Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
Question about Kw, ion product constant of water? ?
At 0C the ion product constant of water, Kw, is 1.2x10^-15. The pH of pure water at this temperature is:
a) 6.88
b) 7
C) 7.46
d) 7.56
I got that the answer was C by looking up Kw at different temperatures. How can I solve the problem without looking up the Kw value?
Thanks!
3 Answers
- Johnny DLv 71 decade agoFavorite Answer
Kw=[H+][OH-]
pure water has [H+]=[OH-] so
Kw=[H+][H+] or Kw=[OH-][OH-] or simply Kw=x²
1.2x10^-15=x²
x=√(1.2x10^-15)
=3.46x10^-8=[H+]
pH=-log[H+]
=-log(3.46x10^-8)
=7.46
- ?Lv 45 years ago
Yes, the ion product constant of water changes with temp. With increase in temp. Kw increases and with decrease in temp.Kw decreases. At 10 deg. C Kw is the square of H+ ion conc.: Kw = 5.431 x 5.431 x 10^-16 Kw = 29.5 x 10^-16 Kw = 2.95 x 10^-15
- ChrisLv 71 decade ago
You have to know, or be given, the fact that at 0ºC, the Kw is 1.2x10^-15. Then you say that x² = 1.2x10^-15 or x = 3.5x10^-4. Taking logs of both sides gives you lg(x) = -7.46 and pH = 7.46
