Question
Question: At what partial pressure, oxygen will have solubility of \(2.6 \times {10^{ - 4}}\) \(mol{\text{ d}}...
At what partial pressure, oxygen will have solubility of 2.6×10−4 mol dm−3 in water of 25∘C ? Henry constant for O2 in water at 25∘C is 1.3×10−3 mol dm−3 atm−1
Solution
Henry’s law constant is a proportionality constant for the relationship between the amount of gas that is dissolved in a liquid and the partial pressure of the gas above the liquid when the temperature is kept constant. Moreover, according to Henry’s law at constant temperature, the amount of gas dissolved in a liquid is directly proportional to the partial pressure of the gas above the liquid. Henry’s law constant is inversely proportional to the solubility of the gas in solution.
Complete answer:
Step 1: Identifying the values given in the question:
So, according to the question, Solubility of the oxygen at 25∘C is 2.6×10−4mol dm−3 and Henry’s constant for oxygen at 25∘C is 1.3 \times {10^{ - 3}}$$$$mol{\text{ }}d{m^{ - 3}}at{m^{ - 1}}.
Step 2: Substituting the values in the mathematical formula of the Henry’s law:
Mathematically, Henry’s law is given by the equation:
p=KH.S
Where p is the partial pressure of the gas in the atmosphere
S is the solubility of the dissolved gas
KH is the Henry’s constant
Now, p=2.6×10−4×1.3×10−3
p=3.38×10−7atm
Therefore, at 3.38×10−7 atm partial pressure, oxygen will have a solubility 2.6×10−4mol dm−3 in water at 25∘C.
Note:
The nature of gas, nature of solvent, temperature, and pressure are the factors on which the value of Henry’s law constant is dependent. According to Henry's law as there is an increase in temperature the value of Henry’s law constantly increases and the solubility of the gas in liquid decreases. The solubility of carbon dioxide in an unopened carbonated drink is the result of Henry’s law.