Solubility Euilibria calculation

Question

Q1 )
Calculate the final [Pb2+] in solution after mixing 20 drops of 0.3 M Pb(NO3)2 with 30 drops of 0.4 M FeCl3
Ksp for PbCl2 = 1.7 *10-5

Answer : 7.4*10-5 mole dm-3

Q2 )
Starting with saturated solution of BaSO4, CaSO4 and PbSO4
which pairs will give a ppt on mixing equal volumes?
Ksp for BaSO4 = 1.0*10-10
Ksp for CaSO4 = 6.0*10-5
Ksp for PbSO4 = 1.0*10-8

STEP PLZZ =]

Answer 1

Q1)
The solutions are diluted in mixing. After mixing :
[Pb(NO3)2]o = 0.3 x (20/50) = 0.12 M
Hence, [Pb2+]o = 0.12 M
[FeCl3]o = 0.4 x (30/50) = 0.24 M
Hence, [Cl-]o = 0.24 x 3 = 0.72 M

The formation of PbCl2 solution :
Pb2+(aq) + 2Cl-(aq) = PbCl2(s)
Mole ratio Pb2+ : Cl- = 1 : 2
Ksp = [Pb2+][Cl-]2 = 1.7 x 10-5 M3

Cl- is in excess.
Pb2+ ions are almost completely reacted.
At eqm :
[Cl-] = 0.72 - 0.12x2 = 0.48 M
[Pb2+] = Ksp/[Cl-]2 = (1.7 x 10-5)/(0.48)2 = 7.4 x 10-5 M

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Q.2)
In the saturated solution of BaSO4 :
BaSO4(s) = Ba2+(aq) + SO42-(aq)
Ksp(BaSO4) = [Ba2+][SO42-] = 1.0 x 10-10 M2
Hence , [Ba2+] = [SO42-] = 1.0 x 10-5 M

In the saturated solution of CaSO4 :
CaSO4(s) = Ca2+(aq) + SO42-(aq)
Ksp(CaCO3) = [Ca2+][SO42-] = 6.0 x 10-5 M2
Hence , [Ca2+] = [SO42-] = 7.7 x 10-3 M

In the saturated solution of PbSO4 :
PbSO4(s) = Pb2+(aq) + SO42-(aq)
Ksp(PbSO4) = [Pb2+][SO42-] = 1.0 x 10-8 M2
Hence , [Pb2+] = [SO42-] = 1.0 x 10-4 M

The concentrations will be reduced to half when mixing equal volumes of solutions.

Case 1 : Mixing equal volumes of BaSO4 and CaSO4
[Ba2+] = (1.0 x 10-5)/2 = 5.0 x 10-6 M
[Ca2+] = (7.7 x 10-3)/2 = 3.9 x 10-3 M
[SO42-] = (1.0 x 10-5 + 7.7 x 10-3)/2 = 3.9 x 10-3 M
[Ba2+][SO42-] = (5.0 x 10-6) x (3.9 x 10-3) = 2.0 x 10-8 M2 > Ksp(BaSO4)
BaSO4 precipitate is formed.
[Ca2+][SO42-] = (3.9 x 10-3) x (3.9 x 10-3) = 1.5 x 10-5 M2 < Ksp(CaSO4)
There is no precipitate of CaSO4.

Case 2 : Mixing equal volumes of BaSO4 and PbSO4
[Ba2+] = (1.0 x 10-5)/2 = 5.0 x 10-6 M
[Pb2+] = (1.0 x 10-4)/2 = 5.0 x 10-5 M
[SO42-] = (1.0 x 10-5 + 1.0 x 10-4)/2 = 5.5 x 10-5 M
[Ba2+][SO42-] = (5.0 x 10-6) x (5.5 x 10-5) = 2.8 x 10-10 M2 > Ksp(BaSO4)
BaSO4 precipitate is formed.
[Pb2+][SO42-] = (5.0 x 10-5) x (5.5 x 10-5) = 2.8 x 10-9 M2 < Ksp(PbSO4)
There is no precipitate of PbSO4.

Case 3 : Mixing equal volumes of CaSO4 and PbSO4
[Ca2+] = (7.7 x 10-3)/2 = 3.9 x 10-3 M
[Pb2+] = (1.0 x 10-4)/2 = 5 x 10-5 M
[SO42-] = (7.7 x 10-3 + 1.0 x 10-4)/2 = 3.9 x 10-3 M
[Ca2+][SO42-] = (3.9 x 10-3) x (3.9 x 10-3) = 1.5 x 10-5 M2 < Ksp(CaSO4)
There is no precipitate of CaSO4.
[Pb2+][SO42-] = (5 x 10-5) x (3.9 x 10-3) = 2.0 x 10-7 M2 > Ksp(PbSO4)
PbSO4 precipitate is formed.