Annoying Kw question

1. If the PH of pure water at 5c is 7.6 calculate the value of KW at this temperature.

so for this question i went kw=[H+] [OH-] and the since i know ph i shift logged 7.6 to give me a value of 2.51 * 10^-8. I know the H+ ions are equal to OH- ions in water so H+^2 = KW = 2.51*10^-8 and then i square rooted it to find out the Kw value so square root 2.51*10^-8 is 6.31*10^-16. The textbook said i was right so i moved onto my next question.

2. At 338k the ph of pure water is 6.5. calculate Kw at this temperature.

So i started as the same as question 1. I shift logged 6.5 to get H+ and it was 3.16*10^-4, then knowing h+=oh- in pure water i kept 3.16*10-4 and square rooted but the textbook says im wrong :frown:

can someone explain.

Reply 1

oceane

pH = -log [H+]
[H+] = 10^(-6.5) You shift log minus 6.5
[H+] = 3.16 *10^(-7) Which is what you got but 3 powers of 10 out
Kw = [H+] [OH-] and [H+]=[OH-] so you square the concentration of H+ to give you 1 *10^3, not the square root (although as you got the first bit of the Q right I take it you meant square)

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