Confusing Ksp/Common Ion Effect question

[kkentm]

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The Ksp of Mg(OH)2 in water is 1.2 × 10–11 mol3/L3. If the Mg2+ concentration in an acid solution is 1.2 × 10–5, what is the pH at which Mg(OH)2 just begins to precipitate?

A.3
B.4
C.5
D. 11

Heres the answer and the explanation:

Question 13 is appropriately answered by choice D. This question deals with solubility constants and the common ion effect. The first step to solving this problem is to express the solubility product constant, which is the ion product of the saturated solution, as the product of the concentration of Mg2+ ion and the concentration of the hydroxide ion squared. Second, you must determine the minimum concentration of hydroxide necessary to precipitate the Mg(OH)2. From the equation just given, you can solve for the concentration of hydroxide, which is the square root of the Ksp divided by the concentration of magnesium. Substituting in the values provided in the question, you should find that the concentration is equal to ten to the minus 3 moles per liter. That means that the pOH of the solution, the negative log of the hydroxide ion concentration, will be 3. Since the pOH plus the pH of any solution must equal 14, the pH must be 11. This corresponds to choice D.




So even after reading the explanation, I dont really understand the question. My confusion stems from the second sentence, which gives the concentration of Mg ions in an acidic solution and then asks for the pH of when the compound begins to precipitate. Can someone who understands this please give me a brief explanation? Thanks

 

[rpdnathan]

When you write the equilibrium reaction: Mg(OH)2 --> Mg2+ + 2OH-, you would write the solubility expression as [Mg2+][OH-]^2 = Ksp and in the question, they give you the concentration of Magnesium and the value of Ksp and so you solve for the concentration of [OH-]. Once you find that, you take -log[OH-] to find the pOH. Remember pOH + pH = 14 and so you can solve for pH.

 

[kkentm]

ok i have a complete understanding of everything you wrote but my confusion is really in the phrasing of the question:

1. is the acidity of the solution relevant? shouldnt the dissociation of mgoh2 result in a basic solution?
2. how do i incorporate the common ion effect into this question?
3. how do i evaluate for when mgoh2 "just begins to precipitate"

 

[kkentm]

When you write the equilibrium reaction: Mg(OH)2 --> Mg2+ + 2OH-, you would write the solubility expression as [Mg2+][OH-]^2 = Ksp and in the question, they give you the concentration of Magnesium and the value of Ksp and so you solve for the concentration of [OH-]. Once you find that, you take -log[OH-] to find the pOH. Remember pOH + pH = 14 and so you can solve for pH.

if i were to evaluate this answer backwards (ie. writing out the appropriately relevant question), my question would be something like: a certain concentration of Mg(OH)2 is added to water, if the Ksp = whatever Ksp is, and [mg2+] = whatever conc of mg2+ is, what is the ph of the solution? i assume the question given correlates with the situation i just described in my own constructed question (since the answer and steps to derive the answer would be the same) so how come the wording of the given question is so different?

 

[rpdnathan]

if i were to evaluate this answer backwards (ie. writing out the appropriately relevant question), my question would be something like: a certain concentration of Mg(OH)2 is added to water, if the Ksp = whatever Ksp is, and [mg2+] = whatever conc of mg2+ is, what is the ph of the solution? i assume the question given correlates with the situation i just described in my own constructed question (since the answer and steps to derive the answer would be the same) so how come the wording of the given question is so different?

It is different because the person who wrote the question wants the test taker to recognize the fact that when Q = Ksp, then the product will just begin to precipitate. Finding the pH is a matter of calculation.

 

[rpdnathan]

ok i have a complete understanding of everything you wrote but my confusion is really in the phrasing of the question:

1. is the acidity of the solution relevant? shouldnt the dissociation of mgoh2 result in a basic solution?
2. how do i incorporate the common ion effect into this question?
3. how do i evaluate for when mgoh2 "just begins to precipitate"

The dissociation of Mg(OH)2 should result in a basic solution. The acidity is relevant because this affects the equilibrium of the dissolution of Magnesium Hydroxide. Mg(OH)2 will begin to precipitate when Q, constant of the equilibrium equals Ksp. This is when the solution is saturated and when Q > Ksp, then precipitation will occur. If Q < Ksp, then precipitation will not occur and solution is not saturated.

As I wrote before, Mg(OH)2 --> Mg2+ + 2OH- (assume that there is double arrows for equilibrium) and the common-ion effect says that if you increase concentration of Mg2+ for example, then there would be a disturbance in the products and so the equilibrium will shift to the reactants side thereby precipitating more Mg(OH)2. On the other hand, if you increase the concentration of H+, then it will combine with 2OH- to form water and this will again shift the equilibrium to the left thereby precipitating more of Mg(OH)2. The common ion effect really just describes how the equilibrium can shift to reduce the disturbance on the system.