Question

Calculate the solubility of hydrogen in water at 25 degree Celsius if its partial pressure above the solution is 1 bar. Given that Henry's constant for hydrogen at 25 degree Celsius is 71.18 K bar.

Answer 1

Step 1:

As we know, $P_{gas}=(K_H\times x_{gas})$

1 $=(71.18\times {10}^3)\times x_{gas}$

$x_{gas}=\frac{1}{71.18\times {10}^3}=1.404\times {10}^{-5}$

$\frac{n_{gas}}{n_{gas}+n_{water}}=1.404\times {10}^{-5}$

Step 2:

$n_{gas}\ll n_{water}$

$\therefore n_{gas}+n_{water}\approx n_{water}$

$n_{water}=\frac{1000}{18}$

$n_{gas}=1.404\times {10}^{-5}\times \frac{1000}{18}$

$n_{gas}=7.8\times {10}^{-4}mol{L}^{-1}$

Hence, the solubility of hydrogen gas in water is $n_{gas}=7.8\times {10}^{-4}mol{L}^{-1}$