Final answer:
The concentration of oxygen in a fresh water stream in equilibrium with air at 25°C and 1.0 atm is 1.27 × 10-³ M.
Step-by-step explanation:
Henry's law relates the solubility of a gas in a liquid to its partial pressure. According to Henry's law, the concentration of oxygen in water can be calculated using the formula:
[O2] = k * P
Where [O2] is the concentration of oxygen in the water, k is the Henry's law constant, and P is the partial pressure of oxygen.
Given that the pressure of oxygen in the atmosphere is 1.0 atm, we need to use the Henry's law constant to find the solubility of oxygen in water at 25°C.
The Henry's law constant for O2 at 25°C is 1.27 × 10-³ M/atm.
[O2] = (1.27 × 10-³ M/atm) * (1.0 atm)
= 1.27 × 10-³ M
Therefore, the concentration of oxygen in a fresh water stream in equilibrium with air at 25°C and 1.0 atm is 1.27 × 10-³ M.