Question

Carbon dioxide gas has a Henry's law constant of 4.48×10-5 M/mmHg at 30.0 °C when dissolving in water. If the total pressure of gas (CO2 gas plus water vapor) over water is 1.00 atm, what is the concentration of CO2 in the water in grams per milliliter?Pressure of the water vapor at 30.0 °C = 31.8 mmHg. g/mL

Answer 1

Given:

Henry's law constant at 30 deg. celcius =4.48x10^-5 M/mmHg

Total pressure of the gases= 1 atm= 760 mmHg

Pressure of water vapor=31.8 mmHg

We will firstly determine the pressure of the carbon dioxide gas by using Dalton's Law of partial pressures:

`\begin{gathered} P_T=P_(CO_2)+P_(H_2O) \\ 760mmHg=P_(CO_2)+31.8mmHg \\ P_(CO_2)=760mmHg-31.8mmHg \\ P_(CO_2)=728.2\text{ }mmHg \end{gathered}`

By determining the pressure of the carbon dioxide gas we can now use the Henry's Law of Gas Solubility to detemine the molar concentration:

`\begin{gathered} C_(CO_2)=k_H* P_(CO_2) \\ C_(CO_2):molarity\text{ }of\text{ }gas \\ k_H:the\text{ }Henry\text{ }Law\text{ }Constant=4.48*10^(-5)MmmHg^(-1) \\ P_(CO_2):the\text{ }partial\text{ }pressure\text{ }of\text{ }the\text{ }gas=728.2mmHg \\ \\ C_(CO_2)=4.48*10^(-5)MmmHg^(-1)*728.2mmHg \\ C_(CO_2)=0.033M \end{gathered}`

Now thaat we have the molar concentration we will convert this mass concentration:

`\begin{gathered} mass\text{ }concentration=molar\text{ }mass* molar\text{ }concentration \\ mass\text{ }concentration=44.01gmol^(-1)*0.033molL^(-1) \\ mass\text{ }concentration=1.45gL^(-1) \end{gathered}`

We convert this mass concentration to g/mL:

`1.45(g)/(L)*(1L)/(1000mL)=1.45*10^(-3)gmL^(-1)`

Answer: The concentration of the CO2 is 1.45x10^-3 g/mL.