Question

The solubility products of PbSO4 and SrSO4 are 6.3×10−7 and 3.2×10−7, respectively. Part A What are...

The solubility products of PbSO4 and SrSO4 are 6.3×10−7 and 3.2×10−7, respectively.

Part A What are the values of [SO2−4], [Pb2+], and [Sr2+] in a solution at equilibrium with both substances?

[SO2−4], [Pb2+], [Sr2+]= ?

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Answer #1

If both are in equilibrium

[SO4-2] = same for both

Ksp1 = [Pb2+][SO4-2]

[SO4-2] = Ksp/Pb+2

Ksp2 = [Sr+2][SO4-2]

[SO4-2] = Ksp2 / Sr+2

Ksp1/Pb+2 =  Ksp2 / Sr+2

(6.3*10^-7) / ([Pb+2]) = (3.2*10^-7)/[Sr+2]

[Sr+2]/[Pb+2] = (3.2*10^-7)/(6.3*10^-7) = 0.5079

[Sr+2] = 0.5079*[Pb+2]

there are infinite amount of equilbiriums between Sr2+ and Pb2+, since both depend on each other

note that the system msut be "fixed" tha tis, we will need more data, volume, or concentrations to set [Sr+2] and [PB2+]

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