1.) Calculate the solubility of lead(II) iodate, Pb(IO3)2 in pure water and in a 0.50M solution of sodium iodate. The Ksp = 2.6 × 10‑13.
Ksp = 2.6 x10^-13
a) In pure water
Pb(IO3)2(s) ---------------------------- Pb+2(aq) + 2 IO3-(aq)
S mol/L 2S mol/L where S= molar solubility
Ksp = [Pb+2] [IO3-]^2
Ksp = S x (2S)^2
Ksp = 4S^3
Ksp/4 = S^3
S^3 = Ksp / 4 = 2.6 x10^-13/4 = 0.65 x 10^-13
S = 4.02 x10^-5 mol/L
Solubility of Lead Iodate = S = 4.02 x10^-5 M
b) in 0.50M solution of Sodium Iodate ( NaIO3)
NaIO3 ------------------- Na+ + IO3-
0.50M 0.50M
Pb(IO3)2(s) -------------------------- Pb+2(aq) + 2 IO3-(aq)
S mol/L ( 2S + 0.50)mol/L
Ksp = [Pb+2][IO3-]^2
Ksp = S x (2S + 0.50)^2
Ksp = S x [ 4S^2 + 0.25 + 2S]
Ksp = 4S^3 + 0.25S + 2S^2
Ksp = 0.25S where 4S^3 and 2S^2 is neglected
2.6 x10^-13 = 0.25 S
S= 2.6 x10^-13 / 0.25 = 10.4 x10^-13
S = 1.04 x10^-12 M
Solubility of Lead iodate = 1.04 x 10^-12M
1.) Calculate the solubility of lead(II) iodate, Pb(IO3)2 in pure water and in a 0.50M solution...
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