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Question
H2S, a toxic gas with rotten egg like smell, is used for the quantitative analysis. If the solubility of H2S in water at STP is 0.2 m, then Henry’s law constant for H2S in water at 273 K is
Options
A.3.6 × 108 Pa
B.5.0 × 108 Pa
C.5.0 × 105 Pa
D.2.78 × 107 Pa
Solution
$P = K_{H}.X \Rightarrow 1\text{ bar} = K_{H} \times \frac{0.2}{0. + \frac{1000}{18}}$
$\therefore K_{H} = 277.78\text{ bar} \simeq \text{2.78} \times \text{1}\text{0}^{7}\text{ Pa}$
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