Compute [Mg^2+] and [OH-] for a solution that is prepared by dissolving 3.2*10^-5 mole of magnesium hydroxide
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Compute [Mg^2+] and [OH-] for a solution that is prepared by dissolving 3.2*10^-5 mole of magnesium hydroxide

[From: ] [author: ] [Date: 11-05-06] [Hit: ]
1 L = 3.2 X 10^-4 M. Since you produce 2 OH- ions for each Mg2+ ion (magnesium hydroxide is Mg(OH)2, its concentration will be twice that of the Mg2+.......
Magnesium hydroxide is a strong base. Compute [Mg^2+] and [OH-] for a solution that is prepared by dissolving 3.2*10^-5 mole of magnesium hydroxide in 100.0 mL of water.

a) [Mg^2+] = 3.2*10^-4M, [OH-] = 3.2*10^-4M
b) [Mg^2+] = 3.2*10^-5M, [OH-] = 3.2*10^-5M
c) [Mg^2+] = 1.6*10^-3M, [OH-] = 3.2*10^-3M
d) [Mg^2+] = 3.2*10^-4M, [OH-] = 6.4*10^-4M
e) None of the above are correct

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d) is the correct answer. The molarity of Mg2+ is 3.2 X 10^-5 mol / 0.1 L = 3.2 X 10^-4 M. Since you produce 2 OH- ions for each Mg2+ ion (magnesium hydroxide is Mg(OH)2, its concentration will be twice that of the Mg2+.
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