Question

When solid quartz "dissolves" in water, it combines with water molecules in the reaction

${\mathrm{SiO}}_2(\mathrm{s})+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)⟷{\mathrm{H}}_4{\mathrm{SiO}}_4\left(\mathrm{aq}\right)$

(a) Use this data in the back of this book to compute the amount of silica dissolved in water in equilibrium with solid quartz, at 25° C

(b) Use the van't Hoff equation (Problem 5.85) to compute the amount of silica dissolved in water in equilibrium with solid quartz at 100°C.

Short answer

Therefore,

$$\begin{gathered} \text{(a)}{M}_{{\mathrm{H}}_4{\mathrm{SiO}}_4}=8.472\times {10}^{-5}\mathrm{mol}/\mathrm{kg} \\ \text{(b)}K(373\mathrm{K})=1.2577\times {10}^{-3} \end{gathered}$$

Step-by-step solution

Given information

When solid quartz "dissolves" in water, it combines with water molecules in the reaction

${\mathrm{SiO}}_2(\mathrm{s})+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)⟷{\mathrm{H}}_4{\mathrm{SiO}}_4\left(\mathrm{aq}\right)$

Explanation

(a) Consider the following reaction, which depicts the quartz dissolving in water:

${\mathrm{SiO}}_2+2{\mathrm{H}}_2\mathrm{O}\leftrightarrow {\mathrm{H}}_4{\mathrm{SiO}}_4$

The equilibrium constant for the reaction is:

${\mu }_{{\mathrm{SiO}}_2}+2{\mu }_{{\mathrm{H}}_2\mathrm{O}}={\mu }_{{\mathrm{H}}_4{\mathrm{SiO}}_4}$

For standard condition, this can be written as:

${\mu }_{{\mathrm{SiO}}_2}°+2{\mu }_{{\mathrm{H}}_2\mathrm{O}}°={\mu }_{{\mathrm{H}}_4{\mathrm{SiO}}_4}\left(1\right)$

During this reaction, the chemical potentials of water and quartz remain unchanged, but the chemical potential of ${\mathrm{H}}_4{\mathrm{SiO}}_4$can be expressed in terms of molality

${\mu }_{{\mathrm{H}}_4{\mathrm{SiO}}_4}={\mu }_{{\mathrm{H}}_4{\mathrm{SiO}}_4}°+kT\ln \left(M_{{\mathrm{H}}_4{\mathrm{SiO}}_4}\right)$

Substitute into (1) and get

${\mu }_{{\mathrm{SiO}}_2}°+2{\mu }_{{\mathrm{H}}_2\mathrm{O}}°={\mu }_{{\mathrm{H}}_4{\mathrm{SiO}}_4}°+kT\ln \left(M_{{\mathrm{H}}_4{\mathrm{SiO}}_4}\right)\left(2\right)$

This can be written as:

$$\begin{gathered} \ln \left(M_{{\mathrm{H}}_4{\mathrm{SiO}}_4}\right)=-\frac{\Delta G°}{RT} \\ {M}_{{\mathrm{H}}_4{\mathrm{SiO}}_4}=\exp \left(-\frac{\Delta G°}{RT}\right) \end{gathered}$$

Calculations

We need to find the change of the Gibbs free energy to find the concentration:

$\begin{array}{cc}& G°\left(\mathrm{kJ}\right)\\ {\mathrm{H}}_4{\mathrm{SiO}}_4& -1307.67\\ {\mathrm{H}}_2\mathrm{O}& -237.13\\ {\mathrm{SiO}}_2& -856.64\end{array}$

The change in Gibbs free energy is

$$\begin{gathered} \Delta G°=G{°}_{{\mathrm{H}}_4{\mathrm{SiO}}_4}-2G{°}_{{\mathrm{H}}_2\mathrm{O}}-G{°}_{{\mathrm{SiO}}_2} \\ =-1307.67\mathrm{kJ}+2(237.13\mathrm{kJ})+856.64\mathrm{kJ} \\ =23.23\mathrm{kJ} \end{gathered}$$

The concentration is:

$$\begin{gathered} M_{{\mathrm{H}}_4{\mathrm{SiO}}_4}=\exp \left(-\frac{23.23\times {10}^3\mathrm{J}}{(8.314\mathrm{J}/\mathrm{mol}·\mathrm{K})(298\mathrm{K})}\right) \\ {M}_{{\mathrm{H}}_4{\mathrm{SiO}}_4}=8.472\times {10}^{-5}\mathrm{mol}/\mathrm{kg} \end{gathered}$$

(b)The equilibrium constant of temperature T₂is

$$\begin{gathered} \ln \left(K\left(T_2\right)\right)=\ln \left(K\left(T_1\right)\right)+\frac{\Delta H°}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) \\ K\left(T_2\right)=\exp \left(\ln \left(K\left(T_1\right)\right)+\frac{\Delta H°}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)\right) \end{gathered}$$

But,

$\ln \left(K\left(T_1\right)\right)=\ln \left(M_{{\mathrm{H}}_4{\mathrm{SiO}}_4}\right)$

Therefore,

$K\left(T_2\right)=\exp \left(\ln \left(M_{{\mathrm{H}}_4{\mathrm{SiO}}_4}\right)+\frac{\Delta H°}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)\right)$

The enthalpies are given by: therefore the change is:

$\begin{array}{cc}& H°\left(\mathrm{kJ}\right)\\ {\mathrm{H}}_4{\mathrm{SiO}}_4& -1449.36\\ {\mathrm{H}}_2\mathrm{O}& -285.83\\ {\mathrm{SiO}}_2& -910.94\end{array}$

The change is:

$$\begin{gathered} \Delta H°=H_{{\mathrm{H}}_4{\mathrm{SiO}}_4}°-2H_{{\mathrm{H}}_2\mathrm{O}}°-H_{{\mathrm{SiO}}_2}° \\ =-1449.36\mathrm{kJ}+2(285.83\mathrm{kJ})+910.94\mathrm{kJ} \\ =33.24\mathrm{kJ} \end{gathered}$$

Calculations

Substitute the values in the equation:

$$\begin{gathered} K(373\mathrm{K})=\exp \left(\ln \left(8.472\times {10}^{-5}\mathrm{mol}/\mathrm{kg}\right)+\frac{33.24\times {10}^3\mathrm{J}}{8.314\mathrm{J}/\mathrm{mol}·\mathrm{K}}\left(\frac{1}{298\mathrm{K}}-\frac{1}{373\mathrm{K}}\right)\right) \\ K(373\mathrm{K})=1.2577\times {10}^{-3} \end{gathered}$$