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The Solubility Product Constant Ksp

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I have this question that I can't answer. It's for homework and I need your help.

 

Q. The solubility product of a saturated magnesium hydroxide solution is 1.2 x 10^-11 (mol/L)^3 at 18 celsusis degree. Predict the pH of a saturated magnesium hydroxide solution at 18 celsusis degree.

 

Please answer back as soon as possible. Thanks :)

Please show us what you have tried. We can help you but we are not here to do your homework for you.

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Yes P-man that's the problem. My chemistry teacher warned us that we wouldn't be able to answer the question because we never learned about the concepts that need to be know yet. So any suggestions would be wonderful.

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My last question was: What would be the normal (non-acidic) concentration of calcium ions in a lake lined by limestone?

 

My answer was:

 

CaCO3(s) = (Ca 2+)(aq) + (CO3 2-)(aq)

Ksp = [Ca 2+ (aq)][CO3 2- (aq)] = 9.9 x 10^-9 mol^2/L^2

 

[Ca 2+] = sqroot(Ksp)

=sqroot(9.9 x 10^-9 mol^2/L^2)

=9.9 x 10^-5 mol/L

 

That was my last question and I answered it. Now for the question that I requested, I know I have to 1/3 sq root of the magnesium hydroxide solution. But how am I supposed to predict the pH value???

That was my last question and I answered it. Now for the question that I requested, I know I have to 1/3 sq root of the magnesium hydroxide solution. But how am I supposed to predict the pH value???

If you have a saturated solution of Mg(OH)2, how much Mg(OH)2 will there be in solution, how many moles per liter?

 

From this you can compute the concentration of OH(-) ions.

 

Next, you can compute the concentration of H(+) ions, using the identity [H(+)][OH(-)]=10^(-14). That should give you the pH.

 

Beware, in this computation I made a BIG simplification. It only works if [OH(-)] >> 10^(-7). For very low concentrations of OH(-) you need to solve a set of equations, resulting in a single quadratic equation. For your problem, however, the simplification is OK.

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