Question

What is the molar solubility (in mol/L) of PbCl₂ (Ksp = 2.01⋅10-7) in a 0.46 M NaCl solution.

What is the molar solubility (in mol/L) of Cu₃PO₄ (Ksp = 4.10⋅10-5) in a 0.33 M CuNO₃ solution

What is the molar solubility (in mol/L) of AgCl (Ksp = 5.32⋅10-14) in a 0.39 M NaCl solution.

Answer 1

1)

NaCl here is Strong electrolyte
It will dissociate completely to give [Cl-] = 0.46 M
At equilibrium:
PbCl2     <---->     Pb2+     +         2 Cl-   
                     s                  0.46 + 2s
Ksp = [Pb2+][Cl-]^2
2.01*10^-7=(s)*(0.46+ 2s)^2
Since Ksp is small, s can be ignored as compared to 0.46
Above expression thus becomes:
2.01*10^-7=(s)*(0.46)^2
2.01*10^-7= (s) * 0.2116
s = 9.50*10^-7 M
Answer: 9.50*10^-7 M

2)

CuNO3 here is Strong electrolyte
It will dissociate completely to give [Cu+] = 0.33 M
At equilibrium:
Cu3PO4    <---->    3 Cu+     +          PO43-  
                     0.33 +3s           s       
Ksp = [Cu+]^3[PO43-]
4.1*10^-5=(0.33 + 3 s)^3*(s)
Since Ksp is small, s can be ignored as compared to 0.33
Above expression thus becomes:
4.1*10^-5=(0.33)^3*(s)
4.1*10^-5= 3.594*10^-2 * 1(s)^1
s = 1.141*10^-3 M
Answer: 1.14*10^-3 M

3)

NaCl here is Strong electrolyte
It will dissociate completely to give [Cl-] = 0.39 M
At equilibrium:
AgCl      <---->     Ag+      +          Cl-    
                     s                  0.39 + s
Ksp = [Ag+][Cl-]
5.32*10^-14=(s)*(0.39+ s)
Since Ksp is small, s can be ignored as compared to 0.39
Above expression thus becomes:
5.32*10^-14=(s)*(0.39)
5.32*10^-14= (s) * 0.39
s = 1.364*10^-13 M
Answer: 1.36*10^-13 M