# Homework Help: Buffer problem or not?

1. Oct 9, 2007

### kdpointer

Buffer problem or not??

I have this problem and I worked it out completely, got my answers, and breathed a sigh of relief. Then, when I was looking through my notes and the book I noticed that the problem may have to be done completely differently because it might be a buffer problem. Can someone please tell me if it is a buffer problem?

A weak acid HA (pKa=5.00) was titrated with 1.00M KOH. The acid solution had a volume of 100 mL and a molarity 0.100M. Find the pH at the following volumes of base added and make a graph of pH versus Vb: 0, 1, 5, 9, 9.9, 10, 10.1, and 12 mL.

I did the whole problem assuming it was a simple pH problem. Do I need to treat it as a buffer since it is a titration of a weak acid with a strong base? This is all I need to know.

2. Oct 9, 2007

### Gannon

This can be done with the buffer equation, since you have the pKa and a volume and molarity for both the acid and base. As you plug in the series of volumes they give you, you should begin to see a trend as the base volume increases. It's going to be tedious performing that equation 9 times, so be careful with your calculations!

3. Oct 9, 2007

### symbolipoint

Yes, weak acid and buffer situations. There are two important regions for titrant quantity. The region before the equivalence point, and the region after the equivalence point. Become familiar with this reasonably general form for a monoprotic weak acid:

K = [H][A + H]/[HA - H]

The typesetting is not too good there, so what I'm trying to say is:

(hydronium molarity) multiplied by (formality of the conjugate base plus the molarity of hydronium ion) divided by (formality of the weak acid minus the molarity of hydronium ion).

Kw = 1 x 10^(-14) = Ka*Kb comes from
Kw is for dissociation constant for water;
Ka is for HA <=> H + A
Kb is for NaA + H2O <=> NaOH + HA

You can first find the pH for the equivalence point, which is like the solution created by dissolving the sodium (or potassium) salt of the weak acid in water.