# Determining Molar Solubility

1. Dec 11, 2008

1. The problem statement, all variables and given/known data

What is the molar solubility of Mg(OH)2 (Ksp1.8*10-11) in8.62*10-2M MgCl2 (aq)?

2. Relevant equations Chemistry is awwweeesssommmmeee....

3. The attempt at a solution

Okay. Here goes. But mind you, I have not put my all into this course, so I will probably need a good slap in the face every now and again when I make a silly assumption or ask a stupid question

The first thing I need to do is write the reaction, which I am guessing is

$Mg(OH)_2\rightarrow 2OH+Mg$ but I am not entirely sure why. That is, don't I need to include the water or something? Or do we just assume it does not participate?

Next I will construct an ice table.. arrg what's that array command again...

$$\left[\begin{array}{ccc}Mg(OH)_2 & 2OH & Mg \\8.62*10^{-2} & 0 & 0\\ -s & +2s & +s\\ 8.62*10^{-2}-s & 2s & s\end{array}\right]$$

This may be wrong (probably). Maybe I did need to include the water. I assumed that the concentration of the Mg(OH)2 was equal to that of the MgCl2

I don't think that is correct. Can I get some guidance from here?

Thanks,
Casey

2. Dec 11, 2008

### Staff: Mentor

You starting concentration of Mg(OH)2 is 0.

However, you already have dissolved magnesium in the solution, so initial concentration of Mg2+ is not 0.

Using ICE table is not a bad idea, but it will be tricky. Your first column - the one with dissolved solid - will remain unchanged. You will have only two columns - one for magnesium and one for OH-. That's the easy part. It will yield third degree polynomial, that's the hard part.

However, you can start assuming that you know concentration of Mg2+ - after all it should not change substantially. Use it to calculate concentration of OH- in saturated solution, that will give you information about amount of dissolved Mg(OH)2. Finally check, if your initial assumption (that concentration of Mg2+ won't change substantially) was valid.

3. Dec 11, 2008

So then my reaction is not written correctly then? That is what is most difficult for me, is setting up the reaction.

If the [Mg(OH)2]0=0 then the rxn $Mg(OH)_2\rightarrow 2OH+Mg$ can have no meaning.

So what is the correct rxn ?

4. Dec 11, 2008

### Staff: Mentor

Mg(OH)2(s) <-> Mg2+(aq) + 2 OH-(aq)

Don't ignore charges when dealing with ions.

5. Dec 11, 2008

I do not see how this reaction makes sense (but that's why I suck at chemistry).

Why isn't the MgCl2 included in the rxn ?

6. Dec 11, 2008

### Staff: Mentor

Because it doesn't react, it is just dissolved.

However, solubility product tells you when Mg(OH)2 will start to precipitate - when [Mg2+][OH-]2 will be higher than the given value. So [Mg2+][OH-]2 product can't be never higher than Kso. When there was already Mg2+ there is the same limit - just now there exist some initial concentration of Mg2+, so when Mg(OH)2 dissolves you should take this original concentration into account.