Question

1. A river contains a relatively low concentration of dissolved salts, so that the ideal assumption applies. The river has the same \(\mathrm{pH}(8.1)\) and total carbonate concentration \(\left(2 \times 10^{-3} \mathrm{M}\right)\) as bulk seawater. The work of Garrels and Thompson (1962) indicates that the total ion activity coefficients of \(\mathrm{H}^{+}, \mathrm{OH}^{-}\), and various carbonate-containing species in seawater are as follows:

$$ \begin{array}{|c|c|} \hline & \begin{array}{c} \text { activity } \\ \text { coefficient } \\ \text { lon } \end{array} & \gamma_{\mathrm{T}} \\ \hline \mathrm{H}^{+} & 0.59 \\ \mathrm{OH} & 0.244 \\ \mathrm{H}_{2} \mathrm{CO}_{3} & 1.13 \\ \mathrm{HCO}_{3}^{-} & 0.42 \\ \mathrm{CO}_{3}^{-2} & 0.018 \\ \hline \end{array} $$

The activity coefficient of \(\mathrm{H}_{2} \mathrm{O}\) is assigned a value of 1.0 in both freshwater and seawater. The equilibrium constants shown below are based on the infinite dilution reference state.

-Compute the values of the same equilibrium constants using major ion seawater as the reference state.

\(\mathrm{H}_{2} \mathrm{O}=\mathrm{H}^{+}+\mathrm{OH}^{-} \quad \log \mathrm{K}_{\mathrm{w}}=-14.0\) \(\mathrm{K}_{1}=-6.35\) \(\begin{array}{ll}\mathrm{H}_{2} \mathrm{CO}_{3}=\mathrm{H}^{+}+\mathrm{HCO}_{3}^{-} & \log \mathrm{K} \\ \mathrm{HCO}_{3}^{-}=\mathrm{H}^{+}+\mathrm{CO}_{3}^{-2} & \log \mathrm{K}\end{array}\) \(\mathrm{g} \mathrm{K}_{2}=-10.33\) [Based on a problem from Benjamin (2002) Water Chemistry]