Question

Can Ba2+ be separated from Pb2+ by the addition of Na2SO4? Assume the starting solution is 0.10 M in both cations and use whatever concentration of sulfate you like. (Ksp of PbSO4 = 1.3 x 10-8), while that of( BaSO4 = 1.5 x 10-9). Calculate how much of the first cation to precipitate is left in solution when the second starts to fall out of solution. Is it less than 1% of the initial present?

Answer 1

Since both are 0.1 M to start with as per the individual Ksp Ba²⁺ must precipitate out first since its Ksp is lower

Ksp = [Ba²⁺][SO₄^2-]

1.5 x 10^-9 = 0.1 x [SO₄^2-]

[SO₄^2-] = 1.5 x 10^-9/0.1

[SO₄^2-] = 1.5 x 10^-8 M

so when 1.5 x 10^-8 M SO₄^2- is added Ba²⁺ starts to precipitate out

For Pb²⁺ to precipitate out the calculation will be

Ksp = [Pb²⁺][SO₄^2-]

1.3 x 10^-8 = 0.1 x [SO₄^2-]

[SO₄^2-] = 1.3 x 10^-8/0.1

[SO₄^2-] = 1.3 x 10^-7 M

so when 1.3 x 10^-7 M SO₄^2- is added Pb²⁺ starts to precipitate out

At this concentration the amount of Ba²⁺ in solution is

1.5 x 10^-9 = [Ba²⁺] x 1.3 x 10^-7

[Ba²⁺] = 1.5 x 10^-9/1.3 x 10^-7

[Ba²⁺] =  0.0115 M

So the Ba²⁺ in solution is 0.0115 M when Pb²⁺ starts to precipitate out, this is about 11.5% of the original.

It is not less than 1% of the initial.