Error in calculating H+ activity I'm starting to study thermodynamics of minerals and rocks, so sorry for the basic question. I am trying to work out some exercises of a book and I found this:

Calculate the activity of $\ce{H+}$ when the activity of $\ce{Sn^4+}$ is $10^{-10}$ for the reaction $\ce{SnO2 + 4H+ &lt;=&gt; Sn^4+ + 2H2O}$. Standard free energies of formations $\Delta$Fof are: SnO2 = -123.20 kcal; H2O = -56.69 kcal; Sn4+ = 0.65 kcal; H+ = 0.

I used the following equations:

$\Delta$For = ($\Sigma$$\Delta$Fof,products) - ($\Sigma$$\Delta$Fof,reactants)
$\Delta$For = -RTln(K), where R = 0.001987 Kcal/molK and T = 298.15K
K = $\frac{[Sn^{4+}]}{[H^{+}]^4}$, given that the activities of SnO2 and H2O are unit.

The result I found was $3.82 \cdot 10^{-5}$, but the book says the result is $10^{-0.40}$. I assume this is a fundamental error, but given it is also a fundamental principle I need to understand it to proceed in my study.
Thanks in advance for any help.
$$ \Delta G = \Delta G^\circ + RT \ln{\frac{a_{\ce{Sn^{4+}}}a_{\ce{H2O}}^{\;\;\;\;2}}{a_{\ce{SnO2}}a_{\ce{H+}}^{\;\;4}}} $$

At equilibrium, $\Delta G = 0$.

For $\Delta G^\circ = 2\times(-56.69) + 0.65 - (-123.20 - 4\times0) = +10.47\mathrm{\frac{kcal}{mol}}$ 

The question says "activity", but in the absence of any other information, let's assume that "activity" is equal to "concentration".  That is, let's assume activity coefficients are 1.

Thus, we get $$ \Delta G = 0 = +10.47  + RT \ln{\frac{a_{\ce{Sn^{4+}}}\cdot1^2}{1\cdot a_{\ce{H+}}^{\;\;4}}} $$

Thus, we get $$ {\left( \frac{-10.47}{RT}\right)}  = \ln{{10^{-10}} a_{\ce{H+}}^{\;\;-4}} = -4\ln{a_{\ce{H+}}} -10 \ln{10} $$

$$\frac{\frac{-10.47}{RT} + 10 \ln{10}}{-4} = \ln{a_{\ce{H+}}}$$