In: Chemistry
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 25oC.
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.
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