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Chemistry Acid Base HELP PLEASE

  1. Dec 19, 2006 #1
    1. The problem statement, all variables and given/known data

    A vessel contains 500mL of .1 molar H2S solution. For H2S Ka1 = 1 x 10^-7
    Ka2 = 1.3 x 10^-13.

    What will the pH be when 800mL of .1molar NaOH has been added.

    2. Relevant equations

    ph = -log([H])
    ph = pka + log(B/A)

    3. The attempt at a solution
    This is how I started, I found the moles of each. I found I started with .08 moles of NaOH and .05 moles of H2S, subtracted them to get .03 moles of NaOH. Then .03 / (.5 + .8) to get the concentration (total volume), then because all thats left is strong base I took the -log(.023) and then subtracted that from 14 to get the pH. I keep ending up with 12.36 but the answer is 13.1. I need to be able to do this for my exam tomarrow, can anyone help?
    Last edited: Dec 19, 2006
  2. jcsd
  3. Dec 19, 2006 #2


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    What are k1 and k2?
  4. Dec 19, 2006 #3
    Ka values, it is a polyprotic acid meaning its Hydrogens leave in steps and it has a different Ka value for each Hydrogen.
  5. Dec 19, 2006 #4


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    Well, I've not studied chemistry for quite a while now, but your reasoning seems fine to me. Sorry I can't be anymore help, but perhaps someone with a Chemistry background will read this and be able to help!
  6. Dec 19, 2006 #5
    No - the reasoning is incorrect. H2S is not a strong base. It is a weak acid. Therefore it will disassociate in two steps, each one with a different Ka value.

    Write an equation for equilibrium concentrations for the reaction which produces a hydronium ion and a deprotonated hydrogen sulfide. Then write another one which produces the hydronium ion and the sulfide ion. (Remember that in between these two reactions, the hydronium concentration is retained.)

    It's a polyprotic disassociation, as you said.
  7. Dec 19, 2006 #6
    H2S <==> HS + H

    HS <==> H + S

    I did this and I calculated the concentration of H at both steps
    I don't know how this helps me though the concentration of the H ends up being around 1 x 10^-4.
  8. Dec 19, 2006 #7


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    You have .1 mole equivalent amount of acid and .08 moles of NaOH, in relative terms, which mean that after the NaOH has been consumed, you're going to be left with concentrations of S 2- and HS-.......use the Henderson Hasselbach equation using the second Ka to find the pH of the resulting solution.
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