20.0 mLs of 0.10 M NaI and 20.0 mLs of 0.10 M Pb(NO3)2 are mixed together. Calculate the final molar concentrations of Pb2 and I–. Ksp(PbI2) = 7.9x10–9.
20.0 mL of 0.10 M NaI and 20.0 mL of 0.10 M Pb(NO3)2 are mixed together.
0.02 L*0.10 M of NaI = 0.002 mol I- per 0.04 L
0.02 L*0.10 M of Pb(NO3)2 = 0.002 mol Pb^2+ per 0.04 L
Initial : PbI2 = Pb^2+ + 2I-
CONC.N 0.002---0.002 M
Q= [Pb^2+][I-]^2
Q= (0.002)(0.002)^2 = 8.0*10^-9
Ksp(PbI2) = 7.9x10–9. since Q = Ksp solution is saturated
Thus concentrations of Pb^2+ and I- are 0.002 M
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