Question

Solubility products for a series of iodides are:
AgI Ksp = 8.3 x 10-17
PbI2 Ksp = 7.1 x 10-9
BiI3 Ksp = 8.1 x 10-19
List these three compounds in order of decreasing (from highest to lowest) molar
solubility in:
a. Water
b. 0.10 M NaI

Answer 1

The molar solubility of the compound, s, represents the number of moles of the substance that will dissolve in aqueous solution at a particular temperature.

For AgI s=[Ag⁺] and K_sp=[Ag⁺][I^-]=s*s=s², then s= $\sqrt{Ksp}$ =$\sqrt{8.3\cdot10^{-17} }$ =$9.3\cdot10^{-9}$

for PbI₂ s=[Pb²⁺] and K_sp=[Pb²⁺][I^-]²=s*(2s)²=4s³, then s=$\sqrt[3]{Ksp\over4}=\sqrt[3]{7.1\cdot10^{-9}\over{4}}=1.2\cdot10^{-3}$ ,

for BiI₃ s=[Bi³⁺] and K_sp=[Bi³⁺][I^-]³=s*(3s)³=27s⁴, then s=$\sqrt[4]{Ksp\over27}=\sqrt[4]{8.1\cdot10^{-19}\over{27}}=1.3\cdot10^{-5}$ .

From these considerations we get a list in order of decreasing molar solubility in water:

PbI₂$\rightarrow$ BiI₃$\rightarrow$ AgI

For 0.10 M NaI solution [I^-]=0.1mol*L^-1. So, we get:

For AgI s=[Ag⁺] and K_sp=[Ag⁺][I^-]= s*0.1=0.1s, then s=10$\cdot$ K_sp=8.3⋅10⁻¹⁶

for PbI₂ s=[Pb²⁺] and K_sp=[Pb²⁺][I^-]²=s*(0.1)²=0.01s, then s=100$\cdot$ Ksp=7.1⋅10⁻⁹=7.1⋅10⁻⁷

for BiI₃ s=[Bi³⁺] and K_sp=[Bi³⁺][I^-]³=s*(0.1)³=0.001s, then s=1000$\cdot$ Ksp=8.1⋅10−16

From these considerations we get a list in order of decreasing molar solubility in 0.10 M NaI solution:

PbI₂→AgI→BiI₃

Answer:

a. a list in order of decreasing molar solubility in water: PbI₂→BiI₃→AgI

b. a list in order of decreasing molar solubility in 0.10 M NaI solution: PbI₂→AgI→BiI₃