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Question: \[{{\text{H}}_2}{\text{S}}\] is a toxic gas used in qualitative analysis, if solubility of \[{{\text...

H2S{{\text{H}}_2}{\text{S}} is a toxic gas used in qualitative analysis, if solubility of H2S{{\text{H}}_2}{\text{S}} in water at STP is 0.19m, What is the value of KH{K_H}?
A ) 0.0263 bar
B ) 69.16 bar
C ) 192 bar
D ) 282 bar

Explanation

Solution

Molality is the number of moles of solute (hydrogen sulphide) present in one kilogram of the solvent (water).
The expression for the Henry’s law constant is given below KH=PX{K_H} = \dfrac{P}{X}

Complete answer:
To solve this question we are using Henry’s law which states that:
According to Henry’s law, at constant temperature, the amount of the gas that dissolves in the given volume of solvent, is directly proportional to the partial pressure of the gas.
The molality of hydrogen sulphide in water is 0.195m. 1 kg (or 1000 g ) of water contains 0.195 moles of hydrogen sulphide. The molecular weight of water is 18 g/mol. Calculate the number of moles of water.
Number of moles = Given  massMolar  mass{\dfrac{Given\; mass}{Molar\; mass}}
= 1000 g18 g/mol\dfrac{{1000{\text{ g}}}}{{18{\text{ g/mol}}}} = 55.56 mol
Now, calculate the mole fraction of hydrogen sulphide,
Mole fraction is defined as the moles of a particular compound divided by the total moles of solution.
So, Mole fraction of Hydrogen sulphide in the mixture is given as,
= 0.195 mol55.56 mol + 0.195 mol=0.0035\dfrac{{0.195{\text{ mol}}}}{{55.56{\text{ mol + 0}}{\text{.195 mol}}}} = 0.0035
Under conditions of standard temperature and pressure (STP), the pressure is 0.987 bar.
According to Henry’s law,

{K_H} = \dfrac{P}{X} \\\\$$ Substitute values in the above expression and calculate pressure

{K_H} = \dfrac{P}{X} \\
= \dfrac{{0.987{\text{ bar}}}}{{0.0035}} \\
= 282{\text{ bar}} \\

Thus,thevalueoftheHenry’slawconstantis282bar.∗∗Hence,thecorrectoptionistheoptionD).∗∗∗∗Note:∗∗Sinceinthegivenoptionstheunitofhenry’slawconstantisbar,whichistheunitofpressure,theconcentrationusedisexpressedintermsofmolefraction.Donotdirectlyusethemolalityasconcentrationintheexpressionforthehenry’slawconstant.Firstconvertmolalityintomolefraction.Otherwisethewronganswerwillbeobtained.Thus, the value of the Henry’s law constant is 282 bar. _**Hence, the correct option is the option D ).**_ **Note:** Since in the given options the unit of henry’s law constant is bar, which is the unit of pressure, the concentration used is expressed in terms of mole fraction. Do not directly use the molality as concentration in the expression for the henry’s law constant. First convert molality into mole fraction. Otherwise the wrong answer will be obtained.