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April 20, 2014

Homework Help: Chemistry, #1

Posted by Taasha on Friday, August 3, 2007 at 4:10pm.

At 40 degrees C, the value of Kw is 2.92 x 10^-14.
a)Calculate the [H+] and [OH-] in pure water at the same temperature.

b)What is the pH of pure water at that temperature?

I know how to do this with Kw being 1.0 x 10^-14 at 25 degrees C, but how do I do this with a different temperature?

Same process but use the NEW Kw listed at that temperature. pH should be about 6.77. pOH also will be about 6.77.

pH and pOH added together =14.
6.77+6.77=13.54

But, using a pH of 6.77, you get an [H+] of 1.698 and an [OH-] of 1.72 x 10^-14. Is that what you get?

Concept of this is that ..
since you know how to get the H+ concentration at 25 degrees ...
Both H+ concentration and OH- concentration should be equivalent that is why if you square root the Kw of water at 25 degrees you would get 1.0x 10^-7 and thus (1.0x10^-7)2 = 1.0 x10 ^-14 right?

In your case you have a Kw of 2.92x 10^ -14 so....if you square root that ... you get...1.708 x10^-7 for BOTH OH- and H+ concentrations...and the pH of course would be -log (1.708x10^-7)= I trust you can get this now so I'll leave that up to you....

hope it helps...=)

To add to Christina's post, pKw = -log Kw = -log 2.92 x 10^-14 = 13.53 so at pH = pOH = 6.77 then pH + pOH = 6.77 + 6.77 = 13.54. You WON'T get 14 for pKw because it isn't at room temp. That's why Kw isn't listed as 1 x 10^-14.

Thanks so much! I'm having a really hard time with this acid/base stuff!

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