Question

When carbon dioxide "dissolves" in water, essentially all of it reacts to form carbonic acid, H2CO3:

${\mathrm{CO}}_2(\mathrm{g})+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)⟷{\mathrm{H}}_2{\mathrm{CO}}_3\left(\mathrm{aq}\right)$

The carbonic acid can then dissociate into H* and bicarbonate ions,

${\mathrm{H}}_2{\mathrm{CO}}_3\left(\mathrm{aq}\right)⟷{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{HCO}}_3^{-}\left(\mathrm{aq}\right)$

(The table at the back of this book gives thermodynamic data for both of these reactions.) Consider a body of otherwise pure water (or perhaps a raindrop) that is in equilibrium with the atmosphere near sea level, where the partial pressure of carbon dioxide is 3.4 x 10^-4 bar (or 340 parts per million). Calculate the molality of carbonic acid and of bicarbonate ions in the water, and determine the pH of the solution. Note that even "natural" precipitation is somewhat acidic.

Short answer

Therefore,

$$\begin{gathered} m_{{\mathrm{H}}_2{\mathrm{CO}}_3}=1.141\times {10}^{-5}\mathrm{mol}/\text{Liter} \\ pH=5.67 \end{gathered}$$

Step-by-step solution

Given information

When carbon dioxide "dissolves" in water, essentially all of it reacts to form carbonic acid, ${\mathrm{H}}_2{\mathrm{CO}}_3$

${\mathrm{CO}}_2(\mathrm{g})+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)⟷{\mathrm{H}}_2{\mathrm{CO}}_3\left(\mathrm{aq}\right)$

The carbonic acid can then dissociate into H* and bicarbonate ions,

${\mathrm{H}}_2{\mathrm{CO}}_3\left(\mathrm{aq}\right)⟷{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{HCO}}_3^{-}\left(\mathrm{aq}\right)$

Consider a body of otherwise pure water (or perhaps a raindrop) that is in equilibrium with the atmosphere near sea level, where the partial pressure of carbon dioxide is 3.4x 10 bar (or 340 parts per million).

Explanation

Consider the following two reactions, each of which symbolises the carbon dioxide dissolving in water:

$$\begin{gathered} {\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}\leftrightarrow {\mathrm{H}}_2{\mathrm{CO}}_3 \\ {\mathrm{H}}_2{\mathrm{CO}}_3\leftrightarrow {\mathrm{H}}^{+}+{\mathrm{HCO}}_3^{-} \end{gathered}$$

The concentration of carbon acid in terms of partial pressure and the change in the Gibbs free energy can be calculated using Henry's law:

$$\begin{gathered} \frac{m_{{\mathrm{H}}_2{\mathrm{CO}}_3}}{P_{{\mathrm{CO}}_2}/P°}=\exp \left(-\frac{\Delta G°}{RT}\right) \\ {m}_{{\mathrm{H}}_2{\mathrm{CO}}_3}=\left(\frac{P_{{\mathrm{CO}}_2}}{P°}\right)\exp \left(-\frac{\Delta G°}{RT}\right)\left(1\right) \end{gathered}$$

We need to find the change in the Gibbs free energy

$\begin{array}{cc}& G°\left(\mathrm{kJ}\right)\\ {\mathrm{H}}_2{\mathrm{CO}}_3& -623.08\\ {\mathrm{H}}_2\mathrm{O}& -237.13\\ {\mathrm{CO}}_2& -394.36\end{array}$

Change in Gibbs free energy is:

$$\begin{gathered} \Delta G°=G_{{\mathrm{H}}_2{\mathrm{CO}}_3}°-G_{{\mathrm{H}}_2\mathrm{O}}°-G_{{\mathrm{CO}}_2}° \\ =-623.08\mathrm{kJ}+237.13\mathrm{kJ}+394.36\mathrm{kJ} \\ =8.41\mathrm{kJ} \end{gathered}$$

Calculations

Substitute the values in the equation

$$\begin{gathered} m_{{\mathrm{H}}_2{\mathrm{CO}}_3}=\left(\frac{3.4\times {10}^{-4}\mathrm{bar}}{1\mathrm{bar}}\right)\exp \left(-\frac{8.41\times {10}^3\mathrm{J}}{(8.314\mathrm{J}/\mathrm{mol}·\mathrm{K})(298\mathrm{K})}\right) \\ {m}_{{\mathrm{H}}_2{\mathrm{CO}}_3}=1.141\times {10}^{-5}\mathrm{mol}/\text{Liter} \end{gathered}$$

Using the law of mass action for the second reaction, we can write the concentration as:

$\frac{m_{{\mathrm{H}}^{+}}{m}_{{\mathrm{HCO}}_3^{-}}}{m_{{\mathrm{H}}_2{\mathrm{CO}}_3}}=e^{-\Delta G°/RT}\left(2\right)$

To find $\Delta G°$

$\begin{array}{cc}& G°\left(\mathrm{kJ}\right)\\ {\mathrm{H}}_2{\mathrm{CO}}_3& -623.08\\ {\mathrm{H}}^{+}& 0\\ {\mathrm{HCO}}_3^{-}& -586.77\end{array}$

Change in Gibbs energy is:

$$\begin{gathered} \Delta G°=G_{{\mathrm{H}}^{+}}°+G_{{\mathrm{HCO}}_3^{-}}°-G_{{\mathrm{H}}_2{\mathrm{CO}}_3}° \\ =-586.77\mathrm{kJ}+623.08\mathrm{kJ} \\ =36.31\mathrm{kJ} \end{gathered}$$

Calculations

The molarity of the HT ions equals the molarity of the bicarbonate ions, so equation (2) will become:

$$\begin{gathered} \frac{m_{{\mathrm{HCO}}_3^{-}}^2}{m_{{\mathrm{H}}_2{\mathrm{CO}}_3}}=e^{-\Delta G°/RT} \\ {m}_{{\mathrm{HCO}}_3^{-}}^2=m_{{\mathrm{H}}_2{\mathrm{CO}}_3}{e}^{-\Delta G°/RT} \\ {m}_{{\mathrm{HCO}}_3^{-}}=\sqrt{m_{{\mathrm{H}}_2{\mathrm{CO}}_3}{e}^{-\Delta G°/RT}} \end{gathered}$$

Substituting the values,

$$\begin{gathered} m_{{\mathrm{HCO}}_3^{-}}=\sqrt{\left(1.141\times {10}^{-5}\mathrm{mol}/\mathrm{Liter}\right)e^{-36310\mathrm{J}/(8.314\mathrm{J}/\mathrm{mol}·\mathrm{K})(298\mathrm{K})}} \\ =2.22\times {10}^{-6}\mathrm{mol}/\text{Liter} \end{gathered}$$

The molarity of the H⁺ ions is:

$m_{{\mathrm{H}}^{+}}=m_{{\mathrm{HCO}}_3^{-}}=2.22\times {10}^{-6}\mathrm{mol}/\mathrm{Liter}$

The pH of solution is:

$$\begin{gathered} \mathrm{pH}=-{\log }_{10}\left(m_{{\mathrm{H}}^{+}}\right) \\ =-{\log }_{10}\left(2.22\times {10}^{-6}\mathrm{mol}/\mathrm{Liter}\right) \\ =5.67 \end{gathered}$$