Question

(a) If the molar solubility of ZnSe at 25 oC is 1.90e-13 mol/L, what is the Ksp at this temperature?

Ksp =

(b) It is found that 1.77e-05 g of PbCrO4 dissolves per 100 mL of aqueous solution at 25 oC. Calculate the solubility-product constant for PbCrO4.

Ksp =

(c) The Ksp of Y2(CO3)3 at 25 oC is 1.03e-31. What is the molar solubility of Y2(CO3)3?

solubility = mol/L

Answer 1

a)

At equilibrium:
ZnSe      <---->     Zn2+     +          Se2-   

                     s                  s       

Ksp = [Zn2+][Se2-]
Ksp = (s)*(s)
Ksp = 1(s)^2
Ksp = 1(1.9*10^-13)^2
Ksp = 3.61*10^-26
Answer: 3.61*10^-26

b)
S = 1.77*10^-5 g / 100 mL
= 1.77*10^-5 g / 0.100 L
= 1.77*10^-4 g / L

Molar mass of PbCrO4,
MM = 1*MM(Pb) + 1*MM(Cr) + 4*MM(O)
= 1*207.2 + 1*52.0 + 4*16.0
= 323.2 g/mol


Molar mass of PbCrO4= 323.2 g/mol
s = 1.77*10^-4 g/L
To covert it to mol/L, divide it by molar mass
s = 1.77*10^-4 g/L / 323.2 g/mol
s = 5.476*10^-7 mol/L

At equilibrium:
PbCrO4    <---->     Pb2+     +          CrO42-

                     s                  s       

Ksp = [Pb2+][CrO42-]
Ksp = (s)*(s)
Ksp = 1(s)^2
Ksp = 1(5.476*10^-7)^2
Ksp = 3.0*10^-13
Answer: 3.0*10^-13

c)

At equilibrium:
Y2(CO3)3 <---->    2 Y3+     +         3 CO32-

                     2s                 3s      

Ksp = [Y3+]^2[CO32-]^3
1.03*10^-31=(2s)^2*(3s)^3
1.03*10^-31= 108(s)^5
s = 2.488*10^-7 M
Answer: 2.49*10^-7 M