what is the solublity of lead (II) iodide (PbI2) in a solution containing 0.050 M lead (II) nitrate ((Pb(NO3)2)? the ksp value for lead (II) iodide is 9.8x10^-9
Pb(NO3)2 here is Strong electrolyte
It will dissociate completely to give [Pb2+] = 0.050 M
At equilibrium:
PbI2 <===>
Pb2+ + 2
I-
0.050
+s
2s
Ksp = [Pb2+][I-]^2
9.8*10^-9 = (0.050 + s)*(2s)^2
Since Ksp is small, s can be ignored as compared to 0.050
Above expression thus becomes:
9.8*10^-9 = (0.050)*(2s)^2
9.8*10^-9 = 0.050 * 4(s)^2
s^2 = 9.8*10^-9/0.2
s = 2.21*10^-4 M
Answer: s = 2.21*10^-4 M
what is the solublity of lead (II) iodide (PbI2) in a solution containing 0.050 M lead...
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