# 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?

##### 1 Answer

#"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

#bb(PV = nRT)# where:

#P# is thepressurein#"atm"# .#V# is thevolumein#"L"# .#n# is themols of gasin#"mol"# s.#R = "0.082057 L"cdot"atm/mol"cdot"K"# is theuniversal gas constantin the appropriate units.#T# is thetemperaturein#"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

#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

(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.)