Le Chatelier's Principle

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Magnesium hydroxide is only very slightly soluble in water. The reaction by which it goes into so­lution is: Mg(OH)2 (s) ⇌ Mg2+ (aq) + 2 OH- (aq) 

It's possible to dissolve significant amounts of Mg(OH)2 in solutions in which the concentration of either Mg2+ or OH- is very, very small. Explain, using Ksp, why this is the case. 

Explain why Mg(OH)2 might have very appreciable solubility in 1 M HCl.

Oct 3rd, 2015

Thank you for the opportunity to help you with your question!

For the first part of your question:

Mg(OH)2 (s) ⇌ Mg2+ (aq) + 2 OH- (aq) 

This means that if the concentration of Mg2+ or OH- is very very small in the solution, we can assume that their product will be much less than the Ksp, where

Ksp = Conc(Mg2+)*Conc(OH-)*Conc(OH-)/(Conc(Mg(OH)2)

So, to maintain Ksp constant(which is a requirement), the concentration of Mg2+ or of OH- needs to increase, hence it's possible to dissolve significant amounts of Mg(OH)2

For the second part of your question:

HCl ⇌ H+ + Cl-

H+ + OH- ⇌ H2O

The H+ ions react with the OH- ions formed from the dissolution of Mg(OH)2, which leads to a decrease in the concentration of OH- ions. This decreases the value of Ksp, so to maintain Ksp as constant, the concentration must increase, and hence Mg(OH)2 might have very appreciable solubility in 1 M HCl.

Please let me know if you need any clarification. I'm always happy to answer your questions! :)
Sep 22nd, 2015

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Oct 3rd, 2015
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