12.23  [ENDORSED]

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Abigail Volk 1F
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12.23

Postby Abigail Volk 1F » Tue Dec 05, 2017 11:35 pm

The question asks, " The value of Kw for water at body temperature (37 degrees Celcius) is 2.1x10^-14. what is the molarity of H3O+ ions and the pH of neutral water at 37 degrees Celcius? what is the molarity of OH- in neutral water at 37 degrees Celcius?
I'm having a hard time understanding the question. Please help.
Also I thought we only had to do questions with room temperature (25 degrees Celcius). Don't we have to change the way we do this types of problems to satisfy this? thanks

Jasmine Botello 2F
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Re: 12.23

Postby Jasmine Botello 2F » Wed Dec 06, 2017 8:34 am

I think that as long as the temp stays the same throughout the problem then it should not cause any changes in the way that you do the problem.

Madelyn Gehrich 1E
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Re: 12.23  [ENDORSED]

Postby Madelyn Gehrich 1E » Wed Dec 06, 2017 1:15 pm

Kw = [H3O+][OH-]. So if Kw=2.1 x 10^-14, this is equal to [H3O+][OH-]. This comes from the equation 2H2O (l) --> H3O+ (aq) + OH- (aq), we take x^2 = 2.1 x 10^-14, so x=1.4 x 10^-7. You then take the -log(1.4 x 10^-7) to get a pH of 6.8. Once you have the pH you can calculate the pOH knowing that pH + pOH = 14.

elliekauffman1I
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Re: 12.23

Postby elliekauffman1I » Thu Dec 07, 2017 5:31 pm

Why is Kw=x^2? Doesn't that imply that H3O and OH are the same value? And if so, then how are we supposed to know that?

Mishta Stanislaus 1H
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Re: 12.23

Postby Mishta Stanislaus 1H » Sat Dec 09, 2017 5:11 pm

elliekauffman1I wrote:Why is Kw=x^2? Doesn't that imply that H3O and OH are the same value? And if so, then how are we supposed to know that?


I also need help with this question

Grace Ramey 2K
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Re: 12.23

Postby Grace Ramey 2K » Sat Dec 09, 2017 6:32 pm

[H3O+] = [OH-] when a solution is neutral, so if you set [H3O+] = x, then [OH-] = x. Therefore, if [H3O+][OH-] = 2.1 x 10^-14, then x^2 = 2.1 x 10^-14.

Hope this helps!

Timothy Kim 1B
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Re: 12.23

Postby Timothy Kim 1B » Sat Dec 09, 2017 7:09 pm

So does the Equilibrium constant still equal 10^-14 even if the temperature does not equal 25 degrees?

Katie Lam 1B
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Re: 12.23

Postby Katie Lam 1B » Sat Dec 09, 2017 8:41 pm

The equilibrium constant K changes with temperature so Kw is 10^-14 only at 25 degrees Celsius.


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