[lena]

Please help me!
The concentration of Mg2+ ions in seawater is about 1.3 µg’L-1. In commercial (Dow Chemical) recovery process, the magnesium is precipitated as the hydroxide. At what pH does magnesium hydroxide precipitate?

[Donaldson Tan]

Ksp = [Mg2+][OH-]²

pKsp = -lg[Mg2+] + 2pOH

pH + pOH = 14

[lena]

How do you think Can I do it this way?

 Ksp Mg(OH)2 =1.8*10-11
 Ksp=[Mg2+] [OH-]
1.8*10^-11=[1.3*10^-6] [OH]^2
2x^2=1.8*10^-11/1.3*10^-6
x=square root of 1.38*10^-5/2
x=2.63*10^3= [OH]

pOH=log[OH]=2.58
pH=14-2.58=11.42
Thank you!

[AWK]

You nee molar concentration of Mg(2+)
Method 2x^2 does not wotk with external OH(-) ions.

[lena]

But the molar concentration of Mg(2+) given 1.3 µg’L-1,isn't it? I was trying to find [OH]
I stuck!
Can't find mistake!

[Demotivator]

You have
Mg(2+) given 1.3 µg'L-1 =[1.3*10^-6]
But that is grams/liter, not moles/liter as the ksp equation requires.

2) solve for [OH-]^2, not [2xOH]^2

[lena]

Thank you very much! I got 5.4*10^-8 M. Is it correct?
2. Could you please explain why I need to solve for [OH]^2? The equasion Mg(OH)2->Mg2+ +2OH
and at euilibrium we got  [Mg]=5.4*10^-8, [OH] =2x+5.4*10^-8. Am I wrong?

[Demotivator]

yes, mg2+ is  5.4*10^-8 M.

2) If you were asked for the solubility of Mg(OH)2, then you would use x * (2x)^2 since 1 molecule dissociates to produce 2 OH.
But that is not the problem here. All you have is Mg in seawater. The Mg could come from anywhere, not necessarily Mg(OH)2. The OH can come from anywhere as well, not from Mg(OH)2. The question is what is the[OH] the Mg must see in order to precipitate?
the equation is simply
ksp = [Mg2+][OH-]^2

[lena]

Thank you very much!!! You really help me!!!
Would you mind to help me with another problem?
A 3.00L reaction vessel is filled with 0.150mol CO, 0.0900mol H2 and 0.180 mol CH3OH. Equilibrium is reached in the presence of a Zink oxide –chromium (lll) oxide catalyst, and at 300C, Kc=1.1 *10-2 for the reaction, CO (g) + 2H2 (g) óCH3OH (g)
a)As the reaction approaches equilibrium, will the molar concentration of  CH3OH increase, decrease or remain the same? Explain.
What is the equilibrium composition of the mixture?

1) I found the molar concentration of each :
CO=0.05M
H2=0.03M
CH3OH=0.06M
2) Kc=[CH3OH]/[CO][H2]
What should I do next?

[Demotivator]

1) plug in the initial molar concentrations into
 [CH3OH]/[CO][H2]
and see if the number you get is bigger or smaller than Kc. Since you know that at equilibrium it must equal Kc, you can deduce if there's too much of CH3OH to start with or too little.

[lena]

 :)Thanks a lot for your *delete me*!!

[lena]

Can I bother you with the last one? I stuck in a half way.
 Let the equilibrium constants for reactions 2H2O (g)<-> 2H2 (g) + O2(g) and 2CO2 (g) <->2CO(g)+ O2 (g)be Kp1 and Kp2 respectively. Show that equilibrium constant for reaction CO2(g) + H2(g) <->H2O(g) + CO(g) Is Kp3 = (Kp2/Kp1)in 1/2(power), and evaluate it at 1565 K, at which temperature Kp1 +1.6*10-11 and Kp2 = 1.3*10-10.

Here what I got so far.
I reversed the equasion and get rid of 2s and O2.
2H2O (g)<-> 2H2 (g) + O2(g) a     Kp1
 2CO(g)+ O2 (g) <->2CO2 (g)      Kp2
H2O+CO<->H2+CO2                  Kp3

Kp1*Kp2=Kp3
Kp1Kp2=[H2][CO]^2/[H2O][CO2]
Is it correct?
What shouls I do next?
I think I made a mistake when I reversed the equation.
Thank you very mush for all your *delete me*

[Demotivator]

You are told that the answer is kp3= (kp2/kp1)^1/2 and to prove it, so
Kp1*Kp2=Kp3 is not correct.

2H2O (g)<-> 2H2 (g) + O2(g)     Kp1  
but you need to reverse the above equation (not the second) in order to get 2H2 + 2CO2 <->2H2O+ 2CO :

so,
2H2 (g) + O2(g) <-> 2H2O (g)  1/kp1  (notice that when an equation is reversed, the K becomes the inverse 1/K)
2CO2 (g) <->2CO(g)+ O2 (g)     Kp2
     
adding:
2H2 + 2CO2 <-> 2H2O+ 2CO
   therefore kp3 = (1/kp1)*kp2 = kp2/kp1

Now, can you see how the above becomes
(kp2/kp1)^1/2 ?

[Tetrahedrite]

Because sea water is so "salty", using molar concentrations will lead to large errors. In these situtions, to be accurate you need to use activity and ionic strength

[Demotivator]

True, but  the student must work within the context of the course.