Chapter 5, Problem 5A.10AE

(i)

Interpretation Introduction

Interpretation: Under the given conditions, the solubility of ${\text{CO}}_{\text{2}}$ gas in water has to be calculated.

Concept introduction: Henry’s law states that the partial vapor pressure of a solute $\text{B}$ is directly proportional to the mole fraction of solute $\text{B}$.  Hence, for dilute solution the compositions can be determined by Henry’s law, which is expressed as,

  $p_{\text{B}}=K_{\text{B}}\times {\chi }_{\text{B}}$

Answer to Problem 5A.10AE

The solubility of ${\text{CO}}_{\text{2}}$ in water at $p_{{\text{CO}}_{\text{2}}}=0.10\text{atm}$ is $\underset{\_}{3.37\times 10^{-3}k{g}^{-1}mol}$.

Explanation of Solution

Carbon dioxide is dissolving in water, which is in small amount.  Therefore, ${\text{CO}}_{\text{2}}$ acts as a solute in water.  Since the amount of ${\text{CO}}_{\text{2}}$ in water is small, it means solution is very dilute.  The Henry’s law for ${\text{CO}}_{\text{2}}$ is expressed as,

  $p_{{\text{CO}}_{\text{2}}}=K_{{\text{CO}}_{\text{2}}}\times {\chi }_{{\text{CO}}_{\text{2}}}$                                                                                         (1)

Where,

• $p_{{\text{CO}}_{\text{2}}}$ is the partial vapor pressure of ${\text{CO}}_{\text{2}}$.
• ${\chi }_{{\text{CO}}_{\text{2}}}$ is the mole fraction of ${\text{CO}}_{\text{2}}$.
• $K_{{\text{CO}}_{\text{2}}}$ is the Henry’s law constant.

Since, ${\chi }_{{\text{CO}}_{\text{2}}}$ is the mole fraction of ${\text{CO}}_{\text{2}}$, it represents the number of moles or amount of ${\text{CO}}_{\text{2}}$ dissolved in water.  Therefore solubility of ${\text{CO}}_{\text{2}}$ in water can be expressed in terms of mole fraction of ${\text{CO}}_{\text{2}}$.

Therefore, the equation (1) can be written as,

  $p_{{\text{CO}}_{\text{2}}}=K_{{\text{CO}}_{\text{2}}}\times b_{{\text{CO}}_{\text{2}}}$                                                                                         (2)

Where,

• $b_{{\text{CO}}_{\text{2}}}$ is the solubility of ${\text{CO}}_{\text{2}}$ gas in water.

Rearrange the equation (2) in terms of $b_{{\text{CO}}_{\text{2}}}$.

  $b_{{\text{CO}}_{\text{2}}}=\frac{p_{{\text{CO}}_{\text{2}}}}{K_{{\text{CO}}_{\text{2}}}}$                                                                                                   (3)

It is given that,

The value of Henry’s law constant, $K_{{\text{CO}}_{\text{2}}}$ is $\text{3}\text{.01}\times {\text{10}}^{\text{3}}\text{kPa}\text{kg}\text{}{\text{mol}}^{-\text{1}}$.

The Partial pressure, $p_{{\text{CO}}_{\text{2}}}$ is $0.1\text{atm}$.

The conversion of $\text{atm}$ to $\text{kPa}$ is done as,

  $\text{1}\text{atm}=\text{101}\text{.325}\text{kPa}$

Therefore the conversion of $0.1\text{atm}$ to $\text{kPa}$ is done as,

  $\begin{array}{c}\text{0}\text{.1}\text{atm}=\text{0}\text{.1}\times \text{101}\text{.325}\text{kPa}\\ \text{=10}\text{.1325}\text{kPa}\end{array}$

Substitute $K_{{\text{CO}}_{\text{2}}}=\text{3}\text{.01}\times {\text{10}}^{\text{3}}\text{kPa}\text{kg}\text{}{\text{mol}}^{-\text{1}}$ and $p_{{\text{CO}}_{\text{2}}}=\text{10}\text{.1325}\text{kPa}$ in equation (3)

   $\begin{array}{c}{b}_{{\text{CO}}_{\text{2}}}=\frac{10.1325\text{}\text{kPa}}{3.01\times {10}^3\text{kPa}\text{}\text{kg}{\text{mol}}^{-1}}\\ =\underset{\_}{3.37\times 10^{-3}k{g}^{-1}mol}\end{array}$

Therefore, the solubility of ${\text{CO}}_{\text{2}}$ in water at $p_{{\text{CO}}_{\text{2}}}=0.1\text{atm}$ is $\underset{\_}{3.37\times 10^{-3}k{g}^{-1}mol}$.

(ii)

Interpretation Introduction

Interpretation: Under the given conditions, the solubility of ${\text{CO}}_{\text{2}}$ gas in water has to be calculated.

Concept introduction: Henry’s law states that the partial vapor pressure of a solute $\text{B}$ is directly proportional to the mole fraction of solute $\text{B}$.  Hence, for dilute solution the compositions can be determined by Henry’s law, which is expressed as,

  $p_{\text{B}}=K_{\text{B}}\times {\chi }_{\text{B}}$

Answer to Problem 5A.10AE

The solubility of ${\text{CO}}_{\text{2}}$ in water at $p_{{\text{CO}}_{\text{2}}}=1\text{atm}$ is $\underset{\_}{33.7\times 10^{-3}k{g}^{-1}mol}$.

Explanation of Solution

Carbon dioxide is dissolving in water, which is in small amount.  Therefore, ${\text{CO}}_{\text{2}}$ acts as a solute in water.  Since the amount of ${\text{CO}}_{\text{2}}$ in water is small, it means solution is very dilute.  The Henry’s law for ${\text{CO}}_{\text{2}}$ is expressed as,

  $p_{{\text{CO}}_{\text{2}}}=K_{{\text{CO}}_{\text{2}}}\times {\chi }_{{\text{CO}}_{\text{2}}}$                                                                                         (1)

Where,

• $p_{{\text{CO}}_{\text{2}}}$ is the partial vapor pressure of ${\text{CO}}_{\text{2}}$.
• ${\chi }_{{\text{CO}}_{\text{2}}}$ is the mole fraction of ${\text{CO}}_{\text{2}}$.
• $K_{{\text{CO}}_{\text{2}}}$ is the Henry’s law constant.

Since, ${\chi }_{{\text{CO}}_{\text{2}}}$ is the mole fraction of ${\text{CO}}_{\text{2}}$, it represents the number of moles or amount of ${\text{CO}}_{\text{2}}$ dissolved in water.  Therefore solubility of ${\text{CO}}_{\text{2}}$ in water can be expressed in terms of mole fraction of ${\text{CO}}_{\text{2}}$.

Therefore, the equation (1) can be written as,

  $p_{{\text{CO}}_{\text{2}}}=K_{{\text{CO}}_{\text{2}}}\times b_{{\text{CO}}_{\text{2}}}$                                                                                         (2)

Where,

• $b_{{\text{CO}}_{\text{2}}}$ is the solubility of ${\text{CO}}_{\text{2}}$ gas in water.

Rearrange the equation (2) in terms of $b_{{\text{CO}}_{\text{2}}}$.

  $b_{{\text{CO}}_{\text{2}}}=\frac{p_{{\text{CO}}_{\text{2}}}}{K_{{\text{CO}}_{\text{2}}}}$                                                                                                   (3)

It is given that,

The value of Henry’s law constant, $K_{{\text{CO}}_{\text{2}}}$ is $\text{3}\text{.01}\times {\text{10}}^{\text{3}}\text{kPa}\text{kg}\text{}{\text{mol}}^{-\text{1}}$.

The Partial pressure, $p_{{\text{CO}}_{\text{2}}}$ is $1\text{atm}$.

The conversion of $\text{atm}$ to $\text{kPa}$ is done as,

  $\text{1}\text{atm}=\text{101}\text{.325}\text{kPa}$

Substitute the value of $K_{{\text{CO}}_{\text{2}}}=\text{3}\text{.01}\times {\text{10}}^{\text{3}}\text{kPa}\text{kg}\text{}{\text{mol}}^{-\text{1}}$ and $p_{{\text{CO}}_{\text{2}}}=\text{101}\text{.325}\text{kPa}$ in equation (3)

   $\begin{array}{c}{b}_{{\text{CO}}_{\text{2}}}=\frac{101.325\text{}\text{kPa}}{3.01\times {10}^3\text{kPa}\text{}\text{kg}{\text{mol}}^{-1}}\\ =\underset{\_}{33.7\times 10^{-3}k{g}^{-1}mol}\end{array}$

Therefore, the solubility of ${\text{CO}}_{\text{2}}$ in water at $p_{{\text{CO}}_{\text{2}}}=1\text{atm}$ is $\underset{\_}{33.7\times 10^{-3}k{g}^{-1}mol}$.