Solubility Ksp question

[StarryNights]

The Ksp of Mg(OH)2 is 1.2 x 10^-11. If the Mg2+ concentration in an acid solution is 1.2 x 10^-5 M, what is the pH at which Mg(OH)2 just begins to precipitate?

a. 3
b. 4
c. 5
d. 11

I basically set up Ksp = [Mg2+][OH-][OH-] and solved for [OH-][OH-] = Ksp/[Mg2+]. This gives me 1.0 x 10^-6 as the concentration for [OH-][OH-]. So then I thought pOH would be 6 and pH would be 9, but this was wrong. The solution actually took the square root to get [OH-] = 1.0 x 10^-3 for pOH of 3 and pH of 11 as in the bolded answer choice.

I understand what they did, but I don't understand why they only took into account the concentration of one when there would be two OH- for each Mg(OH)2 that dissolves in solution?? Can anyone help??

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[Isoprop]

They did take that into account by square rooting the answer. You solved for [OH][OH] not for [OH].

If x = [OH], then the equation would be Ksp = [Mg] * x^2. You want to solve for x, not x^2.

Does that make sense?

 

[StarryNights]

Yea, I get that. But when you're figuring out the pOH, wouldn't you use the overall OH- concentration, which would be 2[OH-]? I don't get why they used 1.0 x 10^-3, which is just one OH-?

Or to clarify, I get that [OH-] is equal to 1.0 x 10^-3, but in solution there are actually twice as much because each Mg(OH)2 dissociates into two OH-s. So I guess I don't get why they went straight with 1.0 x 10^-3...maybe 2.0 x 10^-3 doesn't make a big difference to the answer? I totally butchered this problem, sorry, I'm just really confused.

 

[Isoprop]

Okay I see what you are saying. I didn't see how you calculated [OH] so I misinterpreted your reasoning. Let me just tell you how to get to the right answer and see if you have any questions.

When the solution is just about to precipitate (saturated), the concentrations of Mg and OH satisfies the Ksp equilibrium expression. If we put in just the right amount of Mg(OH)2 to reach saturated,

x amount of Mg(OH)2 will create x amount of Mg and 2x amount of OH.

Thus, Ksp = [Mg][OH]^2 = (x)(2x)^2 = 4x^3.

Solve for x gives us (Ksp/4)^(1/3). You have to multiply this by 2 to get [OH], find the pOH, and then calculate the pH.

I haven't calculated the answer, but it should be d because the pH should be greater than 7. When the solution is saturated, there is more OH than H.

 

[boaz]

The Ksp of Mg(OH)2 is 1.2 x 10^-11. If the Mg2+ concentration in an acid solution is 1.2 x 10^-5 M, what is the pH at which Mg(OH)2 just begins to precipitate?

a. 3
b. 4
c. 5
d. 11

I basically set up Ksp = [Mg2+][OH-][OH-] and solved for [OH-][OH-] = Ksp/[Mg2+]. This gives me 1.0 x 10^-6 as the concentration for [OH-][OH-]. So then I thought pOH would be 6 and pH would be 9, but this was wrong. The solution actually took the square root to get [OH-] = 1.0 x 10^-3 for pOH of 3 and pH of 11 as in the bolded answer choice.

I understand what they did, but I don't understand why they only took into account the concentration of one when there would be two OH- for each Mg(OH)2 that dissolves in solution?? Can anyone help??

They did take that into account by square rooting the answer. You solved for [OH][OH] not for [OH].

If x = [OH], then the equation would be Ksp = [Mg] * x^2. You want to solve for x, not x^2.

Does that make sense?

x is our unknown for which we want to solve.

This is correct.

Yea, I get that. But when you're figuring out the pOH, wouldn't you use the overall OH- concentration, which would be 2[OH-]? I don't get why they used 1.0 x 10^-3, which is just one OH-?

Or to clarify, I get that [OH-] is equal to 1.0 x 10^-3, but in solution there are actually twice as much because each Mg(OH)2 dissociates into two OH-s. So I guess I don't get why they went straight with 1.0 x 10^-3...maybe 2.0 x 10^-3 doesn't make a big difference to the answer? I totally butchered this problem, sorry, I'm just really confused.

[OH-] is indeed the grand total OH- concentration. Make sure you have that clearly defined in your mind.

Okay I see what you are saying. I didn't see how you calculated [OH] so I misinterpreted your reasoning. Let me just tell you how to get to the right answer and see if you have any questions.

When the solution is just about to precipitate (saturated), the concentrations of Mg and OH satisfies the Ksp equilibrium expression. If we put in just the right amount of Mg(OH)2 to reach saturated,

x amount of Mg(OH)2 will create x amount of Mg and 2x amount of OH.

Thus, Ksp = [Mg][OH]^2 = (x)(2x)^2 = 4x^3.

Solve for x gives us (Ksp/4)^(1/3). You have to multiply this by 2 to get [OH], find the pOH, and then calculate the pH.

I haven't calculated the answer, but it should be d because the pH should be greater than 7. When the solution is saturated, there is more OH than H.

This is NOT the correct way to approach this question. The problem with this, which is along the line of thinking of the OP, is that it assumes that all Mg2+ and OH- comes from Mg(OH)2. But it doesn't say this anywhere in the question. Mg2+ could have come from MgCl and OH- could have come from NaOH. This is what equilibrium expressions are for: they are constant, ALWAYS true, no matter what else is in the solution and where the species came from.

The equilibrium expression is this:

K=(TOTAL concentration of OH-)^2/(TOTAL cocentration of Mg2+)

Since pOH is -log(TOTAL concentration of OH-),

it is (TOTAL concentration of OH-) that we are looking for,

NOT (TOTAL concentration of OH-)^2

pH=11 is the correct result.

Hope this helps.

 

[Isoprop]

This is NOT the correct way to approach this question. The problem with this, which is along the line of thinking of the OP, is that it assumes that all Mg2+ and OH- comes from Mg(OH)2. But it doesn't say this anywhere in the question. Mg2+ could have come from MgCl and OH- could have come from NaOH. This is what equilibrium expressions are for: they are constant, ALWAYS true, no matter what else is in the solution and where the species came from.

The equilibrium expression is this:

K=(TOTAL concentration of OH-)^2/(TOTAL cocentration of Mg2+)

Since pOH is -log(TOTAL concentration of OH-),

it is (TOTAL concentration of OH-) that we are looking for,

NOT (TOTAL concentration of OH-)^2

pH=11 is the correct result.

Hope this helps.

You're absolutely right. I didn't know why the [Mg] was important. Thanks for correcting me.

 

[StarryNights]

Hey you guys, thanks a lot for your help!

After reading your comments and looking through my old gen chem textbooks for similar problems, I see that Ksp = [Mg2+][OH-]^2. You guys are completely right that [OH-] is the total concentration in solution, but it is however = 2[Mg+] so that the Ksp = 4x^3, with x = [Mg2+]. The textbook problem actually solved it for PbCl2, but the set up is exactly the same and that [Cl-] = 2[Pb2+].

Again, thanks!