Question
A Tl+ |Tl couple was prepared by saturating 0.10 M – KBr with TlBr and allowing Tl+ ions form the insoluble bromide to equilibrate. This couple was observed to have a potential −0.444 V with respect to Pb2+|Pb couple in which Pb2+ was 0.10 M. What is the Ksp of TlBr. [Given: $E_{Pb^{2 +}|Pb}^{o}$= − 0.126 V, $E_{Ti^{+}|Ti}^{o}$= −0.336 V, log 2.5 = 0.4, 2.303 RT/F = 0.06]
Options
Solution
$E_{Tl^{+}|Tl} - E_{Pb^{2 +}|Pb} = - 0.444\text{ V}$
Or, $\left( E_{Tl^{+}|Tl} - E_{Pb^{2 +}|Pb} \right) - \frac{0.06}{2}.\log\frac{\left\lbrack Pb^{2 +} \right\rbrack}{\left\lbrack Tl^{+} \right\rbrack^{2}} = - 0.444$
Or, $\left\lbrack ( - 0.336) - ( - 0.126) \right\rbrack - 0.03.\log\frac{0.1}{\left( \frac{K_{sp}}{0.1} \right)^{2}} = - 0.444$
$\therefore K_{sp} = 4 \times 10^{- 6}$
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