4) Hydrogen sulfide H2S is a volatile acid that equilibrates, in closed systems,

Question

4) Hydrogen sulfide H2S is a volatile acid that equilibrates, in closed systems, with aqueous solutions according to the reaction: For a gaseous phase containing H2Sg in equilibrium with water, in a closed vessel, calculate a) The Henry's law constant, KH, for H2S. Note that KH Keq for the two phase b) The partial pressure of H2S) overlying the water if the total soluble sulfide partitioning reaction given above. concentration in water is Cs-[H2Sa HS+IS], is 1x103 M and the solution pH is 8.5. The dissociation reactions of HzSae) in the water phase are described by K.I K,.2

Explanation / Answer

a) According to Henry's law, at a given temperature the concentration of dissolved gas is directly proportional to the partial pressure of the gas( at equilibrium between liquid and gas phase)

[H2S]aq=C(H2S)=KH*P(H2S)

or,KH=[H2S]aq/P(H2S)

Also ,keq=equilibrium constant=[H2S]aq/[H2S]g

KH=Keq=[H2S]aq/P(H2S)=[H2S]aq/[H2S]g   {[ ] denotes concentration)

2)CT,S=[H2S]aq +[HS-]+[S2-]

pH=-log [H+]=8.5

[H+]=10^-8.5=3.162*10^-9

ka1(H2S)=9.1*10^-8=[SH-][H+]/[H2S]aq

ka2(H2S)=1.3*10^-13M=[S2-][H+]''/[HS-]

[H]''=3.162*10^-9 (as calculated above)

[S2-]=[H]''=3.162*10^-9M

So,[HS-]=[S2-][H+]''/ka2=(3.162*10^-9)^2/(1.3*10^-13M)=7.691*10^-5M

now you have ,[H+]''=[S2-]=3.162*10^-9M

and [HS-]=7.691*10^-5M

CT,S=[H2S]aq +[HS-]+[S2-]=[H2S]aq+(7.691*10^-5M)+(3.162*10^-9M)=1*10^-3M

or,[H2S]aq+(7.691*10^-5M)+(3.162*10^-9M)=1*10^-3M

or,[H2S]aq=1*10^-3M (aprrox)

P(H2S)=[H2S]aq/Keq =[H2S]aq/([H2S]aq/[H2S]g)    [KH=Keq]

or P(H2S)=[H2S]g)

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