Question

Calculate Henry's law constant for H₂S, whose solubility in water at STP is assumed to be 0.195 m.

Answer 1

(a) Calculation of mole fraction of H₂S 0.195 m means that 0.195 mole of H₂S dissolves in 1000 g of water.
Number of moles of water in 1000 g, $\left(\left(\text{n}\right)_{\left(\text{H}\right)_{2} \text{O}}\right)$
$=\frac{\left(\text{1000 g}\right)}{\left(1 8 \left(\text{g mol}\right)^{- 1}\right)}=\text{55.55}\text{mol}$
$\text{Mole fraction H}_{2} \text{S} = \frac{\text{n}_{\text{H}_{2} \text{S}}}{\text{n}_{\text{H}_{2} \text{S}} + \text{n}_{\text{H}_{2} \text{O}}}$
$= \frac{\left(0 \cdot 1 9 5 \text{mol}\right)}{\left(0 \cdot 1 9 5 + 5 5 \cdot 5 5\right) \text{mol}}$
$= \frac{\left(0 \cdot 1 9 5 \text{mol}\right)}{\left(5 5 \cdot 7 4 5 \text{mol}\right)} = 0 \cdot 0 0 3 5$
(b) Calculation of Henry's law constant
According to Henry's law
$\text{x}_{\text{H}_{2} \text{S}} = \frac{\text{Partial pressure of H}_{2} \text{S}}{\text{K}_{\text{H}} \text{for} \text{H}_{2} \text{S}} \text{at STP}$
$\text{K}_{\text{H}} \text{for H}_{2} \text{S} = \frac{\text{Partial pressure of H}_{2} \text{S}}{\text{x}_{\text{H}_{2} \text{S}}}$
$= \frac{\left(0 \cdot 9 8 7 \text{bar}\right)}{\left(0 \cdot 0 0 3 5\right)} = 2 8 2 \text{bar}$