Given : concentration of Mg2+ = [Mg2+] = 0.0099 M
concentration of Ca2+ = [Ca2+] = 0.021 M
(a.) The dissociation equation of Mg(OH)2 is :
Mg(OH)2 (s)
Mg2+ (aq) + 2 OH- (aq)
Let molar solubility of Mg(OH)2 = s
[Mg2+] = [Mg(OH)2] + concentration of Mg2+
[Mg2+] = s + 0.0099 M
[OH-] = 2 * [Mg(OH)2]
[OH-] = 2s
Ksp [Mg(OH)2] = [Mg2+][OH-]2
6.3 x 10-10 = (s + 0.0099 M) * (2s)2
Assuming s << 0.0099 M
6.3 x 10-10 = (0.0099 M) * (2s)2
6.3 x 10-10 = (0.0099 M) * (4s2)
s2 = (6.3 x 10-10) / (0.0099 M)
s2 = 6.36 x 10-8
s = (6.36 x 10-8)1/2
s = 2.52 x 10-4 M
[OH-] = 2s
[OH-] = 2 * (2.52 x 10-4 M)
[OH-] = 5.04 x 10-4 M
(b.)
The dissociation equation of Ca(OH)2 is :
Ca(OH)2 (s)
Ca2+ (aq) + 2 OH- (aq)
Let molar solubility of Ca(OH)2 = s
[Ca2+] = [Ca(OH)2] + concentration of Ca2+
[Ca2+] = s + 0.021 M
[OH-] = 2 * [Ca(OH)2]
[OH-] = 2s
Ksp [Ca(OH)2] = [Ca2+][OH-]2
6.5 x 10-6 = (s + 0.021 M) * (2s)2
Solving for x, s = 7.545 x 10-3 M
[OH-] = 2s
[OH-] = 2 * (7.545 x 10-3 M)
[OH-] = 0.0151 M
Since less concentration of OH- is required to precipitate Mg(OH)2, therefore, Mg(OH)2 will precipitate first and Ca(OH)2 will precipitate second.
concentration of OH- when Ca2+ begins to precipitate = 0.0151 M
Ksp Mg(OH)2 = [Mg2+][OH-]2
6.3 x 10-10 = [Mg2+] * (0.0151 M)2
[Mg2+] = (6.3 x 10-10) / (0.0151 M)2
[Mg2+] = 2.8 x 10-6 M
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