Question

At 25 °C and 775 Torr, carbon dioxide has a solubility of 0.0347 M in water. What is its solubility at 25 °C and 1470 Torr?

Answer 1

Answer : The solubility at
`25^oC` is, 0.0658 M

Explanation :

According to the Henry's Law, the solubility of a gas in a liquid is directly proportional to the pressure of the gas.

`S\propto P`

or,

`(S_1)/(S_2)=(P_1)/(P_2)`

where,

`S_1` = initial solubility of carbon dioxide gas = 0.0347 M

`S_2` = final solubility of carbon dioxide gas = ?

`P_1` = initial pressure of carbon dioxide gas = 775 torr

`P_2` = final pressure of carbon dioxide gas = 1470 torr

Now put all the given values in the above formula, we get the final solubility of the carbon dioxide gas.

`(0.0347M)/(S_2)=\frac{775\text{ torr}}{1470\text{ torr}}`

`S_2=0.0658M`

Therefore, the solubility at
`25^oC` is, 0.0658 M