November 26, 2024, 03:51:06 AM
Forum Rules: Read This Before Posting


« previous next »
Pages: [1]   Go Down

Topic: calculating pH  (Read 3545 times)

0 Members and 1 Guest are viewing this topic.

Offline AmyJ

  • New Member
  • **
  • Posts: 7
  • Mole Snacks: +0/-0
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...
Logged

Offline enahs

  • 16-92-15-68 32-7-53-92-16
  • Retired Staff
  • Sr. Member
  • *
  • Posts: 2179
  • Mole Snacks: +206/-44
  • Gender: Male
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 }
Logged

Offline AmyJ

  • New Member
  • **
  • Posts: 7
  • Mole Snacks: +0/-0
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!
Logged

Offline enahs

  • 16-92-15-68 32-7-53-92-16
  • Retired Staff
  • Sr. Member
  • *
  • Posts: 2179
  • Mole Snacks: +206/-44
  • Gender: Male
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.
Logged

Offline AmyJ

  • New Member
  • **
  • Posts: 7
  • Mole Snacks: +0/-0
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.
Logged
Pages: [1]   Go Up
« previous next »
Sponsored Links