Solid PbI2 was added to a 0.030 M NaI solution. Calculate the molar concentration of lead ion in this solution. Ksp = 7.9 x 10-9 (at 25ºC) (Please show work)
PbI2(s) <=> Pb2+ + 2I-
Answer
8.77×10-6M
Explanation
PbI2(s) <------> Pb2+(aq) + 2I-(aq)
Ksp = [Pb2+] [ I-]2 = 7.9×10-9
Initial concentration
[Pb2+] = 0
[I-] = 0.030
change in concentration
[Pb2+ ] = + x
[I-] = +2x
Equillibrium concentration
[Pb2+] = x
[I-] = 0.030 + 2x
so,
x (0.030 + 2x)2 = 7.9 ×10-9
solving for x
x = 8.77×10-6
Therefore
molar concentration of Pb2+ = 8.77×10-6M
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