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Henderson Hasselbalch equation question

by elegysix
Tags: equation, hasselbalch, henderson
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elegysix
#1
Jul14-11, 12:27 PM
P: 314
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-]|
Initial)------|1/80-----|1/80--------||0---|
Change)----|(-u)----- |(-u)-------- ||+u--|
Equilibrium)-|(1/80-u)--|(1/80-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
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