$H_{2} S$ is a toxic gas used in qualitative analysis. If solubility of $H_{2} S$ in water at STP is 0.195 m, what is the value of $K_{H}$ ?

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192 bar

289.4 bar

0.0263 bar

69.16 bar

answer is D.

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Detailed Solution

$P = K_{H} X no . of moles of H_{2} S = 0.195 mol no . of moles of H_{2} O = \frac{1000}{18} = 55.55 X_{H_{2} S} = \frac{0.195}{0.195 + 55.55} = 0.0035 X = \frac{P}{X} = \frac{0.987}{0.0035} = 282 bar$

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