ElectrochemistryMedium

Question

At 25°C, the solubility product of Pb(OH)2 (s) is [Given: $E_{Pb^{2 +}|Pb}^{o} = - 0.13\text{ V;}E_{OH^{-}|Pb(OH)_{2}|Pb}^{o} = - 0.55\text{ V}$and $\frac{2.303RT}{F} = 0.06$]

Options

A.1.36 × 10–5
B.1.0 × 10–7
C.1.0 × 10–14
D.1.25 × 10–15

Solution

$E_{OH^{-}|Pb(OH)_{2}|Pb}^{o} = E_{Pb^{2 +}|Pb}^{o} - \frac{0.06}{2}.\log\frac{1}{K_{sp}}$

$\text{Or, }( - 0.55) = ( - 0.13) - \frac{0.06}{2}.\log\frac{1}{K_{sp}} $$$\therefore K_{sp} = 1.0 \times 10^{- 14}$$

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