Chapter 13, Problem 13.159P

(a)

Interpretation Introduction

Interpretation:

The $k_{\text{H}}$ in $\text{mol/L}\cdot \text{atm}$ 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.

The formula to calculate the density of the substance is as follows:

  $\text{Density of substance}\left(\rho \right)=\frac{\text{Mass of substance}\left(M\right)}{\text{Volume of substance}\left(V\right)}$        (2)

Answer to Problem 13.159P

$0.0212\text{mol/L}\cdot \text{atm}$ is $k_{\text{H}}$ in $\text{mol/L}\cdot \text{atm}$.

Explanation of Solution

Rearrange equation (2) to calculate the mass of the substance as follows:

  $\text{Mass of substance}=\left(\text{Density of substance}\right)\left(\text{Volume of substance}\right)$        (3)

Substitute $\text{1}\text{.674}\text{g/mL}$ for the density of the solution and $\text{1 L}$ for the volume of the solution in equation (3) to calculate the mass of ${\text{C}}_6{\text{F}}_{14}$.

  $\begin{array}{c}{\text{Mass of C}}_6{\text{F}}_{14}=\left(\frac{\text{1}\text{.674}\text{g}}{1\text{mL}}\right)\left(\text{1 L}\right)\left(\frac{1\text{mL}}{{10}^{-3}\text{L}}\right)\\ =1.674\times {10}^{-3}\text{g}\end{array}$

The formula to calculate the moles of ${\text{C}}_6{\text{F}}_{14}$ is as follows:

  ${\text{Moles of C}}_6{\text{F}}_{14}=\frac{{\text{Given mass of C}}_6{\text{F}}_{14}}{{\text{Molar mass of C}}_6{\text{F}}_{14}}$        (4)

Substitute $1.674\times {10}^{-3}\text{g}$ for the given mass and $338\text{g/mol}$ for the molar mass of ${\text{C}}_6{\text{F}}_{14}$ in equation (4).

  $\begin{array}{c}{\text{Moles of C}}_6{\text{F}}_{14}=\left(1.674\times {10}^{-3}\text{g}\right)\left(\frac{1\text{}\text{mol}}{338\text{g}}\right)\\ =4.9527\text{mol}\end{array}$

The formula to calculate the mole fraction of ${\text{O}}_2$ is as follows:

  ${\text{Mole fraction of O}}_2=\frac{{\text{Moles of O}}_2}{{\text{Moles of O}}_2+{\text{Moles of C}}_6{\text{F}}_{14}}$        (5)

Rearrange equation (5) to calculate the moles of ${\text{O}}_2$ as follows:

  ${\text{Moles of O}}_2=\left({\text{Mole fraction of O}}_2\right)\left({\text{Moles of O}}_2+{\text{Moles of C}}_6{\text{F}}_{14}\right)$        (6)

The number of moles of ${\text{O}}_2$ is very small so they can be neglected in the denominator.

Substitute $4.28\times {10}^{-3}$ for the mole fraction of ${\text{O}}_2$ and $4.9527\text{mol}$ for the moles of ${\text{C}}_6{\text{F}}_{14}$ in equation (6).

  $\begin{array}{c}{\text{Moles of O}}_2=\left(4.28\times {10}^{-3}\right)\left(4.9527\text{mol}\right)\\ =0.021198\text{mol}\end{array}$

Since the volume of the solution is considered as $1\text{L}$ so the moles are equal to the molarity. So the molarity or solubility of ${\text{O}}_2$ is $0.021198\text{mol/L}$.

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

  $k_{\text{H}}=\frac{S_{\text{gas}}}{P_{\text{gas}}}$        (7)

Substitute $0.021198\text{mol/L}$ for $S_{\text{gas}}$ and $101325\text{Pa}$ for $P_{\text{gas}}$ in equation (7).

  $\begin{array}{c}{k}_{\text{H}}=\frac{0.021198\text{mol/L}}{\left(101325\text{Pa}\right)\left(\frac{1\text{atm}}{101325\text{Pa}}\right)}\\ =0.021198\text{mol/L}\cdot \text{atm}\\ \approx 0.0212\text{mol/L}\cdot \text{atm}\text{.}\end{array}$

Conclusion

The value of $k_{\text{H}}$ depends on the solubility of gas which in turns depends on the intermolecular forces.

(b)

Interpretation Introduction

Interpretation:

The solubility of ${\text{O}}_2$ in water at $25°\text{C}$ is to be calculated in $\text{ppm}$.

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.

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:

  $\text{ppm}=\left(\frac{\text{mass}\text{of solute}}{\text{mass}\text{of}\text{solution}}\right){10}^6$        (8)

Answer to Problem 13.159P

$\text{8}\text{.86 ppm}$ is the solubility of ${\text{O}}_2$ in $\text{ppm}$.

Explanation of Solution

The value of $k_{\text{H}}$ in $\text{mol/L}\cdot \text{atm}$ is calculated as follows:

  $\begin{array}{c}{k}_{\text{H}}=\frac{1}{756.7\text{}\text{L}\cdot \text{atm/mol}}\\ =1.322\times {10}^{-3}\text{mol/L}\cdot \text{atm}\end{array}$

The formula to calculate the pressure of ${\text{O}}_2$ is as follows:

  $P_{{\text{O}}_2}=\left(\frac{\%{\text{of O}}_2}{\%\text{}\text{of air}}\right)\left(1\text{atm}\right)$        (9)

Substitute $20.95\%$ for % of ${\text{O}}_2$ and $100\%\text{}$ for % of air in equation (9).

  $\begin{array}{c}{P}_{{\text{O}}_2}=\left(\frac{20.95\%}{100\%}\right)\left(1\text{atm}\right)\\ =0.2095\text{atm}\end{array}$

Substitute $1.322\times {10}^{-3}\text{mol/L}\cdot \text{atm}$ for $k_{\text{H}}$ and $0.2095\text{atm}$ for $P_{\text{gas}}$ in equation (1).

  $\begin{array}{c}{S}_{\text{gas}}=\left(\frac{1.322\times {10}^{-3}\text{mol}}{1\text{}\text{L}\cdot \text{atm}}\right)\left(0.2095\text{atm}\right)\\ =2.76959\times {10}^{-4}\text{}\text{mol/L}\end{array}$

The mass of ${\text{O}}_2$ in $1\text{L}$ of solution is calculated as follows:

  $\begin{array}{c}{\text{Mass of O}}_2=\left(\frac{2.76959\times {10}^{-4}\text{}\text{mol}}{1\text{L}}\right)\left(\frac{32\text{g}{\text{O}}_2}{1{\text{mol O}}_2}\right)\\ =8.8627\times {10}^{-3}\text{g/L}\end{array}$

Consider the mass of the solution to be $1000\text{g}$.

Substitute $8.8627\times {10}^{-3}\text{g}$ for the mass of solute and $1000\text{g}$ for the mass of the solution in equation (8) to calculate the solubility in $\text{ppm}$.

  $\begin{array}{c}\text{Solubility}\left(\text{ppm}\right)=\left(\frac{8.8627\times {10}^{-3}\text{g}}{1000\text{g}}\right)\left({10}^6\right)\\ =8.8627\text{ppm}\\ \approx 8.86\text{ppm}\text{.}\end{array}$

Conclusion

The solubility of gas depends on the partial pressure of the gas. Higher partial pressure means more solubility and vice-versa.

(c)

Interpretation Introduction

Interpretation:

The decreasing order of $k_{\text{H}}$ for ${\text{O}}_2$ in water, ethanol, ${\text{C}}_6{\text{F}}_{14}$ and ${\text{C}}_6{\text{H}}_{14}$ is to be determined. Also, its explanation is to be given.

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 13.159P

The decreasing order of $k_{\text{H}}$ of ${\text{O}}_2$ in different solvents is as follows:

    ${\text{C}}_6{\text{F}}_{14}>{\text{C}}_6{\text{H}}_{14}>\text{ethanol}>\text{water}$

Explanation of Solution

The value of $k_{\text{H}}$ is directly proportional to the solubility of the gas. The solubility of a gas in any solvent depends on the intermolecular forces in the solvent molecules. Ethanol and water molecules both forms hydrogen bonding but its extent is more in water as compared to that in ethanol so the solubility of oxygen is the least in water, followed by that in ethanol. ${\text{C}}_6{\text{F}}_{14}$ and ${\text{C}}_6{\text{H}}_{14}$ have weak dispersion forces between their molecules. But due to more electronegativity of fluorine, ${\text{C}}_6{\text{F}}_{14}$ has weaker intermolecular forces than that of ${\text{C}}_6{\text{H}}_{14}$ so the solubility of oxygen is maximum in ${\text{C}}_6{\text{F}}_{14}$, followed by ${\text{C}}_6{\text{H}}_{14}$.

So the decreasing order of $k_{\text{H}}$ of ${\text{O}}_2$ in different solvents is as follows:

    ${\text{C}}_6{\text{F}}_{14}>{\text{C}}_6{\text{H}}_{14}>\text{ethanol}>\text{water}$

Conclusion

The value of $k_{\text{H}}$ depends on the solubility of gas which in turns depends on the intermolecular forces.