Chapter 16, Problem 55A

(a)

Interpretation Introduction

Interpretation: The solubility of the solution at $0.60\text{ atm}$ is to be calculated.

Concept Introduction: Henry’s Law which states that at a constant pressure the solubility of the gas molecule in the liquid is directly proportional to the partial pressure of the gas molecule.

The expression that shows the relationship between pressure and gas concentration according to Henry’s law is as follows:

  $S=K_{H}P$

Here, S is the solubility, P is the pressure, and $K_H$ is Henry’s Constant.

The formula can also be expressed as,

  $\frac{S_1}{S_2}=\frac{P_1}{P_2}$   ...... (1)

Here, ${\text{P}}_1$ is the initial pressure, ${\text{P}}_2$ is the final pressure, ${\text{S}}_1$ is the initial solubility, and ${\text{S}}_2$ is the final solubility.

Answer to Problem 55A

The solubility of the solution $0.60\text{ atm}$ at $0.0156\text{ g/L}$

Explanation of Solution

Given,

  $\begin{array}{l}{P}_1=1.00\text{ atm}\\ P_2=0.60\text{ atm}\\ {\text{S}}_1=0.026\text{ g/L}\end{array}$ .

Temperature is constant.

Henry’s Law states that at a constant pressure the solubility of the gas molecule in the liquid is directly proportional to the partial pressure of the gas molecule.

Substitute $0.60\text{ atm}$ for ${\text{P}}_2$ , $0.026\text{ g/L}$ for ${\text{S}}_1$ , and $1.00\text{ atm}$ for $P_1$ in equation (1).

  $\frac{0.026\text{ g/L}}{S_2}=\frac{1.00\text{ atm}}{0.60\text{ atm}}$

Rearrange the equation,

  $\begin{array}{c}{S}_2=\frac{0.60\text{ atm}}{1.00\text{ atm}}\times 0.026\text{ g/L}\\ =0.0156\text{ g/L}\end{array}$

b)

Interpretation Introduction

Interpretation: The solubility of the solution at $1.80\text{ atm}$ is to be calculated.

Concept Introduction: Henry’s Law which states that at a constant pressure the solubility of the gas molecule in the liquid is directly proportional to the partial pressure of the gas molecule.

Answer to Problem 55A

The solubility of the solution $1.80\text{ atm}$ at $0.0468\text{ g/L}$ .

Explanation of Solution

Given,

  $\begin{array}{l}{P}_1=1.00\text{ atm}\\ P_2=1.80\text{ atm}\\ {\text{S}}_1=0.026\text{ g/L}\end{array}$ .

Temperature is constant.

Substitute $\text{1}\text{.80 atm}$ for ${\text{P}}_2$ , $0.026\text{ g/L}$ for ${\text{S}}_1$ , and $1.00\text{ atm}$ for $P_1$ in equation (1).

  $\frac{0.026\text{ g/L}}{S_2}=\frac{1.00\text{ atm}}{1.80\text{ atm}}$

Rearrange the equation,

  $\begin{array}{c}{S}_2=\frac{1.80\text{ atm}}{1.00\text{ atm}}\times 0.026\text{ g/L}\\ =0.0468\text{ g/L}\end{array}$

The solubility of the solution $1.80\text{ atm}$ at $0.0468\text{ g/L}$ .