Chapter 10, Problem 10.46QP

(a)

Interpretation Introduction

Interpretation: The change that has a greater effect on the solubility of oxygen in water, out of decreasing the temperature and raising the pressure, is to be identified.

Concept introduction: The Henry’s law states that “the amount of gas dissolved in given liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.”

To determine: The effect of decreasing temperature from $20°\text{C}$ to $10°\text{C}$ on the solubility of oxygen in water.

Answer to Problem 10.46QP

Solution

The change of solubility of oxygen in water from $20°\text{C}$ to $10°\text{C}$ is $0.01$.

Explanation of Solution

Explanation

The solubility of oxygen gas in water at $20°\text{C}$ is $0.03\text{mol}/\text{L}$ and at $10°\text{C}$ is $0.04\text{mol}/\text{L}$ as shown in Figure 1.

Figure 1

Therefore, the change in solubility of oxygen from $20°\text{C}$ to $10°\text{C}$ is $0.01\text{mol}/\text{L}$.

(b)

Interpretation Introduction

To determine: The effect of raising the pressure from $1.00\text{atm}$ to $1.25\text{atm}$ on the solubility of oxygen in water.

Answer to Problem 10.46QP

Solution

The change in solubility of oxygen from $1\text{atm}$ to $1.25\text{atm}$ is $0.000617\text{mol}/\text{L}$.

Explanation of Solution

Explanation

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

Pressure is $1\text{atm}$.

The concentration of ${\text{O}}_{\text{2}}$ in water is calculated by the formula,

$C=K_{\text{H}}\times p$ (1)

Where,

• $K_{\text{H}}$ is the Henry’s law constant.
• $p$ partial pressure of ${\text{O}}_{\text{2}}$.

Partial pressure of the ${\text{O}}_2$ gas is calculated by the formula,

$p=m\times P$

Where,

• $m$ is the mole fraction of oxygen.
• $P$ is the atmospheric pressure.

The mole fraction of oxygen is 0.209.

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

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

Partial pressure of ${\text{O}}_2$ gas at $1\text{atm}$ is $0.209\text{atm}$.

Substitute the value of partial pressure and Henry’s law constant in equation (1).

$\begin{array}{c}C=1.3\times {10}^{-3}\text{mol}/\text{L}\cdot \text{atm}\times 0.209\text{atm}\\ =2.717\times {10}^{-4}\text{mol}/\text{L}\end{array}$

The solubility of ${\text{O}}_{\text{2}}$ in water at $1\text{atm}$ is $2.717\times {10}^{-4}\text{mol}/\text{L}$.

The pressure value is raised from $1\text{atm}$ to $1.25\text{atm}$. The partial pressure of ${\text{O}}_{\text{2}}$ will change in increasing temperature.

Partial pressure of the ${\text{O}}_2$ gas is calculated by the formula,

$p=m\times P$

Where,

• $m$ is the mole fraction of oxygen.
• $P$ is the atmospheric pressure.

The mole fraction of oxygen is 0.209.

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

$\begin{array}{c}p=0.209\times 1.25\text{atm}\\ =\text{0}\text{.2575}\text{atm}\end{array}$

Partial pressure of ${\text{O}}_2$ gas at $1.25\text{atm}$ is $0.2575\text{atm}$.

Substitute the value of partial pressure and Henry’s law constant in equation (1).

$\begin{array}{c}C=1.3\times {10}^{-3}\text{mol}/\text{L}\cdot \text{atm}\times 0.2575\text{atm}\\ =3.334\times {10}^{-4}\text{mol}/\text{L}\end{array}$

The solubility of ${\text{O}}_{\text{2}}$ in water at $1.25\text{atm}$ is $3.334\times {10}^{-4}\text{mol}/\text{L}$.

The change in solubility of oxygen from $1\text{atm}$ to $1.25\text{atm}$ is $\begin{array}{c}=\left(3.334-2.717\right)\times {10}^{-3}\text{mol}/\text{L}\\ =\text{0}\text{.617}\times {\text{10}}^{-3}\text{mol}/\text{L}\\ =0.000617\text{mol}/\text{L}\end{array}$

Therefore, the change in solubility of oxygen from $1\text{atm}$ to $1.25\text{atm}$ is $0.000617\text{mol}/\text{L}$.

Conclusion

The change in temperature has great effect on the solubility of oxygen in water than change in pressure.