Comparing a hydronium with a hydroxide in a weak acid solution (0.1M)

[HazyMan]

TL;DR Summary

I am trying to find out why the hydronium molarity is greater than the hydroxide molarity in this solution of a weak acid "HA".

This question says: An HA weak acid solution with a molarity of 0.1M is dissolved in water. In the new solution, is the molarity of OH- greater than the H3O+ molarity, or the opposite? Or are they equal?

I came up with two possible answers:

1. [H3O+]>[OH-] because there are no hydroxides involved at all, but this leads me to the second answer, which IS wrong according to my solution book but i just wanted to mention it..

2 [H3O+]=[OH-]. I came up with this because if you replace "A" with a hydroxide, you will end up with HOH+H2O<->H3O+OH which is basically water self-ionization.

I'm just not sure if my first answer is correct either, because the solution book simply says that [H3O+]>[OH-], implying that hydroxides are indeed involved.

Is my answer (the first one) correct?

 

[Borek]

Are you aware of the water autodissociation? Do you know what the water ion product is?

(if not: google them)

 

[HazyMan]

Are you aware of the water autodissociation? Do you know what the water ion product is?

(if not: google them)

That's what i thought of regarding the second answer i came up with. However if that was the case, wouldn't the hydronium molarity be equal to the hydroxide one? In the book it says that this is not the correct answer.
I didn't think of the ion product. If i substitute that in then i get Kw=([H3O+][OH-])/[HA]

 

[Borek]

wouldn't the hydronium molarity be equal to the hydroxide one?

In a pure water - yes. But you have an acid solution, so there is more than one source of H⁺.

If i substitute that in then i get Kw=([H3O+][OH-])/[HA]

No idea what you did here nor how, but it is definitely wrong.

 

[HazyMan]

In a pure water - yes. But you have an acid solution, so there is more than one source of H⁺.
No idea what you did here nor how, but it is definitely wrong.

Isn't the water ion product notated as Kw? Isn't it an equilibrium constant?

 

[Borek]

Isn't the water ion product notated as Kw? Isn't it an equilibrium constant?

Yes to both, it still doesn't make the formula correct.

If it is equilibrium just for water, why do you put HA there?

 

[HazyMan]

Yes to both, it still doesn't make the formula correct.

If it is equilibrium just for water, why do you put HA there?

i forgot that OH was substituted in A. So it's Kw=[H3O+][OH-] ?

 

[Borek]

Yes.

I feel like you are trying to reinvent the wheel instead of consulting any source.

http://www.chembuddy.com/?left=pH-calculation&right=water-ion-product

https://en.wikipedia.org/wiki/Self-ionization_of_water

 

[HazyMan]

Yes.

I feel like you are trying to reinvent the wheel instead of consulting any source.

http://www.chembuddy.com/?left=pH-calculation&right=water-ion-product

https://en.wikipedia.org/wiki/Self-ionization_of_water

I am familiar with these but yes, i usually tend to try to understand something and its origins instead of simply accepting it. It helps me understand more

 

[HazyMan]

Wait, i think i realized something. All along i thought i had to replace A with OH but now i get what's going on. I'm supposed to compare the overall H3O+ concetration with the OH- concetration which is only produced by the water's self ionization, which as a phenomenon, is known to be quite rare. So it is obvious that H3O has a bigger concetration simply because of the HA acid being dissolved ALONG with the water