Chapter 21, Problem 21.68QP

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 21.1 to calculate the partial pressure of CO₂. Your answer should correspond roughly to the pH of rainwater.)

Interpretation Introduction

Interpretation: The pH of water when equilibrium is reached between atmospheric ${\text{CO}}_{\text{2}}$ and ${\text{CO}}_{\text{2}}$ is dissolved in the water should be determined. Henry’s law constant for   ${\text{CO}}_{\text{2}}$ at $\text{20ºC}$ is $\text{0}\text{.032}\text{mol/L}\text{.atm}$.

Concept Introduction:

• pH is the abbreviation of potential of hydrogen and it is a numerical scale used to specify the acidity and basicity of an aqueous solution.

   $\text{pH= -log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$

• Partial pressure is the pressure which the gas would have if it alone occupied the volume. Partial pressure of one gas (indicating A as the corresponding gas here ) in the total gas can be calculated by using the equation,

${\text{P}}_{\text{A}}\text{=}{\text{X}}_{\text{A}}{\text{P}}_{\text{Total}}$ ${\text{P}}_{\text{A}}$ is the partial pressure of A.

${\text{P}}_{\text{Total}}$ is the partial pressure of total gas.

${\text{X}}_{\text{A}}$ is the mole fraction of A

• Mole Fraction: The mole fraction is given by dividing concentration of one component in the mixture by the total components present in the mixture.
• Balanced chemical equation of a reaction is written according to law of conservation of mass
• Henry’s law : the amount of dissolved gas is proportional to its partial pressure in the gas phase. The proportionality factor in the Henry’s law is known as Henry law constant. 

                          C=kP    where C is the concentration of the dissolved substance.

                                           k is the  Henry law constant.

   P is the partial pressure of the dissolved gaseous

Substance.

• Equilibrium constant ( ${\text{K}}_{\text{C}}$): It is the ratio of product of product concentration raised to powers of their coefficient to product of reactant concentration raised to powers of their coefficient.

  For a general reaction $\text{aA+}\text{}\text{}\text{}\text{bB}\rightleftarrows \text{cC}\text{+}\text{dD}$,

  ${\text{K}}_{\text{C}}\text{=}\frac{{\left[\text{C}\right]}^{\text{c}}{\left[\text{D}\right]}^{\text{d}}}{{\left[\text{A}\right]}^{\text{a}}{\left[\text{B}\right]}^{\text{b}}}$      c, d, a and b are the co-efficient.

• Molarity (M) is the number of moles of solute divided by the number of liters of solution.
• To determine: partial pressure carbon dioxide in the dry air.

Answer to Problem 21.68QP

The pH of water is 5.72.

Explanation of Solution

Partial pressure is the pressure which the gas would have if it alone occupied the volume.

Due to the presence of carbon dioxide the volume of dry air is about $\text{0}\text{.033\%}$.

Partial pressure of one gas (indicating A as the corresponding gas here) in the total gas can be calculated by using the equation

                       ${\text{P}}_{\text{A}}\text{=}{\text{X}}_{\text{A}}{\text{P}}_{\text{Total}}$

Then the partial pressure of carbon dioxide is can be calculated as follows,

      ${\text{P}}_{{\text{CO}}_{\text{2}}}\text{=}{\text{X}}_{{\text{CO}}_{\text{2}}}{\text{P}}_{\text{Total}}$

Mole fraction of carbon dioxide is,

$\begin{array}{l}{\text{X}}_{{\text{CO}}_{\text{2}}}=\frac{0.033}{100}\\ =3.3\times {10}^{-4}\end{array}$

The mole fraction for the given gas is obtained by dividing the amount of carbon dioxide present in the air by 100 which is total percentage of air.

The partial pressure of carbon dioxide is,

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

Concentration of carbon dioxide in water can be calculated by using Henry’s law.

According with Henry’s law the amount of dissolved gas is proportional to its partial pressure in the gas phase. The proportionality factor in the Henry’s law is known as Henry law constant.

                        C=kP    where C is the concentration of the dissolved substance.

                                           k is the  Henry law constant.

   P is the partial pressure of the dissolved Substance.

$\begin{array}{l}\text{c}=\text{kP}\\ {\text{[CO}}_{\text{2}}\text{]}=\text{(0}\text{.032}\text{mol}\text{/L}\text{.atm)(3}{\text{.3×10}}^{\text{-4}}\text{atm)}\\ =\text{1}{\text{.06×10}}^{\text{-5}}\text{mol/L}\end{array}$

Balanced chemical equation of a reaction is written according to law of conservation of mass.

Here assuming that all of the dissolved ${\text{CO}}_{\text{2}}$ is converted to ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$

Therefore,

The chemical equation for the reaction is,

                ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}\rightleftarrows {\text{H}}^{\text{+}}\text{+}{\text{HCO}}_{\text{3}}{}^{\text{-}}$

Balanced chemical equation of a reaction is written according to law of conservation of mass.

Therefore,

The total number of each atom in the reactant side should equal to the total number of each atoms in the product side.

So, in order to balance a chemical equation, the coefficients of compounds or atoms are needed to be changed in such a way that total number of each atoms in the reactant side and the total number of each atoms in the product side is to become equal.

The balanced equations for the given reactions are,

${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}\rightleftarrows {\text{H}}^{\text{+}}\text{+}{\text{HCO}}_{\text{3}}{}^{\text{-}}$

Here, assuming that all the dissolved ${\text{CO}}_{\text{2}}$ is converted to ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$, and thus $\text{1}{\text{.06×10}}^{\text{-5}}\text{mol/L}$ of ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$.

${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$ is a weak acid, the equilibrium constant of this acid in water can be expressed as follows,

The equilibrium expression is,

$\begin{array}{l}{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}\rightleftarrows {\text{H}}^{\text{+}}\text{+}{\text{HCO}}_{\text{3}}{}^{\text{-}}\\ \text{Inital}\text{(M)}\text{:}\text{1}{\text{.06×10}}^{\text{-5}}\text{0}\text{0}\\ \text{Change}\text{(M):}\text{-x}\text{+x}\text{+x}\\ \text{----------------------------------------------------------------------}\\ \text{Equilibrium}\text{(M)}\text{(1}{\text{.06×10}}^{\text{-5}}\text{)-x}\text{x}\text{x}\end{array}$

Equilibrium constant for this reaction is,

$\text{K=}\frac{\left[{\text{H}}^{\text{+}}\right]\left[{\text{HCO}}_{\text{3}}{}^{\text{-}}\right]}{\left[{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}\right]}$

The given K value for the reaction is $\text{4}{\text{.2×10}}^{\text{-7}}$.

Therefore,

         $\text{K}\text{=}\text{4}\text{.2}{\text{×10}}^{\text{-7}}\text{=}\frac{\left[{\text{H}}^{\text{+}}\right]\left[{\text{HCO}}_{\text{3}}{}^{\text{-}}\right]}{\left[{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}\right]}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{(1}{\text{.06×10}}^{\text{-5}}\text{)-x}}$

Solving the quadratic equation then,

 Concentration of ${\text{H}}^{\text{+}}$ $\text{x}\text{=}\text{1}{\text{.9×10}}^{\text{-6}}\text{M}\text{=}{\text{[H}}^{\text{+}}\text{]}$

pH is the abbreviation of potential of hydrogen and it is a numerical scale used to specify the acidity and basicity of an aqueous solution.

   $\text{pH}\text{= -log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$

At equilibrium condition the concentration of ${\text{H}}^{\text{+}}$ is equal to the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$.

Therefore, the pH of water when equilibrium is reached between atmospheric ${\text{CO}}_{\text{2}}$ and ${\text{CO}}_{\text{2}}$ is dissolved in the water is,

                                    $\text{pH}\text{=}{\text{-log[H}}^{\text{+}}\text{]}$

Concentration of ${\text{H}}^{\text{+}}$,         ${\text{[H}}^{\text{+}}\text{]}\text{=}\text{1}\text{.9}\text{×}{\text{10}}^{\text{-6}}\text{M}$

           $\begin{array}{l}\text{pH}\text{=}\text{-log}\text{(1}{\text{.9×10}}^{\text{-6}}\text{)}\\ \text{=}\text{5}\text{.72}\end{array}$

Therefore the pH of water when equilibrium is reached between atmospheric ${\text{CO}}_{\text{2}}$ and ${\text{CO}}_{\text{2}}$ is dissolved in the water is $\text{5}\text{.72}$.

Conclusion

pH of the water under given condition is determined.