Determining Molar Solubility of Mg(OH)2 in MgCl2 Solution

[Saladsamurai]

Homework Statement

What is the molar solubility of Mg(OH)₂ (K_sp1.8*10^-11) in8.62*10^-2M MgCl₂ (aq)?

Homework Equations

Chemistry is awwweeesssommmmeee...

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

 

[Borek]

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.

 

[Saladsamurai]

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 ?

 

[Borek]

Mg(OH)₂(s) <-> Mg²⁺(aq) + 2 OH^-(aq)

You start with a solid hydroxide, so its concentration is 0.

Don't ignore charges when dealing with ions.

 

[Saladsamurai]

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 ?

 

[Borek]

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.