Question

The `[H^(+)]` of 0.10 `MH_(2)S` solution is (given `K_(1)=1.0 xx 10^(-7) , K_(2)=1.0 xx 10^(-14))`
A. `1.0 xx 10^(-7)M`
B. `1.0 xx 10^(-4)M`
C. `1.0 x 10^(-8)M`
D. `1.0 xx 10^(-22)M`

Answer 1

Correct Answer - B
As `K_(2) lt lt lt lt K_(1)` most of the `H^(+)` ions results from primary ionisation
`H_(2)S hArr H^(+)+HS^(-)`
Let `x=[H^(+)]=[HS^(-)]`
`[H_(2)S]=1.0 -x=1.0 mol L^(-1)`
( Since x is very small )
`K_(1)=([H^(+)][HS^(-)])/([H_(2)S])`
or `1.0 xx 10^(-7)=(x^(2))/(0.10)=1.0 xx 10^(-4)M`
`[H^(+)]=x mol L^(-1)`
`x=1.0 xx 10^(-4)`