molarity question

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905084274
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Joined: Fri Sep 28, 2018 12:22 am

molarity question

Postby 905084274 » Mon Jan 14, 2019 6:24 pm

For the 6th edition textbook homework 12.23 part (a), how do you know that since the Kw provided is 2.1x10^-14 that that is equivalent to x^2. So the problem can be set up as (H3O+)(OH-)=x^2. Is this because if you take the square root of 2.1x10^-14 is 1.4x10^-7 mol/L? I wouldn't have thought of this as a way to solve this problem, so some clarification would be nice :)

Elisa Bass 4L
Posts: 61
Joined: Fri Sep 28, 2018 12:23 am

Re: molarity question

Postby Elisa Bass 4L » Mon Jan 14, 2019 6:49 pm

In pure water, the H3O+ and OH- concentrations are equal. Therefore, because they gave you Kw, you can find each of the concentrations by square rooting the given value.

Jennifer Su 2L
Posts: 47
Joined: Wed Nov 21, 2018 12:20 am

Re: molarity question

Postby Jennifer Su 2L » Mon Jan 14, 2019 10:05 pm

Just remember that pure water has the neutral pH of 7, so as the comment above states, [H3O+] and [OH-] should always be equal. If they are not equal, the solution will either be more acidic (higher concentration of H3O+) or more basic (higher concentration of OH-).


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