1)
At equilibrium:
Pb(IO3)2 <---->
Pb2+
+ 2
IO3-
s
2s
Ksp = [Pb2+][IO3-]^2
2.6*10^-13=(s)*(2s)^2
2.6*10^-13= 4(s)^3
s = 4.021*10^-5 M
Answer: 4.0*10^-5 M
2)
NaIO3 here is Strong electrolyte
It will dissociate completely to give [IO3-] = 0.5 M
At equilibrium:
Pb(IO3)2 <---->
Pb2+
+ 2
IO3-
s
0.5 + 2s
Ksp = [Pb2+][IO3-]^2
2.6*10^-13=(s)*(0.5+ 2s)^2
Since Ksp is small, s can be ignored as compared to 0.5
Above expression thus becomes:
2.6*10^-13=(s)*(0.5)^2
2.6*10^-13= (s) * 0.25
s = 1.04*10^-12 M
Answer: 1.0*10^-12 M
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