Question

4. 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 is related to K_SP in the following way:

$S_0 =(\frac{K_{SP}}{x^xy^y})^{1/(x+y)}$ ,

where x and y are the number of cations and anions to which the iodide dissociate, respectively. Therefore, the molar solubilities of AgI, PbI₂ and BiI₃ in water are:

AgI: x=1, y=1, $S_0 = (\frac{8.3·10^{-17}}{1})^{(1/2)} = 9.1·10^{-9}$ M,

PbI₂: x=1, y=2, $S_0 = (\frac{7.1·10^{-9}}{4})^{(1/3)} = 1.2·10^{-3}$ M,

BiI₃: x=1, y=3, $S_0 = (\frac{8.1·10^{-19}}{27})^{(1/4)} = 1.3·10^{-5}$ M.

The list of these three compounds in order of decreasing (from highest to lowest) molar solubility in water is:

PbI₂>BiI₃>AgI.

In 0.10 M NaI, a common ion is present: I^-. The concentration of sodium iodide must be taken in account:

$K_{SP} = [Me]^x(0.1+[Me])^y$ , or (as all x=1) :

$K_{SP} = [Me]·(0.1+[Me])^y$

and the molar solubility equals [Me]. Solving the equation above for each K_SP, we obtain:

AgI: y=1: [Me]<<0.1, [Me] = 8.3·10^-16 M

PbI₂: y=2: [Me]<<0.1, [Me] = 7.1·10^-7 M

BiI₃: y=3: [Me]<<0.1, [Me] =8.1·10^-16 M.

Therefore, the list of these three compounds in order of decreasing (from highest to lowest) molar solubility in 0.10 M NaI is:

PbI₂>AgI<BiI₃.

Answer: water: PbI₂>BiI₃>AgI; 0.10 M NaI: PbI₂>AgI<BiI₃.