# The solubility of carbon dioxide in water is 0.161 g CO_2 in 100 mL of water at 20°C and 1.00 atm. A soft drink is carbonated with carbon dioxide gas at 5.50 atm pressure. What is the solubility of carbon dioxide in water at this pressure?

May 6, 2017

$\text{8.855 g/L}$

This is just asking you to calculate the new molar density (i.e. molar solubility) at a different pressure, i.e. use the ideal gas law. (We'll end up converting it later back to $\text{g/L}$.)

$\boldsymbol{P V = n R T}$

where:

• $P$ is the pressure in $\text{atm}$.
• $V$ is the volume in $\text{L}$.
• $n$ is the mols of gas in $\text{mol}$s.
• $R = \text{0.082057 L"cdot"atm/mol"cdot"K}$ is the universal gas constant in the appropriate units.
• $T$ is the temperature in $\text{K}$.

Solving for molar density, we get:

$\frac{n}{V} = \frac{P}{R T}$

Since we have two solubilities to consider, we must have two states.

${n}_{1} / {V}_{1} = {P}_{1} / \left(R T\right)$

${n}_{2} / {V}_{2} = {P}_{2} / \left(R T\right)$

Therefore, we have:

$R T = \frac{{P}_{1} {V}_{1}}{{n}_{1}} = \frac{{P}_{2} {V}_{2}}{{n}_{2}}$

Solving for the new molar density, we get:

${n}_{2} / {V}_{2} = {n}_{1} / {V}_{1} \left(\frac{{P}_{2}}{{P}_{1}}\right)$

At this point, we can convert the mass-based solubility to mols. We assume that the ${\text{CO}}_{2}$ was as dissolved into the water as possible for the initial pressure, and that it doesn't change the volume of the water.

n_1/V_1 = (0.161 cancel("g CO"_2) xx "1 mol CO"_2/(44.009 cancel("g CO"_2)))/("0.100 L water")

$=$ $\text{0.0366 mol/L}$

Therefore, the new solubility is:

"0.0366 mol/L" xx ("5.50 atm")/("1.00 atm")

$=$ $\text{0.2012 mol/L}$

In the original units, we have:

(0.2012 cancel("mols CO"_2))/"L solution" xx "44.009 g CO"_2/cancel("mol CO"_2)

$=$ $\textcolor{b l u e}{\text{8.855 g/L}}$

Hence, the solubility increased by a factor of $\frac{8.855}{1.61} = 5.5$, which is the ratio of the pressures. Does it make sense that a gas at a higher external pressure is more soluble?

(Note that we could have simply used the initial mass-based solubility as it is and avoided converting to molar solubility. But it is a good exercise to use the molar mass for conversions, so I avoided skipping that step.)