Question

# Can Ba2+ be separated from Pb2+ by the addition of Na2SO4? Assume the starting solution is...

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?

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+][SO42-]

1.5 x 10-9 = 0.1 x [SO42-]

[SO42-] = 1.5 x 10-9/0.1

[SO42-] = 1.5 x 10-8 M

so when 1.5 x 10-8 M SO42- is added Ba2+ starts to precipitate out

For Pb2+ to precipitate out the calculation will be

Ksp = [Pb2+][SO42-]

1.3 x 10-8 = 0.1 x [SO42-]

[SO42-] = 1.3 x 10-8/0.1

[SO42-] = 1.3 x 10-7 M

so when 1.3 x 10-7 M SO42- 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.

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