Question

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

Answer 1

Answer: 282

(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, $\left({\left(\text{n}\right)}_{{\left(\text{H}\right)}_2\text{O}}\right)$
$=\frac{\left(\text{1000 g}\right)}{\left(18{\left(\text{g mol}\right)}^{-1}\right)}=\text{55.55}\text{mol}$
${\text{Mole fraction H}}_2\text{S}=\frac{{\text{n}}_{{\text{H}}_2\text{S}}}{{\text{n}}_{{\text{H}}_2\text{S}}+{\text{n}}_{{\text{H}}_2\text{O}}}$
$=\frac{\left(0\cdot 195\text{mol}\right)}{\left(0\cdot 195+55\cdot 55\right)\text{mol}}$
$=\frac{\left(0\cdot 195\text{mol}\right)}{\left(55\cdot 745\text{mol}\right)}=0\cdot 0035$
(b) Calculation of Henry's law constant
According to Henry's law
${\text{x}}_{{\text{H}}_2\text{S}}=\frac{{\text{Partial pressure of H}}_2\text{S}}{{\text{K}}_{\text{H}}\text{for}{\text{H}}_2\text{S}}\text{at STP}$
${\text{K}}_{\text{H}}{\text{for H}}_2\text{S}=\frac{{\text{Partial pressure of H}}_2\text{S}}{{\text{x}}_{{\text{H}}_2\text{S}}}$
$=\frac{\left(0\cdot 987\text{bar}\right)}{\left(0\cdot 0035\right)}=282\text{bar}$