Chapter 10, Problem 10.60QP

Interpretation Introduction

Interpretation: The solubility of air in water at $20°\text{C}$ and $1\text{atm}$ is given to be $7.9\times {10}^{-4}\text{M}$. The Henry’s law constant of air is to be calculated and the sum of ${\text{K}}_{\text{H}}$ values of ${\text{N}}_{\text{2}}$ and ${\text{O}}_{\text{2}}$ is equal to Henry’s law constant of air is to be predicted.

Concept introduction:

To determine: The Henry’s law constant of air.

Answer to Problem 10.60QP

Solution

The Henry’s law constant of air is $\underset{\_}{3.779\times 10^{-3}mol/L\cdot atm}$ and the sum Henry’s law constant of ${\text{N}}_{\text{2}}$ and ${\text{O}}_{\text{2}}$ is not equal to the Henry’s law constant of air.

Explanation of Solution

Explanation

Given

The solubility of gas in water is $7.9\times {10}^{-4}\text{M}$.

Total pressure of gas is $1\text{atm}$.

The Henry’s constant is calculated by the formula,

$K_{\text{H}}=\frac{C}{p}$

Where,

• $C$ is the concentration of dissolved gas.
• $p$ is the partial pressure of the gas.

Partial pressure of the gas is calculated by the formula,

$p=m\times P$

Where,

• $m$ is the mole fraction of oxygen in atmosphere.
• $P$ is the total pressure of gas.

The mole fraction of oxygen in atmosphere is 0.209.

Substitute the value of mole fraction and total pressure of gas in above formula.

$\begin{array}{c}p=0.209\times 1\text{atm}\\ =\text{0}\text{.209}\text{atm}\end{array}$

Substitute the concentration of gas dissolved in water and partial pressure of the gas in above formula.

$\begin{array}{c}{K}_H=\frac{7.9\times {10}^{-4}\text{M}}{0.209\text{atm}}\\ =\underset{\_}{3.779\times 10^{-3}mol/L\cdot atm}\end{array}$

Therefore, the Henry’s law constant of air is $\underset{\_}{3.779\times 10^{-3}mol/L\cdot atm}$.

The sum of the Henry’s law constant of ${\text{N}}_{\text{2}}$ and ${\text{O}}_{\text{2}}$ is given by the formula,

$k_{\text{H}}=K_{{\text{N}}_2}+K_{{\text{O}}_2}$

Where,

• $K_{{\text{N}}_2}$ is the Henry’s constant of ${\text{N}}_{\text{2}}$.
• $K_{{\text{O}}_2}$ is the Henry’s constant of ${\text{O}}_{\text{2}}$.

The Henry’s constant of ${\text{N}}_{\text{2}}$ is $0.67\times {10}^{-3}\text{mol}/\text{L}\cdot \text{atm}$.

The Henry’s constant of ${\text{O}}_{\text{2}}$ .is $1.3\times {10}^{-3}\text{mol}/\text{L}\cdot \text{atm}$.

Substitute the values of Henry’s law constant of ${\text{N}}_{\text{2}}$ and ${\text{O}}_{\text{2}}$ in the above formula.

$\begin{array}{c}{k}_{\text{H}}=\left(0.67\times {10}^{-3}+1.3\times {10}^{-3}\right)\text{mol}/\text{L}\cdot \text{atm}\\ =\text{1}\text{.97}\times {\text{10}}^{-3}\text{mol/L}\cdot \text{atm}\end{array}$

Therefore, the sum Henry’s law constant of ${\text{N}}_{\text{2}}$ and ${\text{O}}_{\text{2}}$ is not equal to the Henry’s law constant of air.

Conclusion

The Henry’s law constant of air is $\underset{\_}{3.779\times 10^{-3}mol/L\cdot atm}$ and the sum Henry’s law constant of ${\text{N}}_{\text{2}}$ and ${\text{O}}_{\text{2}}$ is not equal to the Henry’s law constant of air.