`H_(2)S`, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of `H_(2)S` I water at STP is 0.195m, calculate Henry's law constant.

Answer

`K_(H)=282 ba r`

Explanation

Using the formula `:`
`m=(CHMi_(2)xx1000)/(CHMi_(1)xxMw_(1))=(CHMi_(2)xx1000)/((1-CHMi_(2))xxMw_(1))`
`0.195m=(CHMi_(2)xx1000)/((1-CHMi_(2))xx18g mol^(-1))`
Solve for `CHMi_(2):`
`0.195xx18(1-CHMi_(2))=1000CHMi_(2)`
`3.51-3.51CHMi_(2)=1000CHMi_(2)`
`CHMi_(2)=(3.51)/(1003.51)=0.00349~~0.0035`
At STP, pressure`=1 atm = 0.987 ` bar
Applying Henry's law,
`p_(H_(2)S)=K_(H)xxCHMi_(2(H_(2S)))`
`(or)K_(H)=(P_(H_(2)S))/(0.0035)=282 ba r`