Finding Concentrations of H3O+ and OH-

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Tessa Lawler 1A
Posts: 57
Joined: Fri Sep 28, 2018 12:18 am

Finding Concentrations of H3O+ and OH-

Postby Tessa Lawler 1A » Mon Jan 21, 2019 11:09 pm

We did an example in class on Monday of last week where we had:
0.0030M Ba(OH)2 (aq) at 25 degrees Celsius
Then we needed to find the concentrations of H3O+ and OH- in the solution.
So, the reaction is as follows:
Ba(OH)2 --> Ba2+(aq) + 2OH-
Since Ba(OH)2 is a strong base, we assume it dissolves completely, and get:
Kw=[H3O+][OH-]
Then, because the stoichiometric ratio of Ba(OH)2 to 2OH- is 1:2, and the concentration of Ba(OH)2 is 0.0030M, we assume 0.0060 as the concentration for 2OH-. We then rearrange the Kw to solve for [H3O+] -
[H3O+]=(1X10-14)/0.0060, and get 1.7x10-12 M for [H3O+].
My question is this: Why is it that we can assume [OH-] is 2x [BaOH2], but we can't assume that [H3O+] is 1x [BaOH2]? Why do we plug in 0.0060 for [OH-] instead of plugging in 0.0030 for [H3O+]?

Ibrahim Malik 1H
Posts: 65
Joined: Fri Sep 28, 2018 12:27 am
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Re: Finding Concentrations of H3O+ and OH-

Postby Ibrahim Malik 1H » Mon Jan 21, 2019 11:15 pm

[H3O+] is not 1x [BaOH2] because the equation we set up for the dissociation of Ba(OH)2 only includes Ba and OH. We can only use mole ratios to determine the concentration of [OH] because it is part of the dissociation equation of Ba(OH)2 whereas [H30] is not part of the equation so we cannot use mole ratios to determine the concentration.

Amar Singh
Posts: 31
Joined: Fri Sep 28, 2018 12:20 am

Re: Finding Concentrations of H3O+ and OH-

Postby Amar Singh » Sun Jan 27, 2019 3:11 pm

As Ibrahim said, [H3O+] is not included in the equation, so we cannot use mole ratios directly to find its concentration. A big mistake that some might do is create some incorrect equation to account for [H3O+] to try and directly solve for it with the initial concentration. We use Kw to solve it because it must remain constant, and if we know that the hydroxide concentration increases, the hydronium concentration decreases to adjust for the ratio.


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