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?

#"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 #"g/L"#.)

#bb(PV = nRT)#

where:

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

Solving for molar density, we get:

#n/V = P/(RT)#

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

#n_1/V_1 = P_1/(RT)#

#n_2/V_2 = P_2/(RT)#

Therefore, we have:

#RT = (P_1V_1)/(n_1) = (P_2V_2)/(n_2)#

Solving for the new molar density, we get:

#n_2/V_2 = n_1/V_1 ((P_2)/(P_1))#

At this point, we can convert the mass-based solubility to mols. We assume that the #"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")#

#=# #"0.0366 mol/L"#

Therefore, the new solubility is:

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

#=# #"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)#

#=# #color(blue)("8.855 g/L")#

Hence, the solubility increased by a factor of #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.)