Henderson Hasselbalch equation question

[elegysix]

Homework Statement

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?

Homework Equations

HH eqn : pH=pKa + log(X-/HX)

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

 

[nate808]

!

Thank you for your post. It seems like you are on the right track with your calculations. You are correct in thinking that the Ka value for acetic acid in water may be different from the Ka value in a solution containing a base. However, in this case, the difference is negligible and you can still use the given Ka value in the Henderson-Hasselbalch equation.

The Henderson-Hasselbalch equation is used to calculate the pH of a solution containing a weak acid and its conjugate base. In this case, you have acetic acid (HX) and its conjugate base (X-). The Ka value for acetic acid is used because it is the dissociation constant for the acid in water. Since you are using the HH equation to calculate the pH of the solution, you are essentially assuming that the solution is in equilibrium. Therefore, the Ka value for acetic acid in water is still applicable.

As for your second question, you are correct in thinking that the Ka value would be different in the I.C.E. table. This is because the solution now contains a base, which will react with the acid to form its conjugate base. However, since the difference in Ka values is small, you can still use the given Ka value in your calculations.

I hope this helps clarify your doubts. If you have any further questions, please let me know.