1. Calculate the pH of the following
0.1 mol dm-3 KOH
0.05 mol dm-3 Ba(oH)2
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How would you work out the pH? watch
- Thread Starter
- 20-03-2016 18:20
Offline22ReputationRep:TSR Support Team
- TSR Support Team
- 20-03-2016 20:02
Posted from TSR Mobile
- 20-03-2016 20:05
As K to OH is 1 to 1.
Wait no yes, pH + pOH = 14
So 14 - pOH = pHLast edited by Dinasaurus; 20-03-2016 at 20:12.
- 20-03-2016 20:17
Exactly what Dinasaurus said.
Antoher method would be the following:
Since Kw = [H+][OH-] and we know that Kw (the ionic product of water) is 1.00 x10^-14 mol^2 dm^-6 and that [OH-] is 0.1 mold dm^-3, then [H+] = Kw / [OH-] = 1.00 x10^-14 / 0.1 = 1.00 x10^-13 and so pH = -log[H+] = -log(1.00 x10^-13) = 13
I prefer the pOH method, but just wanted to show you the other method