[sharbeldam]

Can someone tell me how this solution is acceptable?

We calculated the amount of OH- needed to react with all of the H+ -> 10^-5 moles (1 Litre solution).
After it reacts with all of the protons in the solution we have Mg+2 in the solution which is (10^-5)/2 and now we figure out how much more we can add of the base from this point.

Mg(OH)2<----> Mg+2    +     2OH-
-s                     10^-5/2+s        +2s

And then from the ksp we find s, and the total moles of Mg(OH)2 will be 10^-5/2 +s.

I honestly think that is too complicated, first of all, how could the professor get that the number of moles of Mg+2 at the first part is equal to Mg(OH)2, since it's an equilibrium and not a strong base equation.

Secondly, if someone could simplify the second part, id be greatly appreciative.
Thanks

[chenbeier]

I dont know what the s means.

If you have 10^-5 H⁺ for pH 5 then it correspond to the same amount of OH^- for neutralization. If a hydroxide, which contain 2 OH^- like Mg(OH)₂ is used then the half amount 5 * 10^-6 mol is needed.

For the solubility check the solubility product.

[sharbeldam]

But the thing is, the question is not about how much we need to neutralize the acid, but how much can you dissolve without precipitate given ksp. so first the OH- gets neutralized, and then we can dissolve more of our base... atleast that what the professor said.

[AWK]

Your problem is a bit vague: pH = 5 may show solutions of a strong acid, weak acid, buffer solutions or solutions of some salts. Assuming that it is a strong acid - after its neutralization, you will have a known concentration of magnesium ions (it is counted). By inserting this concentration into the solubility product you will get the cubic equation. After solving it, you'll get the total magnesium concentration as (0.5·10^-5 + x)

[sharbeldam]

Thanks AWK.
Can you try to explain to me simply what actually happens after the reaction of OH- with all of H+.

I got that I have only Mg+2 left, but then what happens? I can add more Mg(OH)2? or because all of the OH- reacted, according to chatelier principle, since OH- concentration went lower, the reaction will go to the right, so I will get this:

Mg(OH)2 <------> Mg+2     + OH-1
-x                         +x             +2x

So what's next, how do i figure out the grams of Mg(OH)2 that was added? And why did the professor say that the ratio of moles between Mg+2 concentration and Mg(OH)2 is 2:1... even though it's equilibrium and not a one way reaction.

[AWK]

Neutralization of a strong base (Mg(OH)₂) with a strong acid (e.g. HCl) will form a neutral solution (although the pH may be slightly different from 7 due to the low ionic strength). Then you will dissolve magnesium hydroxide in a solution containing Mg²⁺ of known concentration. If, as a result of solving the cubic equation, the OH^- ion concentration was less than 20 times lower (or greater) than 10^-7, the calculation result would have to be corrected iteratively.

K_sp=(0.5·10^-5+x)·(2x)²

[sharbeldam]

Okay and one last thing, After i solve for x, i can find the moles of Mg+2 added, would it be equal to the moles of Mg(OH)2?
If the reaction was like that Mg(OH)2-->Mg+2 + 2OH-
then sure it will be equal, but since we have equilibrium here, doesn't that change things??

Thanks for always being a great help AWK

[AWK]

(0.5·10^-5+x)·58.32 if [OH^-] is sufficiently different from 10^-7

[sharbeldam]

(0.5·10-5+x) is the moles of Mg+2, so you are saying that we treat the reaction as if it wasnt equilibrium and the ratios are treated as if it was a one way reaction? hence Mg(OH)2 has the same number of moles?

[AWK]

The neutralization reaction goes to the end. Then you have an equilibrium problem of the solubility product with an excess of a common ion.

[Borek]

After i solve for x, i can find the moles of Mg+2 added, would it be equal to the moles of Mg(OH)2?

How can it be different? This is just a simple stoichiometry that you need to follow.

[sharbeldam]

After i solve for x, i can find the moles of Mg+2 added, would it be equal to the moles of Mg(OH)2?

How can it be different? This is just a simple stoichiometry that you need to follow.

I will elaborate again what's confusing me, sorry if i sound a bit dense in general chem.

A + B <-------> 2C

If you have 2 molar of C in an equilibrium, would it mean you have 1 molar of A? usually no in many problems that I have done.

i.e

A   +   B <--------> 2C
2M      2M               0
-x       -x                 +2x
2-x      2-x               2x

2-x doesn't equal 2x in a usual equilbrium problem.

Back to my problem.
I understand that the OH reacts with H+ not in equilbrium and it's 1:1, but I have some magnesium ions left at the end, but it's equilbrium equation
Mg(OH)2 <-----> Mg+2  + 2OH-

Let's say I have 10 moles left of Mg+2, why should the Mg(OH)2 be 10 moles, still it's still an equilbrium like the example of (a/b/c) above?
Am i having a brain attack ? why is this different?

honestly if it was Mg(OH)2 -----> Mg+2 + 2OH-
i would have no problem

[AWK]

Could you solve the first step of your task: How many grams of Mg(OH)₂ can be dissolved in 1 L of 0.000005 M MgCl₂ solution?

[sharbeldam]

Could you solve the first step of your task: How many grams of Mg(OH)2 can be dissolved in 1 L of 0.000005 M MgCl2 solution?

MgCl2         -->         Mg+2             +                  2Cl-
 0.000005moles    0.000005moles               

Mg(OH)2<----->Mg+2        +         2OH-
XXX-s                  0.000005+s        +2s

ksp=[0.000005+s][2s]^2
[neglecting +s]
ksp=0.000005*4s^2

then i solve for s
so i can find the concentration of Mg+2 and OH-
but i dont know to find the grams of Mg(OH)2 because, should the moles be " s " ? since i added (XXX) and only s dissolved since it's an equilbrium

[AWK]

Next step: 0.000005 M MgCl₂ comes from neutralization of Mg(OH)₂.
Compare 0.000005 with the value of "s".
Maybe the solubility of magnesium hydroxide in pH=5 is nearly the same as in water