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PH calculation

  1. May 24, 2005 #1
    Here is the problem:

    Given pKa=9.3 what is pH of 0.005 mol/l ammonia solution.

    The problem is, I don't get this pKa part. I mean, I know t means Ka=5.0119*10^-10 but ammonia is a base, not acid.

    G
     
  2. jcsd
  3. May 24, 2005 #2

    Borek

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    pKa + pKb = pKw
     
    Last edited by a moderator: Aug 13, 2013
  4. May 24, 2005 #3
    I am a moron :( We spent more than hour discussing it yesterday.

    OK, so Kb=1.9953*10^-5.

    But there is still something wrong.Ammonia is a weak base, so pH=(0.005*Kb)^0.5=3.5? This is way too low.

    G
     
  5. May 24, 2005 #4

    Borek

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    Please do not quote signatures.

    No. What you have calculated is pOH, not pH. Besides, check the degree of dissociation - are you sure you can use simplified equation?
     
    Last edited by a moderator: Aug 13, 2013
  6. May 24, 2005 #5
    Ah, so the pH is 10.5!

    G
     
  7. May 24, 2005 #6

    Borek

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    NO!

    Equation you have used - [OH] = sqrt(C*Kb) is valid only if the degree of dissociation is below 5%. In this case it is higher:

    3.16e-4(dissociated ammonia concentration)/0.005(ammonia concentration) * 100% = 6.3%

    so you have to use full quadratic equation.

    Smarkotan oz gluthozmaz
     
    Last edited by a moderator: Aug 13, 2013
  8. May 27, 2005 #7
    :cry: I missed your last post and my result was wrong :cry:

    G

    PS Check my new problem, please...
     
  9. May 27, 2005 #8

    GCT

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    alright, take it step by step...to go from pKa to Ka, simply negative inverse log of pKa. KaKb=Kw, Kb=Kw/Ka (Kw is the autodissociation of water).

    [tex]Kb=[OH-][NH4+]/[NH3],~Kb=[x][x]/[initial~conc.~NH3-x][/tex] solve for x, and that will be [0H-]. pOH=-log[0H-], pH+p0H=14, solve for pH.

    that's it, you shouldn't be having so much trouble with this
     
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