Chapter 13, Problem 13.155P

(a)

Interpretation Introduction

Interpretation:

The $k_{\text{H}}$ for ${\text{O}}_2$ in acrylic acid is to be calculated.

Concept introduction:

Henry’s law gives the quantitative relationship between the pressure of the gas and its solubility. It states that the amount of gas dissolved in a liquid is proportional to the partial pressure of the gas. Higher the partial pressure of the gas, more will be its solubility and vice-versa.

The formula to calculate the solubility of gases according to Henry’s law is as follows:

$S_{\text{gas}}=k_{\text{H}}\times P_{\text{gas}}$ (1)

Here, $S_{\text{gas}}$ is the solubility of the gas.

$k_{\text{H}}$ is Henry’s constant.

$P_{\text{gas}}$ is the partial pressure of the gas.

Answer to Problem 13.155P

The $k_{\text{H}}$ for ${\text{O}}_2$ in acrylic acid is $7.83\times {10}^{-3}\text{mol/L}\cdot \text{atm}$.

Explanation of Solution

Rearrange equation (1) to calculate $k_{\text{H}}$ as follows:

$k_{\text{H}}=\frac{S_{\text{gas}}}{P_{\text{gas}}}$ (2)

Substitute $1.64\times {10}^{-3}\text{mol/L}$ for $S_{\text{gas}}$ and $0.2095\text{atm}$ for $P_{\text{gas}}$ in equation (2).

$\begin{array}{c}{k}_{\text{H}}=\frac{1.64\times {10}^{-3}\text{mol/L}}{0.2095\text{atm}}\\ =7.8282\times {10}^{-3}\text{mol/L}\cdot \text{atm}\\ =7.83\times {10}^{-3}\text{mol/L}\cdot \text{atm}\end{array}$

Conclusion

The $k_{\text{H}}$ for ${\text{O}}_2$ in acrylic acid is $7.83\times {10}^{-3}\text{mol/L}\cdot \text{atm}$.

(b)

Interpretation Introduction

Interpretation:

The molarity of ${\text{O}}_2$ is to be calculated.

Concept introduction:

Molarity is defined as the number of moles of solute that are dissolved in one litre of solution. It is represented by $M$ and its unit is $\text{mol/L}$.

The formula to calculate the solubility of gases according to Henry’s law is as follows:

$S_{\text{gas}}=k_{\text{H}}\times P_{\text{gas}}$ (1)

Answer to Problem 13.155P

The molarity of ${\text{O}}_2$ is $4\times {10}^{-5}\text{}M$.

Explanation of Solution

Substitute $7.8282\times {10}^{-3}\text{mol/L}\cdot \text{atm}$ for $k_{\text{H}}$ and $\text{0}\text{.005}\text{}\text{atm}$ for $P_{\text{gas}}$ in equation (1).

$\begin{array}{c}{S}_{\text{gas}}=\left(7.83\times {10}^{-3}\text{mol/L}\cdot \text{atm}\right)\left(\text{0}\text{.005}\text{}\text{atm}\right)\\ =3.9141\times {10}^{-5}\text{mol/L}\\ =\text{4}\times {10}^{-5}\text{mol/L}\end{array}$

The solubility is equal to the molarity so the molarity of ${\text{O}}_2$ is $4\times {10}^{-5}\text{}M$.

Conclusion

The molarity of ${\text{O}}_2$ is $4\times {10}^{-5}\text{}M$.

(c)

Interpretation Introduction

Interpretation:

The mole fraction of ${\text{O}}_2$ is to be calculated.

Concept introduction:

The mole fraction is defined as the ratio of the number of moles of solute to the total number of moles in the mixture. It is represented by $X$.

The formula to calculate the mole fraction is as follows:

$\text{Mole}\text{fraction}\left(X\right)=\frac{\text{amount}\left(\text{mol}\right)\text{of}\text{solute}}{\text{amount}\left(\text{mol}\right)\text{of}\text{solute}+\text{amount}\left(\text{mol}\right)\text{of}\text{solvent}}$ (3)

Answer to Problem 13.155P

The mole fraction of ${\text{O}}_2$ is $3\times {10}^{-6}$.

Explanation of Solution

$4\times {10}^{-5}\text{}M$ of ${\text{O}}_2$ indicates that $4\times {10}^{-5}\text{}\text{mol}$ of ${\text{O}}_2$ is dissolved in $1\text{}\text{L}$ of the solution. $14.6\text{}M$ of acrylic acid means that $14.6\text{}\text{mol}$ of acrylic acid are dissolved in $1\text{}\text{L}$ of the solution.

Substitute $4\times {10}^{-5}\text{}\text{mol}$ for the moles of solute and $14.6\text{}\text{mol}$ for the moles of solvent in equation (3).

$\begin{array}{c}\text{Mole}\text{fraction}\left(X\right)=\frac{4\times {10}^{-5}\text{}\text{mol}}{4\times {10}^{-5}\text{}\text{mol}+14.6\text{}\text{mol}}\\ =2.73972\times {10}^{-6}\\ =3\times {10}^{-6}\end{array}$

Conclusion

The mole fraction of ${\text{O}}_2$ is $3\times {10}^{-6}$.

(d)

Interpretation Introduction

Interpretation:

The concentration of ${\text{O}}_2$ in $\text{ppm}$ is to be calculated.

Concept introduction:

The $\text{ppm}$ or parts per million is a concentration term that is defined as the mass of any substance divided by the mass of the solution, multiplied by ${10}^6$.

The formula to calculate the concentration of an ion in $\text{ppm}$ is as follows:

$\text{ppm}=\left(\frac{\text{mass}\text{of solute}}{\text{mass}\text{of}\text{solution}}\right){10}^6$ (4)

Answer to Problem 13.155P

The concentration of ${\text{O}}_2$ is $\text{1 ppm}$.

Explanation of Solution

The formula to calculate the mass of the compound is as follows:

$\text{Mass of compound}=\left(\text{Moles of compound}\right)\left(\text{Molar mass}\right)$ (5)

Substitute $4\times {10}^{-5}\text{}\text{mol}$ for the moles of ${\text{O}}_2$ and $32\text{}\text{g/mol}$ for the molar mass in equation (5) to calculate the mass of ${\text{O}}_2$.

$\begin{array}{c}{\text{Mass of O}}_2=\left(4\times {10}^{-5}\text{}\text{mol}\right)\left(\frac{32\text{}\text{g}}{1\text{}\text{}\text{mol}}\right)\\ =0.00128\text{g}\end{array}$

Substitute $14.6\text{}\text{mol}$ for the moles of acrylic acid and $\text{72}\text{.06}\text{g/mol}$ for the molar mass in equation (5) to calculate the mass of acrylic acid.

$\begin{array}{c}\text{Mass of acrylic acid}=\left(14.6\text{}\text{mol}\right)\left(\frac{\text{72}\text{.06}\text{}\text{g}}{1\text{}\text{}\text{mol}}\right)\\ =1052.076\text{g}\end{array}$

The formula to calculate the mass of the solution is as follows:

$\text{Mass}\text{}\text{of}\text{solution}=\text{Mass}\text{of}\text{solute}+\text{Mass}\text{of}\text{solvent}$ (6)

Substitute $0.00128\text{g}$ for the mass of solute and $1052.076\text{g}$ for the mass of solvent in equation (6).

$\begin{array}{c}\text{Mass}\text{}\text{of}\text{solution}=0.00128\text{g}+1052.076\text{g}\\ \text{=1052}\text{.07728}\text{g}\end{array}$

Substitute $0.00128\text{g}$ for the mass of solute and $\text{1052}\text{.07728}\text{g}$ for the mass of solution in equation (4).

$\begin{array}{c}\text{ppm}=\left(\frac{0.00128\text{g}}{\text{1052}\text{.07728}\text{g}}\right){10}^6\\ =1.2165\text{ppm}\\ =1\text{ppm}\end{array}$

Conclusion

The concentration of ${\text{O}}_2$ is $\text{1 ppm}$.