Question

At a particular temperature, the solubility of O₂ in water is 0.0500 M when the partial pressure is 0.120 atm. What will the solubility be when the partial pressure of O₂ is 2.470 atm?

Answer 1

Answer:

At a partial pressure of 2.470 atm, solubility of $O_{2}$ is 1.03 M

Explanation:

The given problem can be solved by using Henry's law

According to Henry's law at a particular temperature for a particular gas and solvent:                        $\frac{S_{1}}{S_{2}}=\frac{P_{1}}{P_{2}}$

where, $S_{1}$ and $S_{2}$ are solubility of the gas at a partial pressure of $P_{1}$ and $P_{2}$ respectively.

Here, $S_{1}=0.0500M$, $P_{1}=0.120atm$ and $P_{2}=2.470atm$

So, $S_{2}=\frac{S_{1}P_{2}}{P_{1}}=\frac{(0.0500M)\times (2.470atm)}{0.120atm}=1.03M$

Hence, at a partial pressure of 2.470 atm, solubility of $O_{2}$ is 1.03 M