Acid/base equilbrium problems

  1. What is the pH of 0.15 M NH4Cl?
    Okay, I set up my ICE table,

    NH4 + H2O <---> H3O+ + NH3-
    I 0.15 M --- --- ---
    Δ -x --- +x +x
    E 0.15-x --- x x

    KbNH3-=1.8x10-5
    So Kw=1.0e-14=Ka*Kb, so KaNH4=5.56e-10

    So, Ka=[H3O+][A-]/[HA]

    5.56e-10=x^2/0.15-x
    [H+]=[H3O+]=[x]=9.13e-6

    pH=-log[H+]
    pH=5.04

    I understand the work, but my question is why can't I take the -log[x] from kb=[OH-][HA]/? Is it because I have the Kb of NH3 and not the Kb of NH4? Thanks!
     
  2. jcsd
  3. Borek

    Staff: Mentor



    Huh? -log[Kb] is pKb. I suppose that's not what you mean, but what you wrote makes no other sense to me.
     
  4. Oh sorry, I meant to say -log(x) as in taking the -log(H3O+) (from the original ice table)
     
  5. Borek

    Staff: Mentor

    I still have no idea what you mean by "-log(H3O+) (from the original ice table)".

    H3O+ in your ICE table is either 0 (initial value) - you can't take log of zero, or x - which is an unknown, so you can't calculate logarithm of its value before solving the problem.
     
  6. NH4 has no Kb because it is an acid, not a base. I honestly have no idea what your question is even asking.
     
  7. I'm still not sure what you're asking. You take the -log(x) once you solve for it to get the pH. If you used the Kb, then you would do 14.00 + log(x) because x in that case would equal the hydroxide concentration, so to get pH you must subtract the pOH from 14.00.
     
  8. Borek

    Staff: Mentor

    Not that you are wrong, but KaKb=Kw, so you can give both for any acid/conjugate base pair. People often abbreviate it and say Kb of an acid or Ka of a base.

    That beiung said, I doubt that's what the OP meant.

    You answered a spammer quoting a random part of the first post :tongue2:
     
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