Question

The Henry's law constant for O₂ in water at 25 °C is 1.25 × 10^-5 mol L^-1 kPa^-1, and the solubility is 1.5 × 10^-3 mol L^-1. What is the partial pressure of O₂ under these conditions?

Answer 1

Given : Henry's law constant for O₂ in water at 25 °C is (k)1.25 × 10^-5 mol L^-1 kPa^-1

solubility is (C_g) = 1.5 × 10^-3 mol L^-1

According to the Henry's law

For many gaseous solutes, the relation between solubility, C_g, and partial pressure, P_g, is a proportional one: C_g = k* P_g

where k is a proportionality constant that depends on the identities of the gaseous solute and solvent

C_g = k* P_g

1.5 × 10^-3 mol L^-1 = 1.25 × 10^-5 mol L^-1 kPa^-1 *  P_g

P_g = (1 .5 × 10^-3 mol L^-1 ) / (1.25 × 10^-5 mol L^-1 kPa^-1) = 120 kPa

The partial pressure of O₂ under these conditions = 120 kPa