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Homework Help: Henderson Hasselbalch equation question

  1. Jul 14, 2011 #1
    1. The problem statement, all variables and given/known data
    We are given Ka = 1.8x10^-5 for acetic acid.
    We have a weak acid and a strong base, HC2H3O2 (acetic acid) and NaOH (sodium hydroxide).
    We have 5 mL of acid and 5 mL of base, both at .1M, and 30mL of H2O

    If nothing else, how do you know when to use the Henderson-Hasselbalch equation?

    2. Relevant equations
    HH eqn : pH=pKa + log(X-/HX)

    3. The attempt at a solution

    If we have the situation

    HX + H2O <--Ka--> (H30+) + (X-)

    NaOH + H2O <-> (Na+) + (OH-) + H2O

    (Na+) + (OH-) + (H30+) + (X-) + <-> 2*H20 + (X-) + (Na+)

    Na+ has negligible effects on pH

    HX + NaOH + H2O <-> 2*H2O + (X-)

    If I want to find the pH of this solution, can I use the HH equation?
    if so, the Ka that I'm given is for dissociation of HX in water - not the dissocation of HX in a solution containing a base. Can I still use that Ka in the HH eqn? if I can, why does that work?

    pH = pKa + log ( X-/HX)

    pH = -log(1.8x10^-5) + log (X-/HX)

    initial concentrations: [HX], [NaOH] = 5mL*.1M/40mL =1/80 M

    ------------|[HX] ----|[NaOH]------||[X-]|
    Change)----|(-u)----- |(-u)-------- ||+u--|

    using Ka = (aq)products/ (aq)reactants, Ka = 1.8x10^-5, then 1.8x10^-5= u/(1/80-u)^2 -> say we find u= B

    then is the pH given by

    pH= pKa + log( X-/HX ) ?

    pH = -log(1.8x10^-5) + log( B/(1/80-B) ) ?

    What I suspect is wrong:
    1) using the 1.8x10^-5 as the Ka in the HH eqn. I feel like this value would be different since we are putting the acid into a basic solution and not water.
    2) using the 1.8x10^-5 as the Ka to find the change in concentration, u, in the I.C.E. table, for the same reason.

    are these things wrong? what should I do?
    thanks for any help guys
  2. jcsd
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