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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: ghostanime2001 on September 19, 2012, 01:29:20 AM

Title: metal oxide equilibrium
Post by: ghostanime2001 on September 19, 2012, 01:29:20 AM
This is from a worksheet in Grade 12 Chemistry. This is question 11.f) I know how to do all the other compounds except this one:

When the following substances are introduced into water do the resulting equilibria favour products or reactants?
 c)MgO

I know that when I introduce a metal oxide in water, the product will be a metal hydroxide so:

MgO + H2O  :rarrow: Mg(OH)2

I don't know what to do next or how to calculate the equilibrium constant for magnesium oxide, I know the Ka of the reactant acid is H2O but I do not know the Ka of the product acid because it is a base.  So what should I do next ? I have uploaded the worksheet I was given in Grade 12 Chemistry. Thanks so much for anyone's help !!

{MOD Edit: Remove attachments}
Title: Re: metal oxide equilibrium
Post by: Arkcon on September 19, 2012, 08:50:17 AM
Please only ask one question.  You reproduced 4 pages of a textbook, that is getting close to a copyright violation.  Those pages had dozens of questions, none of which matched the one you asked  11.F)  And we'd like to see a little work, that includes typing in the problem into this forum when possible.  If you feel the need to repost the scanned question, then zoom in on the image to ask just the question at hand.
Title: Re: metal oxide equilibrium
Post by: Borek on September 19, 2012, 09:08:37 AM
He meant A-11, point f.
Title: Re: metal oxide equilibrium
Post by: Arkcon on September 19, 2012, 09:11:19 AM
Ah.  OK.  ghostanime2001:, lets try and have the question again.  Sorry.
Title: Re: metal oxide equilibrium
Post by: ghostanime2001 on September 19, 2012, 11:03:31 AM
I have included the original two pages from last time and another 2 pages incase other information need be required to solve.
Title: Re: metal oxide equilibrium
Post by: Borek on September 19, 2012, 11:10:15 AM
As far as I can tell there is no way to answer the question using methods and information from this chapter.
Title: Re: metal oxide equilibrium
Post by: ghostanime2001 on September 19, 2012, 11:44:24 AM
Should I upload page 3 ? lol I did not feel the need to upload page 3 because the information on that page deviates from the method on page 4 and page 5. Also, if anyone knows from what textbook this information comes from, that would really help I could search for the book. My Chemistry teacher told me this is from the 1970's or 80's :S
Title: Re: metal oxide equilibrium
Post by: Xilaim on September 19, 2012, 03:41:08 PM
Determine the equilibrium constant is not necessary:
MgO concentration (solid) = 1
The concentration of H2O (water surplus) = 1
The concentration of Mg(OH)2 (solid) = 1

answer: equilibria favour products.
Title: Re: metal oxide equilibrium
Post by: Borek on September 19, 2012, 03:56:46 PM
Determine the equilibrium constant is not necessary:
MgO concentration (solid) = 1
The concentration of H2O (water surplus) = 1
The concentration of Mg(OH)2 (solid) = 1

answer: equilibria favour products.

Following this logic Al2O3 should react with water producing Al(OH)3. It doesn't.
Title: Re: metal oxide equilibrium
Post by: Xilaim on September 20, 2012, 11:26:24 AM
 ;) Ok Ok
Here's the formula for the solution to the Gibbs energy: lgK = ΔG298/(-2.3*R*T)
R = const = 0.0083
T = 298K
Title: Re: metal oxide equilibrium
Post by: Borek on September 20, 2012, 11:41:08 AM
;) Ok Ok
Here's the formula for the solution to the Gibbs energy: lgK = ΔG298/(-2.3*R*T)
R = const = 0.0083
T = 298K

Yes. And area of the circle is [itex]\pi r^2[/itex]. Neither equation helps answer the question.
Title: Re: metal oxide equilibrium
Post by: Xilaim on September 20, 2012, 12:47:16 PM
Wrong equation?
Title: Re: metal oxide equilibrium
Post by: Borek on September 20, 2012, 02:08:32 PM
Show how you are going to use it to solve the problem. Note that you are not given ΔG, and the chapter is on acid/base equilibrium, so all data you have at hand are Ka values.
Title: Re: metal oxide equilibrium
Post by: ghostanime2001 on September 20, 2012, 02:29:24 PM
would knowing the base dissociation constant for O-2 help ? by the way, what is the Kb value for the oxygen anion ?
Title: Re: metal oxide equilibrium
Post by: Xilaim on September 20, 2012, 03:32:22 PM
Show how you are going to use it to solve the problem. Note that you are not given ΔG, and the chapter is on acid/base equilibrium, so all data you have at hand are Ka values.
ΔG - reference have chemistry.

MgO + H2O  ::equil:: Mg(OH)2

Excess water will shift the equilibrium in the reaction products.
The solubility product = [Mg2+] [OH-]2 = 6.0 x 10 ^ -10
K = [OH-]/Kw
Kw = 1e-14
???
Title: Re: metal oxide equilibrium
Post by: ghostanime2001 on September 20, 2012, 03:40:33 PM
This is a strange question. I wish I had gone to my chemistry teacher for help in grade 12.
Title: Re: metal oxide equilibrium
Post by: Borek on September 20, 2012, 04:51:42 PM
???

You are starting to see it is not that obvious.

Not that I doubt reaction is shifted to products, in general that's how oxides of alkaline earth metals behave. But I don't see how to answer this problem in terms of the chapter content.

O2- is not a bad idea, although I don't remember seeing its Kb value.
Title: Re: metal oxide equilibrium
Post by: ghostanime2001 on September 21, 2012, 12:45:41 AM
the hydroxide Ka value is 1 so shouldn't the Kb value be the inverse of Kw ?
Title: Re: metal oxide equilibrium
Post by: Borek on September 21, 2012, 04:14:51 AM
the hydroxide Ka value is 1

Is it? Ka of 1 means quite strong acid.
Title: Re: metal oxide equilibrium
Post by: Xilaim on September 21, 2012, 11:40:34 AM
Kb - Ionization Constants MgO?
Title: Re: metal oxide equilibrium
Post by: ghostanime2001 on September 21, 2012, 01:09:57 PM
sorry not Ka, Kb
Title: Re: metal oxide equilibrium
Post by: Borek on September 21, 2012, 02:23:45 PM
sorry not Ka, Kb

Then it is not 1, no idea where you got this value from.
Title: Re: metal oxide equilibrium
Post by: ghostanime2001 on October 22, 2012, 03:30:23 AM
I just can't seem to figure out how they got 1022 as their Keq ? Is there any documented Ka/Kb values of strong acids or bases that is not normally given in most textbooks except for "very large". I am looking for numerical values as this would allow me to finally compare with the given value inside the worksheet.
Title: Re: metal oxide equilibrium
Post by: ghostanime2001 on March 22, 2013, 07:42:31 PM
I have bought this textbook from amazon and the book title is Chemistry: Theory and Problems Book Two from James A. Hebden. I found the acid dissociation constant for OH- at the back of the textbook. The value listed is a inequality:
KA (OH-) < 10-36

First of all, MgO is an ionic compound am I correct ? so in the presence of water, it will dissociate completely:

MgO + H2O ::equil:: Mg2+ + O2- then O2-will hydrolyze so..

O2- + H2O  ::equil:: OH- + OH- so the acid on the reactant side is water (red) and the acid on the product side is hydroxide (blue).

so using the method introduced in the uploaded sheets on the first page, finding the equilibrium constant of magnesium oxide in water would be:

Keq = Ka(H2O) / Ka(OH-) = 1 x 10-14 / < 10-36 = ??

I know that if I just calculate 10-14 / 10-36 = 1022 I get the correct answer, but how would I solve the equation with the inequality ? I know I'm very close, please guide me !!
Title: Re: metal oxide equilibrium
Post by: Borek on March 23, 2013, 05:24:44 AM
I have a feeling you are tricking yourself into believing you are solving the problem, just because the numbers seem to be close. I can only repeat what I wrote much, much earlier - I don't see how to solve the problem in terms of acid base equilibria, without seriously stretching the theory.
Title: Re: metal oxide equilibrium
Post by: ghostanime2001 on March 23, 2013, 12:54:38 PM
The value reported in my last reply is the exact one written in the textbook so I am not tricking myself into making numbers work out to the correct answer. Why can't this problem be worked out using acid-base equilibria ? if you are so confident, then show me why it can't work, have you looked at the worksheets I uploaded on the first page ?
Title: Re: metal oxide equilibrium
Post by: Borek on March 23, 2013, 03:00:25 PM
O2- doesn't exist in water solutions, just like any other base stronger than OH- doesn't exist in water solutions, so any talk about its Kb is an abuse of the Bronsted-Lowry theory. Just because you can do the calculations doesn't mean they make sense. It is not addressed at you, but at the book.
Title: Re: metal oxide equilibrium
Post by: ghostanime2001 on March 23, 2013, 03:12:25 PM
Then the only explanation I can come up with is that magnesium oxide dissolves to form magnesium hydroxide and then the hydroxide ions hydrolyze to give 1 as the equilibrium constant.

MgO + H2O  :rarrow: Mg(OH)2

Mg(OH)2  ::equil:: Mg2+ + 2OH-

OH- + H2O  ::equil:: H2O + OH-

Keq = Ka(reactant acid) / Ka(product acid) = Kw(H2O) / Kw(H2O) = 1