Solubility-product constant!?

Question

I need help with this chemistry problem for my AP review. I don't understand how to calculate it with the given information.

A 1 L solution saturated at 25 degrees C with lead(II) iodide contains 0.54 g of PbI2. Calculate the solubility-product constant for this salt at 25 degrees C

Thanks in advance for the help

Answer 1

[Pb][I-]^2 = solubility product (K PbI2)

maolecular weight of PbI2= 207+2*127= 461

So 0.54g of PbI2 has a molarity of 0.54*461 =1.1710^-3

The molarity of ion Pb++is 1.1710^-3
The molarity of ion I- is *2* 1.1710^-3 =2.3410^-3

and the product is
K (pbI2) = 1.17*10^-3 *(2.3410^-3)^2 = 6.41 10^-9

Answer 2

The equilibrium for the saturated solution is
PbI2<> Pb2+ + 2I-
and the equilibrium condition is given by
Ksp = (Pb2+)(I-)^2 ( ) means concentration
One mole of PbI2 weights 461 g
0.54 g is 0.54/461=0.00117 mole
The volume is 1 liter
Concentration of dissolved PbI2 is 0.00117 M
We assume that PbI2 being a strong electrolyte is 100% dissociated into Pb2+ and I-
This means that the dissolving of 0.00117 mole/L will produce
0.00117 mole/L of Pb2+ and (2)(0.00117) mole/L =0.00234 F-
Ksp= (Pb2+)(I-)^=6 10^-9

Answer 3

Ag2CrO4 dissolves in water in accordance to here equation: Ag2CrO4(s) = 2Ag+ + CrO42- a million mole Ag2CrO4 provides 2 Moles of Ag+ and a million mole of CrO42- From the above equation this is clean that [Ag2CrO4]=[CrO42-] and [Ag+] = 2 x [Ag2CrO4]. Substituting into the equation for KSP: KSP = [Ag+]2 [CrO42-] KSP = (2* 7.8 x 10-5 mol L-a million)2 x (7.8 x 10-5 mol L-a million) = a million.87 x 10-12 mol3 L-3