Chapter 20, Problem 20.72QP

A glass of water initially at pH 7.0 is exposed to dry air at sea level at 20°C. Calculate the pH of the water when equilibrium is reached between atmospheric CO₂ and CO₂ dissolved in the water, given that Henry’s law constant for CO₂ at 20°C is 0.032 mol/L · atm. (Hint: Assume no loss of water due to evaporation and use Table 20.1 to calculate the partial pressure of CO₂. Your answer should correspond roughly to the pH of rainwater.)

Interpretation Introduction

Interpretation:

From the given Henry’s law constant, the $\text{pH}$ of water when atmospheric carbon dioxide is equilibrium with carbon dioxide dissolved in water has to be calculated.

Concept Introduction:

Henry law:

Henry law can be defined as the amount of dissolved gas is proportional to its partial pressure in the gas phase.  The proportionality factor is called as Henry’s law constant.

The henry law constant is given by the equation,

${\text{H}}_{\text{cp}}\text{=}\frac{{\text{C}}_{\text{a}}}{\text{p}}$

Where

$\begin{array}{l}{\text{H}}_{\text{p}}\text{=}\text{Henry}\text{law}\text{constant}\left(\text{mole/}\left(\text{L}\text{.atm}\right)\right)\\ {\text{c}}_{\text{a}}\text{=}\text{concentration}\text{of}\text{species}\text{in}\text{aqueous}\text{phase}\\ \text{p}\text{=}\text{partial}\text{pressure}\text{of}\text{the}\text{species}\end{array}$

$pH:$

• $\text{pH}$ is the logarithm of the reciprocal of the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in a solution.
•  $\text{pH}$ is used to determine the acidity of an aqueous solute.
• $\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$

Answer to Problem 20.72QP

The $\text{pH}$ of water when atmospheric carbon dioxide is equilibrium with carbon dioxide dissolved in water is $\text{5}\text{.72}$

Explanation of Solution

Given,

Henry’s law constant $\text{=}\text{0}\text{.032mol/L}\text{.atm}$

$\begin{array}{l}\text{754 mmHg}\text{(atm}\text{. pressure)}\\ \\ \text{0}\text{.033\%}\text{by}\text{volume}\text{of}{\text{CO}}_{\text{2}}\end{array}$

The number of moles and volume are equal for an ideal gas.  The number of moles is calculated from percentage.

$\frac{\text{0}\text{.033\%}}{\text{100\%}}\text{×}\text{100}\text{=}\text{0}\text{.033 moles}$

The mole fraction of carbon dioxide is calculated as

${\text{X}}_{{\text{CO}}_{\text{2}}}\text{=}\frac{\text{0}\text{.033}}{\text{100}}\text{=}\text{3}\text{.3}\text{×}{\text{10}}^{\text{-4}}$

The mole fraction is multiplied by one million to convert into ppm.

$\text{(3}\text{.3}\text{×}{\text{10}}^{\text{-4}}\text{)(1}\text{×}{\text{10}}^{\text{6}}\text{)}\text{=}\text{330ppm}$

The partial pressure of carbon dioxide is calculated as

$\begin{array}{l}{\text{P}}_{{\text{CO}}_{\text{2}}}{\text{=P}}_{\text{total}}{\text{X}}_{{\text{CO}}_{\text{2}}}\\ \\ \text{=}\text{(3}\text{.3}\text{×}{\text{10}}^{\text{-4}}\text{)(754mmHg)}\text{×}\frac{\text{1}\text{atm}}{\text{760mmHg}}\\ \\ \text{=}\text{3}\text{.3}\text{×}{\text{10}}^{\text{-4}}\text{atm}\end{array}$

The concentration of carbon dioxide is calculated using Henry’s law.

$\begin{array}{l}c=kP\\ \\ [C{O}_2]=(0.032mol/L.atm)(3.3\times 10^{-4}atm)\\ \\ =1.06\times 10^{-5}mol/L\end{array}$

Assume that all carbon dioxide is dissolved and converted to ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$  and contains $1.06\times 10^{-5}mol/L$ of ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$.   ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$ is a weak acid and the equilibrium is set up to calculate the concentration of ${\text{H}}^{\text{+}}$

$\begin{array}{l}{H}_{2}CO\rightleftarrows H^{+}+HC{O}_3{}^{-}\\ \\ Inital(M):1.06\times 10^{-5}00\\ \\ Change(M):-x+x+x\\ \\ Equilibrium(M)(1.06\times 10^{-5})-xxx\end{array}$

The equilibrium constant expression for above reaction is given as

$\begin{array}{l}K=\frac{\left[H^{+}\right]\left[HC{O}_3{}^{-}\right]}{\left[H_{2}C{O}_3\right]}\\ \\ K=4.2\times 10^{-7}\\ \\ 4.2\times 10^{-7}=\frac{\left[H^{+}\right]\left[HC{O}_3{}^{-}\right]}{\left[H_{2}C{O}_3\right]}\\ \\ =\frac{x^2}{(1.06\times 10^{-5})-x}\\ \\ 4.452\times 10^{-12}-4.2\times 10^{-7}x-x^2=0\end{array}$

By solving the above quadratic equation, the value of x is determined as

$\mathrm{x}=1.9\times 10^{-6}M=[H^{+}]$

The $\text{pH}$ of water when atmospheric carbon dioxide is equilibrium with carbon dioxide dissolved in water is calculated as

$\begin{array}{l}\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\\ \\ \text{=}\text{-log}\text{(1}{\text{.9×10}}^{\text{-6}}\text{)}\\ \\ \text{=}\text{5}\text{.72}\end{array}$