Question

Consider a solution containing 0.181 M lead ions and 0.365

Miron(II) ions. The Ksp for lead sulfide is 3.4 x 10^ -28 and
thatfor iron(II) sulfide is 3.7 x 10^ -19. Calculate the
[H3O+]required to reduce the concentration of lead ions to 1.0 x
10^-6 Musing H2S.

Answer 1

`[H_3O^+]=6.8*10^(-22)`

Step-by-step explanation:

For a solution with
`[Pb^(+2)]=1*10^(-6)` :

`Kps=[Pb^(+2)]*[S^(-2)]=3.4*10^(-28)`

`1*10^(-6)*[S^(-2)]=3.4*10^(-28)`

`[S^(-2)]=3.4*10^(-22)`

For the iron ions:

`Kps=[0.365]*[S^(-2)]=3.7*10^(-19)`

`[S^(-2)]=1.01*10^(-18)`

Given that the
`[S^(-2)]` concentration required to start precipitating iron is higher than the final concentration for the Pb precipitation (when the desired concentration is reached), the iron will not precipitate.

The concentration of
`H_3O^+`

`H_2S + 2 H_2O \longrightarrow 2 H_3O^+ + S^(-2)`

`[H_3O^+]=2*[S^(-2)]=6.8*10^(-22)`