Question

The solubility of CO2 in water is 0.161 g/100 mL at 20oC and a partial pressure of CO2 of 760 mmHg. What partial pressure of CO2 is necessary in a soft drink canning process in order to allow the solubility of CO2 to equal 0.886 g/100 mL?

Answer 1

Answer: The partial pressure of carbon dioxide having solubility 0.886g/100mL is 4182.4 mmHg

Step-by-step explanation:

Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.

The equation given by Henry's law is:

`C_(CO_2)=K_H* p_(CO_2)` ......(1)

where,

`C_{CO_2` = solubility of carbon dioxide in water = 0.161 g/100 mL

`K_H` = Henry's constant = ?

`p_(CO_2)` = partial pressure of carbon dioxide = 760 mmHg

Putting values in equation 1, we get:

`760mmHg=K_H* 0.161g/100mL\\\\K_H=(760mmHg)/(0.161g/100mL)=4720.5g.mmHg/100mL`

Now, calculating the pressure of carbon dioxide using equation 1, we get:

`C_{CO_2` = solubility of carbon dioxide in water = 0.886 g/100 mL

`K_H` = Henry's constant = 4720.5 g.mmHg/100 mL

`p_(CO_2)` = partial pressure of carbon dioxide = ?

Putting values in equation 1, we get:

`p_(CO_2)=4720.5g.mmHg/100mL* 0.886g/100mL=4182.4mmHg`

Hence, the partial pressure of carbon dioxide having solubility 0.886g/100mL is 4182.4 mmHg