[ghostanime2001]

This is a strange question. I wish I had gone to my chemistry teacher for help in grade 12.

[Borek]

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.

O^2- is not a bad idea, although I don't remember seeing its Kb value.

[ghostanime2001]

the hydroxide Ka value is 1 so shouldn't the Kb value be the inverse of Kw ?

[Borek]

the hydroxide Ka value is 1

Is it? Ka of 1 means quite strong acid.

[Xilaim]

Kb - Ionization Constants MgO?

[ghostanime2001]

sorry not Ka, Kb

[Borek]

sorry not Ka, Kb

Then it is not 1, no idea where you got this value from.

[ghostanime2001]

I just can't seem to figure out how they got 10²² as their K_eq ? Is there any documented K_a/K_b 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.

[ghostanime2001]

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:
K_A (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 Mg²⁺ + O^2- then O^2-will hydrolyze so..

O^2- + H₂O  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:

K_eq = K_a(H₂O) / K_a(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 !!

[Borek]

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.

[ghostanime2001]

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 ?

[Borek]

O^2- 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.

[ghostanime2001]

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  Mg(OH)₂

Mg(OH)₂  Mg²⁺ + 2OH^-

OH^- + H₂O  H₂O + OH^-

K_eq = K_a(reactant acid) / K_a(product acid) = K_w(H₂O) / K_w(H₂O) = 1