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

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:

$C_{CO_2}=K_H\times p_{CO_2}$       ......(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:

$760mmHg=K_H\times 0.161g/100mL\\\\K_H=\frac{760mmHg}{0.161g/100mL}=4720.5g.mmHg/100mL$

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:

$p_{CO_2}=4720.5g.mmHg/100mL\times 0.886g/100mL=4182.4mmHg$

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