Question

The solubility of O2 (g) in water at 0 oC = 14.74 mg/L. The solubility decreases to 7.03 mg/L at 35 oC.

Calculate kH (Henry’s Law Constant) for water at these temperatures.

partial pressure=0.21

Answer 1

Given

At 0^oC solubility of O_2(g) is 14.74 mg/L & molar mass of O_2(g) = 15.99 g/mol

Partial pressure=0.21 atm (unit not given taken as atm)

We have 14.74 mg O_2(g) in 1 L water ( i.e. 14.74 mg /1000 = 0.01474 g )

Calculate moles of O_{2 (g)} in 1 L water = 0.01474 g / 15.99 g/mol =0.000921 mol

moles of O_2(g) in 1 L water = 9.21 x 10^-4 M

Know we know

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

P = K_H X

0.21 = K_H x 9.21 x 10^-4 M

K_H = 0.21atm / 9.21 x 10^-4 mol/L

K_H = 228 atm L mol^-1 (is henry law constant At 0^oC solubility of O_2(g) is 14.74 mg/L in water)

Given

The solubility decreases to 7.03 mg/L at 35 oC.

At 35^oC solubility of O_2(g) is 7.03 mg/L & molar mass of O_2(g) = 15.99 g/mol

Partial pressure=0.21 atm (unit not given taken as atm)

We have 7.03 mg O_2(g) in 1 L water ( i.e. 7.03 mg /1000 = 0.00703 g )

Calculate moles of O_{2 (g)} in 1 L water = 0.00703 g / 15.99 g/mol =0.000439 mol

moles of O_2(g) in 1 L water = 4.39 x 10^-4 M

Know we know

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

P = K_H X

0.21 atm = K_H x 4.39 x 10^-4 M

K_H = 0.21atm / 4.39 x 10^-4 mol/L

K_H = 477.95 atm L mol^-1 (is henry law constant At 35^oC solubility of O_2(g) is 7.03 mg/L in water)

KH​ depends only on nature of gas, nature of liquid and temperature (T). As temperature (T). increases, henrylaw constant (KH​) increases and solubility (X) decreases .