Question

The partial pressure of ${\mathbf{CO}}_{\mathbf{2}}$ gas above the liquid in a bottle of champagne at ${\mathbf{20}}^{\mathbf{o}}\mathbf{C}$is 5.5 atm. What is the solubility of ${\mathbf{CO}}_{\mathbf{2}}$ in champagne? Assume Henry’s law constant is the same for champagne as for water: at ${\mathbf{20}}^{\mathbf{o}}\mathbf{C}$, $\mathbf{KH}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{mol}\text{/}\mathbf{L}·\mathbf{atm}$.

Short answer

The concentration of the ${\text{CO}}_2$gas is $\mathbf{20}\mathbf{.}\mathbf{35}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}{\mathbf{molL}}^{\mathbf{-}\mathbf{1}}$

Step-by-step solution

Definition

The solubility can be defined as the amount of the solute which can be gas, solid or liquid dissolve in the solvent.

Henry’s Law,

$\mathbf{Solubility}\mathbf{=}{\mathbf{K}}_{\mathbf{H}}\mathbf{\times }{\mathbf{P}}_{\mathbf{gas}}$

${\mathbf{P}}_{\mathbf{gas}}\mathbf{=}\mathbf{Pressure} \mathbf{of} \mathbf{Gas}$

${\mathbf{K}}_{\mathbf{H}}\mathbf{=}\mathbf{Henry}\mathbf{\text{'}}\mathbf{s} \mathbf{Law} \mathbf{Constant}$

Numerical Explanation

Henry’s constant of gas,${\text{K}}_{\text{H}}=3.7\times {10}^{-2}{\text{molL}}^{-1}{\text{atm}}^{-1}$.

Pressure of the${\text{CO}}_2$gas is$5.5\text{atm}$ .

As per Henry’s law, the pressure of the gas increases with the increasing solubility of the solute.

$$\begin{gathered} \mathbf{Solubility}\mathbf{=}{\mathbf{K}}_{\mathbf{H}}\mathbf{\times }{\mathbf{P}}_{\mathbf{gas}} \\ \mathbf{Solubility}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}{\mathbf{molL}}^{\mathbf{-}\mathbf{1}}{\mathbf{atm}}^{\mathbf{-}\mathbf{1}}\mathbf{\times }\mathbf{5}\mathbf{.}\mathbf{5}\mathbf{atm} \\ \mathbf{Solubility}\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{35}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}{\mathbf{molL}}^{\mathbf{-}\mathbf{1}} \end{gathered}$$

The concentration of the ${\text{CO}}_2$gas is $\mathbf{20}\mathbf{.}\mathbf{35}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}{\mathbf{molL}}^{\mathbf{-}\mathbf{1}}$.