A lead salts is dissolved in `HC1` which si `95%` ionised. It is found to have `0.1M Pb^(2+)` and `0.28M H^(o+)` ions. The solution is satured with `H_(2)S(g)`. Calculate the amount of `Pb^(2+)` ions that remains unprecipitated.
`K_(sp)` of `PbS = 4 xx 10^(-29)`,
`K_(sp)` of `H_(2)S = 1.1 xx 10^(-22)`

Answer

A, B

Explanation

`[H^(o+)] = (0.28xx94)/(100) = 0.2632M`
`H_(2)S hArr 2H^(o+) + S(^2-)`
`K_(sp) = [H^(o+)]^(2) [S^(2-)]`
`:. [S^(2-)] = (K_(sp))/([H^(o+)]^(2)) = (1.1 xx 10^(-22))/(0.2632 xx 0.2632) = 1.59 xx 10^(-21)M`
Calculation of `[Pb^(2+)]` ion:
`PbS hArr Pb^(2+) S^(2-)`
`K_(sp) = [Pb^(2+)][S^(2-)]`
`[Pb^(2+)] = (K_(sp))/([S^(2-)]) = (4 xx 10^(-29))/(1.59 xx 10^(-21)) = 2.516 xx 10^(-8)M`