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Topic: calculating pH  (Read 3441 times)

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Offline AmyJ

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calculating pH
« on: August 04, 2007, 11:48:11 AM »
I'm having a hard time calculating some pH values.  If given M of KOH and M of Ba(OH)2, how do you get the pH?  They are both strong bases...
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Offline enahs

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Re: calculating pH
« Reply #1 on: August 04, 2007, 11:58:20 AM »
Kw=[H+][OH-] = 1.0 x 10-14 @ 25oC

pH = -log([H+]) = -log([Kw /[OH-])         { log(A/B) = log(A) - log(B) }


pOH = -log([OH-]) -> pH = 14 - pOH         { -log(1.0 x 10-14 = Kw) = 14 }
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Offline AmyJ

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Re: calculating pH
« Reply #2 on: August 04, 2007, 12:13:09 PM »
That is extremely helpful.  It is a solution of the two components though, so I am still unsure how to get the [OH] of the solution (KOH has a different concentration that Ba(OH)2).
Thanks!
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Offline enahs

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Re: calculating pH
« Reply #3 on: August 04, 2007, 12:25:38 PM »
Both KOH and Ba(OH)2 are strong bases, that is, they completely dissociate; right?

KOH -> K+ + OH-

Ba(OH)2 ->  ? + ?

What is the total concentration of OH- from the two bases.
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Offline AmyJ

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Re: calculating pH
« Reply #4 on: August 04, 2007, 12:42:43 PM »
Thank you, I got the correct answer.  I did not realize I had to add the [OH] of each component.
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