At 353 kelvin the ph of 0.001 M KOH will be

1.	At 353 kelvin the ph of 0.001 M KOH will be
Answer» At 363, Kw = 3.80X10^-13 So, Kw = [H+][OH-] = 3.80X10^-13 In this SOLUTION, [OH-] = 0.001, so: [H+] = 3.8X10^-13 / 0.001 = 3.80X10^-10 PH = -log [H+] = 9.4 :)

Answer»

At 363, Kw = 3.80X10^-13

So, Kw = [H+][OH-] = 3.80X10^-13

In this SOLUTION, [OH-] = 0.001, so:

[H+] = 3.8X10^-13 / 0.001 = 3.80X10^-10

PH = -log [H+] = 9.4

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