Question

One millimole $O_2$ gas is dissolved in $5400$ ml water at ${25}^o$C. Calculate the Henry's law constant $K_H$ for this solution. $p_{O_2}=2\times {10}^{-8}$ bar.

Answer 1

According Henry's law,

$p_{O_2}=K_H.x_{O_2}$

$\Rightarrow 2\times {10}^{-8}bar=K_H\times \frac{n_{O_2}}{n_{H_{2}O}}$ [Since $n_{O_2}+n_{H_{2}O}\approx n_{H_{2}O}$]

$\Rightarrow 2\times {10}^{-8}bar=K_H\times \frac{{10}^{-3}}{\frac{5400}{18}\times 1}$

$\Rightarrow K_H=2\times {10}^{-8}\times 300\times {10}^3=600\times {10}^{5}bar$