Question

Henry's law constant for ${\mathrm{CO}}_2$ in water is $1.67\times {10}^8\mathrm{Pa}$ at $298\mathrm{K}$. Calculate the quantity in $500\mathrm{mL}$ of soda water when packed under $2.5\mathrm{atm}$ pressure at $298\mathrm{K}$.

Answer 1

Step 1 Mole fraction of ${\mathrm{CO}}_2$

Henry's law constant for ${\mathrm{CO}}_2$ is $1.67\times {10}^8\mathrm{Pa}$

The partial pressure of ${\mathrm{CO}}_2$ is,

$$\begin{gathered} {\mathrm{p}}_{{\mathrm{CO}}_2}=2.5\mathrm{atm} \\ {\mathrm{p}}_{{\mathrm{CO}}_2}=2.5\times 101325\mathrm{Pa} \\ {\mathrm{p}}_{{\mathrm{CO}}_2}=2.533125\times {10}^5\mathrm{Pa} \end{gathered}$$

According to henry's law, in a solution, “the partial pressure of the gas in the vapour phase ($\mathrm{p}$) is proportional to the mole fraction of the gas ($x$) in the solution.

That is,

$$\begin{gathered} {\mathrm{p}}_{{\mathrm{co}}_2}={\mathrm{K}}_{\mathrm{H}}\mathrm{x} \\ \mathrm{x}=\frac{{\mathrm{p}}_{{\mathrm{co}}_2}}{{\mathrm{K}}_{\mathrm{H}}} \end{gathered}$$

By substituting the values, we get

$$\begin{gathered} \mathrm{x}=\frac{{\mathrm{p}}_{{\mathrm{co}}_2}}{{\mathrm{K}}_{\mathrm{H}}} \\ \mathrm{x}=\frac{2.533125\times {10}^5}{1067\times {10}^8} \\ \mathrm{x}=0.00152 \end{gathered}$$

Step 2: Quantity of ${\mathrm{CO}}_2$ in $500\mathrm{mL}$

The volume of water is $500\mathrm{mL}$

The molar mass of water is $18\mathrm{g}$

The number of moles of water is,

$$\begin{gathered} \mathrm{Number}\mathrm{of}\mathrm{moles}=\frac{\mathrm{Given}\mathrm{mass}}{\mathrm{Molar}\mathrm{mass}} \\ =\frac{500}{18} \\ =27.78\mathrm{moles} \end{gathered}$$

Mole fraction is,

$$\begin{gathered} \mathrm{x}=\frac{{\mathrm{n}}_{{\mathrm{co}}_2}}{{\mathrm{n}}_{{\mathrm{co}}_2}+{\mathrm{n}}_{{\mathrm{H}}_2\mathrm{o}}}\approx \frac{{\mathrm{n}}_{{\mathrm{co}}_2}}{{\mathrm{n}}_{{\mathrm{H}}_2\mathrm{o}}} \\ x=\frac{{\mathrm{n}}_{{\mathrm{co}}_2}}{{\mathrm{n}}_{{\mathrm{H}}_2\mathrm{o}}} \end{gathered}$$

Where ${\mathrm{n}}_{{\mathrm{co}}_2}$is the number of moles of carbon dioxide, ${\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}}$ is the number of moles of water

$$\begin{gathered} x=\frac{{\mathrm{n}}_{{\mathrm{co}}_2}}{{\mathrm{n}}_{{\mathrm{H}}_2\mathrm{o}}} \\ 0.00152=\frac{{\mathrm{n}}_{{\mathrm{co}}_2}}{27.78} \\ {\mathrm{n}}_{{\mathrm{co}}_2}=0.00152\times 27.78 \\ {\mathrm{n}}_{{\mathrm{co}}_2}=0.042\mathrm{mole}s \end{gathered}$$

So, the number of carbon dioxide moles is ${\mathrm{n}}_{{\mathrm{co}}_2}=0.042\mathrm{mole}s$

Hence, the amount of ${\mathrm{CO}}_2$ in $500\mathrm{mL}$ of soda water is $0.042\times 44=1.848g$