Question

Sea water contains many solutes, Ca²⁺ and Mg²⁺ among them. A sample of seawater has a Ca²⁺ concentration of 0.070M and Mg²⁺ concentration of 0.050M. Assume that Ca²⁺ and Mg²⁺ are the only reactive species in this experiment.

a) Which ion will precipitate first if solid NaOH is added. Assume constant volume and a temperature of 25^oC.

b) Calculate the hydroxide ion concentration at the point when each hydroxide (Ca(OH)₂ and Mg(OH)₂) begins to precipitate.

c) If the first species to precipitate is X(OH)_2, calculate the concentration of X²⁺ ion remaining in solution when the second ion precipitates.

Answer 1

a) Ksp of Ca(OH)2 is 5.5 × 10^-6

Ksp of Mg(OH)2 is 1.8×10^-11

Ksp of Mg(OH)2 is less comparing to Ksp of Ca(OH)2 , So Mg(OH)2 will precipitate first

b) Ca(OH)2 (s) <-------> Ca2+(aq) + 2OH-(aq)

Ksp = [Ca2+][OH-]^2

5.5 ×10^-6 = 0.070M × [OH-]^2

[OH-]^2 = 7.85× 10^-5

[OH-]=8.86 ×10^-3M

when [OH-] is > 8.86× 10^-3 Ca(OH)2 will start to precipitate

Mg(OH)2 (s) <------> Mg2+(aq) + 2OH-(aq)

Ksp = [Mg2+] [ OH-]^2

1.8×10^-11 = 0.050M × [OH-]^2

[OH-] = 1.89 × 10^-5

When [OH-] is > 1.89 × 10^-5M ,Mg(OH)2 start to precipitate

c) When Ca(OH)2 start to precipitate [OH-] is 8.86×10^-3M

Ksp of Mg(OH)2 = [Mg2+][OH-]^2

1.8×10^-11 = [Mg2+] × (8.86×10^-3)^2

[Mg2+] = 7.84 × 10^-4M

When Ca(OH)2 start to precipitate , [Mg2+] = 7.84 × 10^-4M