ produced = neq OH– produced = Q/F
Or, $n_{H^{+}} \times 1 = n_{OH^{-}} = \frac{1.25 \times 965 \times 60}{96500} = 0.75$
Anode: $HPO_{4}^{2 -} + H^{+} \rightleftharpoons H_{2}PO_{4}^{-}$
1.0 M 0.75 M 1.0 M
Final 0.25 M 1.75 M
$\therefore P^{H} = P^{K_{a}} + \log\frac{\left\lbrack HPO_{4}^{2 -} \right\rbrack_{0}}{\left\lbrack H_{2}PO_{4}^{-} \right\rbrack} = 2.15 + \log\frac{0.25}{0.75} = 1.30$
Cathode: $H_{2}PO_{4}^{-} + OH^{-} \rightleftharpoons HPO_{4}^{2 -} + H_{2}O$
1.0 M 0.75 M 1.0 M
Final 0.25 M 0 1.75 M
$\therefore P^{H} = P^{K_{a}} + \log\frac{\left\lbrack HPO_{4}^{2 -} \right\rbrack_{0}}{\left\lbrack H_{2}PO_{4}^{-} \right\rbrack} = 2.15 + \log\frac{1.75}{0.25} = 3.0$
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