The Henry's law constant for O₂ in water at 25 °C is 1.26 × 10^-3 M/atm.What partial pressure of O₂ is necessary to achieve an equilibrium concentration of 1.5 × 10^-3 M O₂?

Question

Quizplus · Accepted Answer

Using Henry's Law, $ C = kP $, where $ C $ is the concentration, $ k $ is the Henry's law constant, and $ P $ is the partial pressure. Solving for $ P $, $ P = \frac{C}{k} = \frac{1.5 \times 10^{-3} \, \text{M}}{1.26 \times 10^{-3} \, \text{M/atm}} = 1.2 \, \text{atm} $.

The Henry's Law Constant for O2 in Water at 25

The Henry's Law Constant for O₂ in Water at 25