Question

a)A solution of saturated PbBr₂ is found to contain 2.4 ✕ 10⁻²M bromide ion. Calculate the K_sp of PbBr₂.

b) A 40.0-mL solution contains 0.029 M barium chloride (BaCl₂). What is the minimum concentration of sodium sulfate (Na₂SO₄) required in the solution to produce a barium sulfate (BaSO₄) precipitate? The solubility product for barium sulfate is

K_sp = 1.1 ✕ 10⁻¹⁰.

c)The solubility product, K_sp, for magnesium hydroxide, Mg(OH)₂, is 5.6 ✕ 10⁻¹² at 25°C. What is the molar solubility of magnesium hydroxide at 25°C?

Answer 1

a)

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

                     s                  2s      

Ksp = [Pb2+][Br-]^2
Ksp = (s)*(2s)^2
Ksp = 4(s)^3
Ksp = 4(2.4*10^-2)^3
Ksp = 5.53*10^-5
Answer: 5.5*10^-5

b)

BaCl2 here is Strong electrolyte
It will dissociate completely to give [Ba2+] = 0.029 M
At equilibrium:
BaSO4     <---->     Ba2+     +          SO42-  
                     2.9*10^-2 +s           s       
Ksp = [Ba2+][SO42-]
1.1*10^-10=(2.9*10^-2 + s)*(s)
Since Ksp is small, s can be ignored as compared to 2.9*10^-2
Above expression thus becomes:
1.1*10^-10=(2.9*10^-2)*(s)
1.1*10^-10= 2.9*10^-2 * 1(s)^1
s = 3.793*10^-9 M

[Na2SO4] = [SO42-] = 3.793*10^-9 M
Answer: 3.79*10^-9 M

c)

At equilibrium:
Mg(OH)2   <---->     Mg2+     +         2 OH-   

                     s                  2s      

Ksp = [Mg2+][OH-]^2
5.6*10^-12=(s)*(2s)^2
5.6*10^-12= 4(s)^3
s = 1.119*10^-4 M
Answer: 1.12*10^-4 M