Question

(a) If the molar solubility of Ag₃AsO₄ at 25 ^oC is 1.40e-06 mol/L, what is the K_sp at this temperature?

K_sp =  




(b) It is found that 1.75e-06 g of Cu₃(AsO₄)₂ dissolves per 100 mL of aqueous solution at 25 ^oC. Calculate the solubility-product constant for Cu₃(AsO₄)₂.

K_sp =  




(c) The K_sp of PbBr₂ at 25 ^oC is 6.60e-06. What is the molar solubility of PbBr₂?

solubility =  mol/L

Answer 1

a)

At equilibrium:
Ag3AsO4   <---->    3 Ag+     +          AsO43-

                     3s                 s       

Ksp = [Ag+]^3[AsO43-]
Ksp = (3s)^3*(s)
Ksp = 27(s)^4
Ksp = 27(1.4*10^-6)^4
Ksp = 1.037*10^-22
Answer: 1.04*10^-22

b)
S = 1.75*10^-6 g / 100 mL
= 1.75*10^-6 g / 0.1 L
= 1.75*10^-5 g /L

Molar mass of Cu3(AsO4)2,
MM = 3*MM(Cu) + 2*MM(As) + 8*MM(O)
= 3*63.55 + 2*74.92 + 8*16.0
= 468.49 g/mol


Molar mass of Cu3(AsO4)2= 468.49 g/mol
s = 1.75*10^-5 g/L
To covert it to mol/L, divide it by molar mass
s = 1.75*10^-5 g/L / 468.49 g/mol
s = 3.735*10^-8 mol/L

At equilibrium:
Cu3(AsO4)2<---->    3 Cu+     +         2 AsO43-

                     3s                 2s      

Ksp = [Cu+]^3[AsO43-]^2
Ksp = (3s)^3*(2s)^2
Ksp = 108(s)^5
Ksp = 108(3.735*10^-8)^5
Ksp = 7.854*10^-36
Answer: 7.85*10^-36

c)

At equilibrium:
PbBr2     <---->     Pb2+     +         2 Br-   

                     s                  2s      

Ksp = [Pb2+][Br-]^2
6.6*10^-6=(s)*(2s)^2
6.6*10^-6= 4(s)^3
s = 1.182*10^-2 M
Answer: 1.18*10^-2 M