What is the concentration of $\ce{Pb^2+}$ in a lake that has a pH of $6.0$ and is in equilibrium with $\ce{PbO2(s)}$ and atmospheric oxygen?

\begin{align} \ce{PbO2(s) + 4H+ + 2e- &-> Pb^2+ + 2H2O} & \log K &= 49.2 \end{align}

I spent so much time to solve this, but I could not, my final answer does not appear to make sense: \begin{align} \mathrm{pH} &= 6 & \implies&& [\ce{H+}] &=10^{-6}\\ \log K &= 49.2 & \implies&& K &= 10^{49.2}\\ K &= \frac{[\ce{Pb^2+}]}{[\ce{H+}]^4} & \implies&& 10^{49.2} &=\frac{[\ce{Pb^2+}]}{(10^{-6})^4} \\ &&\implies&& [\ce{Pb^2+}] & =10^{25.2} \end{align}

What is my mistake?


closed as off-topic by Klaus-Dieter Warzecha, Wildcat, ringo, getafix, Jannis Andreska Nov 5 '16 at 14:47

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I think that the problem here is one of notation. I'm pretty sure that the logarithm intended to be used is the natural logarithm, $\log_e(K)$ also represented as $\ln(K)$, and not the base 10 one. Otherwise your method is good.

If you use this, you get a much more reasonable answer (work it out). It also makes sense that the question was worded in this way, as the equations in which one tends to use the equilibrium constant $K$ actually use its natural logarithm, for example the Gibbs's free energy and the Nernst equation.

By the way, this reaction is also part of the reduction process at the cathode of a lead-acid battery.


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