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Example · Example 3

Q.At 298 K298\ \text{K}, the Henry's law constant for oxygen gas dissolved in water is KH=4.259×104 barK_H = 4.259 \times 10^4\ \text{bar}. Calculate the mole fraction of dissolved O2\text{O}_2 in water that is in equilibrium with air, in which the partial pressure of O2\text{O}_2 is 0.20 bar0.20\ \text{bar}.

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By Henry's law, x=p/KH=0.20 bar/(4.259×104 bar)=0.20/42590≈4.696×10−6x = p/K_H = 0.20\ \text{bar} / (4.259 \times 10^4\ \text{bar}) = 0.20/42590 \approx 4.696 \times 10^{-6}. [!ANSWER] The mole fraction of dissolved O2\text{O}_2 in water in equilibrium with air is x≈4.70×10−6x \approx 4.70 \times 10^{-6} — an extremely small, but life-sustaining, quantity of dissolved oxygen.

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