# solubility/pH question

#### happygirl

##### Member
10+ Year Member
5+ Year Member
Can someone help me with this solubility/pH question??

"The Ksp of Mg(OH)2 in water is 1.2x10^-11. If the Mg2+ concentration is 1.2x10^-5, what is the pH at which Mg(OH)2 just begins to precipitate?"

So, I get that I need to find out the [OH-]. So, since the Mg2+ concentration is already given and we know that Mg(OH)2 dissociates into 1 Mg and 2OH's, I multiplied the [Mg2+] by 2 since there are two OH's for every one Mg2+. Essentially:
Mg(OH)2 --> Mg2+ + 2OH-
x --> x + 2x

But according to Kaplan's explanation, the correct way to do it is: [OH-] = square root of Ksp/[Mg2+]. I see how this is derived from
Ksp = [Mg2+][OH-]^2, but don't you still have to account for the 2:1 ratio of OH to Mg?

I solved the problem similarly to if the question were asking for the concentrations of the ions of a saturated solution of Mg(OH)2:
You'd figure out the Mg2+ concentration by solving for 'x' in Ksp= 1.2x10^-11 = 4x^3. Once you get x, you plug it back in to find [OH-] knowing that
x ---> x + 2x.

I feel like I must be missing something really basic =( Help!!!

***I understand that after you find the [OH], you have to find the pOH and then from there find the pH. But my problem lies in the first step of calculating the OH concentration.

And the correct answer is pH 11. I guessed that since I figured OH is being dissolved, the solution will probably be basic, but I still don't really understand hoow to do it.

Thanks!!!!!!!!!

#### BloodySurgeon

Moderator Emeritus
10+ Year Member
Can someone help me with this solubility/pH question??

"The Ksp of Mg(OH)2 in water is 1.2x10^-11. If the Mg2+ concentration is 1.2x10^-5, what is the pH at which Mg(OH)2 just begins to precipitate?"

So, I get that I need to find out the [OH-]. So, since the Mg2+ concentration is already given and we know that Mg(OH)2 dissociates into 1 Mg and 2OH's, I multiplied the [Mg2+] by 2 since there are two OH's for every one Mg2+. Essentially:
Mg(OH)2 --> Mg2+ + 2OH-
x --> x + 2x

But according to Kaplan's explanation, the correct way to do it is: [OH-] = square root of Ksp/[Mg2+]. I see how this is derived from
Ksp = [Mg2+][OH-]^2, but don't you still have to account for the 2:1 ratio of OH to Mg?

I solved the problem similarly to if the question were asking for the concentrations of the ions of a saturated solution of Mg(OH)2:
You'd figure out the Mg2+ concentration by solving for 'x' in Ksp= 1.2x10^-11 = 4x^3. Once you get x, you plug it back in to find [OH-] knowing that
x ---> x + 2x.

I feel like I must be missing something really basic =( Help!!!

***I understand that after you find the [OH], you have to find the pOH and then from there find the pH. But my problem lies in the first step of calculating the OH concentration.

And the correct answer is pH 11. I guessed that since I figured OH is being dissolved, the solution will probably be basic, but I still don't really understand hoow to do it.

Thanks!!!!!!!!!
Mg(OH)2 <-----> Mg2+ + 2OH-

Ksp = [Mg2+][OH-]^2

[Mg2+] = x
[OH-] = 2x

Ksp = (x) (2x)^2 = 4x^3

(1x10^-11/4) = x^3 = 25 x 10^-9

x = 5x10^-3
[OH-] = 2(5x10^-3) = 1x10^-2

pOH=2
pH=12 (I estimated the Ksp which is probably why im off by 1)

OP
H

#### happygirl

##### Member
10+ Year Member
5+ Year Member
Mg(OH)2 <-----> Mg2+ + 2OH-

Ksp = [Mg2+][OH-]^2

[Mg2+] = x
[OH-] = 2x

Ksp = (x) (2x)^2 = 4x^3

(1x10^-11/4) = x^3 = 25 x 10^-9

x = 5x10^-3
[OH-] = 2(5x10^-3) = 1x10^-2

pOH=2
pH=12 (I estimated the Ksp which is probably why im off by 1)

Hi, thanks for the reply. Yeah, I see how that works! Thanks! I guess I had some math errors when I did it that way. But, I still dont understand why just multiplying the 1.2x10^-5 by 2 doesn't work. When I do it that way, I get a pH that is around 9 something. Obviously I must be missing something since your calculated x = 5x10^-3 and not 1.2x10^-5.

#### BloodySurgeon

Moderator Emeritus
10+ Year Member
It would be logical to use the concentration of Mg2+ to be the molar concentration and just multiply by 2 to make the concentration of OH-, however we do not have enough information to make that speculation. Maybe there was Mg2+ ions already in solution or better yet there were already OH- ion with a higher pH already in solution making the concentrations of OH- : Mg2+ not 2:1 (this would be the common ion effect). Therefore we must assume nothing.

We will use the concentration that is given for Mg2+ and predict what is the amount of OH- in THIS particular solution to make it precipitate.

OP
H

#### happygirl

##### Member
10+ Year Member
5+ Year Member
ok, makes sense. Thanks =)

#### TawMus

10+ Year Member
5+ Year Member
I actually solved this using a different method...

Mg(OH)2 ----> Mg2+ + 2OH-

so Ksp = [Mg2+][OH]2

so 1.2x10^-11= 1.2x10^-5 [OH]2

1.2x10^-11/1.2x10^-5 = 1.0 x 10^-6 = [OH]2

Then you sq root 1.0x10^-6 giving you 1.0x 10^-3 = [OH]

the pOH of that value is 3 and if you subtract 14-3= 11 = pH!

#### werd

##### Senior Member
15+ Year Member
Mg(OH)2 <-----> Mg2+ + 2OH-

Ksp = [Mg2+][OH-]^2

[Mg2+] = x
[OH-] = 2x

Ksp = (x) (2x)^2 = 4x^3

(1x10^-11/4) = x^3 = 25 x 10^-9

x = 5x10^-3
[OH-] = 2(5x10^-3) = 1x10^-2

pOH=2
pH=12 (I estimated the Ksp which is probably why im off by 1)
this method isn't correct, which is why the answer isn't the correct answer of pH=11. as you state below, you cannot assume that "Mg=x and OH=2x" because there may be other sources of Mg than the Mg(OH)2 compound. it is also unnecessary to define Mg concentration as "x" because you are given its value of 1.2*10^-5 in the question stem.
the solution given by TawMus above is the correct way to solve this equation.

#### BloodySurgeon

Moderator Emeritus
10+ Year Member
this method isn't correct, which is why the answer isn't the correct answer of pH=11. as you state below, you cannot assume that "Mg=x and OH=2x" because there may be other sources of Mg than the Mg(OH)2 compound. it is also unnecessary to define Mg concentration as "x" because you are given its value of 1.2*10^-5 in the question stem.
the solution given by TawMus above is the correct way to solve this equation.
Ouch! your absolutely right! I typed it one way and explained it another (this is why I hate doing math on the computer). I just hope I don't make any silly mistake like that on the real thing.

#### TawMus

10+ Year Member
5+ Year Member
Ouch! your absolutely right! I typed it one way and explained it another (this is why I hate doing math on the computer). I just hope I don't make any silly mistake like that on the real thing.

You will probably do great man, you really know you stuff. #### tweaked17

10+ Year Member
I would do a backflip if this showed up on my MCAT haha.. more complex Ksp problems trip me up sometimes.