Chapter 21, Problem 68AP

A glass of water initially at pH 7.0 is exposed to dry air at sea level at $\text{20}°\text{C}\text{.}$ Calculate the pH of the water when equilibrium is reached between atmospheric ${\text{CO}}_{\text{2}}{\text{ and CO}}_{\text{2}}$ dissolved in the water, given that Henry's law constant for ${\text{CO}}_{\text{2}}\text{ at 20}°\text{C is 0}\text{.032 mol/L }·\text{ atm}\text{.}$ (Hint: Assume no loss of water due to evaporation, and use Table 21.1 to calculate the partial pressure of ${\text{CO}}_{\text{2}}$. Your answer should correspond roughly to the pH of rainwater.)

TABLE 21.1 Composition of Dry Air at Sea Level
Gas	Composition (% by Volume)
${\text{N}}_{\text{2}}$	78.03
${\text{O}}_{\text{2}}$	20.99
$\text{Ar}$	0.94
${\text{CO}}_{\text{2}}$	0.033
$\text{Ne}$	0.0015
$\text{He}$	0.000524
$\text{Kr}$	0.00014
$\text{Xe}$	0.000006

Interpretation Introduction

Interpretation:

The pH of water at the equilibrium state is to be calculated.

Concept introduction:

Henry’s law is a gas law that states that the partial pressure of a gas and the concentration of a dissolved gas are directly proportional to each other.

Henry’s law is represented by the following expression:

$P=X\times P_{\text{T}}$

Here, $X$

is the mole fraction and $P_{\text{T}}$ is the atmospheric pressure.

Equilibrium constant is expressed as follows:

$K=\frac{[\text{products}]}{[\text{reactants]}}$

Answer to Problem 68AP

Solution: pH is $5.72$.

Explanation of Solution

The partial pressure of carbon dioxide is calculated as follows:

$P_{{\text{CO}}_2}=X_{{}_{{\text{CO}}_2}}\times P_{\text{T}}$

Here, $X_{{}_{{\text{CO}}_2}}$

is the mole fraction of ${\text{CO}}_2$ and $P_{\text{T}}$ is the atmospheric pressure.

Substitute the value of $X_{{}_{{\text{CO}}_2}}$

(from table $21.1$

) and $P_{\text{T}}$

$\left(754\text{ mm Hg}\right)$ in the above equation:

$\begin{array}{l}{P}_{{\text{CO}}_2}=\left(\left(3.3\times {10}^{-4}\right)\left(754\cancel{\text{ mm Hg}}\right)\times \frac{1\text{ atm}}{760\cancel{\text{mm Hg}}}\right)\\ P_{{\text{CO}}_2}=3.3\times {10}^{-4}\text{ atm}\end{array}$

Therefore, the partial pressure of carbon dioxide is $3.3\times {10}^{-4}\text{ atm}$.

The concentration of ${\text{CO}}_2$

in water is calculated using Henry’s law as follows:

$c=\text{k}P$

Here, $c$ concentration of gas in the given solution, $\text{k}$ is the Henry’s law constant, and $P$ is the partial pressure of ${\text{CO}}_2$.

Substitute the values of $\text{k}$

and $P$

in the above equation:

$\begin{array}{l}[{\text{CO}}_2]=\left(0.032\text{ mol/L}\text{.}\cancel{\text{atm}}\right)\left(3.3\times {10}^{-4}\cancel{\text{atm}}\right)\\ [{\text{CO}}_2]=1.06\times {10}^{-5}\text{ mol/L}\end{array}$

Therefore, the concentration of ${\text{CO}}_2$ in water is $1.06\times {10}^{-5}\text{ mol/L}$.

Let all of the dissolved ${\text{CO}}_2$ get converted into carbonic acid and $1.06\times {10}^{-5}\text{ mol/L}$

of carbonic acid is formed. Carbonic acid is a weak acid.

The equilibrium expression for carbonic acid in water is as follows:

$\begin{array}{l}\underset{\_}{\begin{array}{l}{\text{ H}}_2{\text{CO}}_3\left(aq\right)\rightleftharpoons {\text{ H}}^{+}\left(aq\right)+{\text{ HCO}}_3^{-}\left(aq\right)\\ \text{Initial}\left(\text{M}\right):\text{ 1}\text{.06}\times {\text{10}}^{-5}\text{ 0 0}\\ \text{Change}\left(\text{M}\right):-x+x+x\end{array}}\\ \text{Equilibrium}\left(\text{M}\right):\left(1.6\times {10}^{-5}\right)-xxx\end{array}$

The equilibrium constant is $4.2\times {10}^{-7}$ (from table $16.8$

).

The $[{\text{H}}^{+}]$ is calculated with the help of equilibrium constant, as follows:

$K=\frac{[{\text{H}}^{+}][{\text{HCO}}_3^{-}]}{[{\text{H}}_2{\text{CO}}_3]}$

Substitute the values of equilibrium concentration in the above equation:

$4.2\times {10}^{-7}=\frac{x^2}{\left(1.06\times {10}^{-5}\right)-x}$

This is a quadratic equation, by solving:

$\begin{array}{c}x=1.9\times {10}^{-6}\text{ M}\\ {\text{[H}}^{+}]=1.9\times {10}^{-6}\text{ M}\\ \text{pH }=-\text{log}\left(1.9\times {10}^{-6}\right)\\ \text{pH}=5.72\end{array}$

Conclusion

The pH of water at equilibrium is $5.72$.