Question

The Henry's law constant for O_{2} 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 O_{2} under these conditions?

Answer #1

**Given : Henry's law constant
for O _{2} 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
O _{2} under these conditions = 120 kPa**

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