Oxygen gas makes up 21 % of the atmosphere by volume. What is the solubility of O2(g) in water at 25°C if the atmospheric pressure is 741 mmHg? The Henry's law constant for oxygen gas at 25°C is 1.3 x 10–3 mol/L·atm.
Question
Oxygen gas makes up 21 % of the atmosphere by volume. What is the solubility of O2(g) in water at 25°C if the atmospheric pressure is 741 mmHg? The Henry's law constant for oxygen gas at 25°C is 1.3 x 10–3 mol/L·atm.
Solution
To solve this problem, we need to use Henry's Law, which states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. The formula for Henry's Law is:
S = kH * P
where: S is the solubility of the gas, kH is the Henry's Law constant, and P is the partial pressure of the gas.
Step 1: Calculate the partial pressure of oxygen. The partial pressure of a gas is the pressure that the gas would have if it alone occupied the volume. Since oxygen makes up 21% of the atmosphere, the partial pressure of oxygen is 21% of the total atmospheric pressure.
P(O2) = 0.21 * 741 mmHg = 155.61 mmHg
Step 2: Convert the pressure from mmHg to atm. 1 atm = 760 mmHg, so
P(O2) = 155.61 mmHg * (1 atm / 760 mmHg) = 0.20475 atm
Step 3: Use Henry's Law to calculate the solubility of oxygen. Now we can use Henry's Law to calculate the solubility of oxygen in water at 25°C.
S(O2) = kH * P(O2) = 1.3 x 10–3 mol/L·atm * 0.20475 atm = 0.000266175 mol/L
So, the solubility of O2(g) in water at 25°C and an atmospheric pressure of 741 mmHg is approximately 0.000266175 mol/L.
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