# Adding Acids or Bases to Buffered Solutions

1. Aug 1, 2006

### quicknote

A beaker with 110 mL of an acetic acid buffer has a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.1 M. A student adds 6.90 mL of a 0.480M HCl solution to the beaker. How much will the pH change? The pK_a of acetic acid is 4.76.

My steps:

H + CH3COO <--> CH3COOH

1. find the initial mols of the conjugate acid (Not sure this is right):
mols =0.1*0.110
= 0.011 moles of acetic acid

2. inital mols of the conjugate base would be the same 0.011 moles because of a one to one ratio.

3. amount of H+ ions added:
0.480*0.0069 = 0.003312 moles

4. Final moles of conjugate base:
0.011-0.003312 = 0.007688 moles

5. final moles of acid
0.011+0.003312 = 0.014312

6. calculation of new pH...used Hasselbach equation:

$$pH = pK_a + log \frac{[Conjugate Base]}{[Conjugate Acid]}$$

$$pH = 4.76 + log \frac{[0.007688]}{[0.014132]}$$
=4.50

Therefore change in pH is 0.50

I think there is something wrong with my initial mols. "The total molarity of acid and conjugate base in this buffer is 0.1 M." I'm not sure how to calculated it based on this info.
Any help will be much appreciated. Thanks!

2. Aug 2, 2006

### sdekivit

you know the buffer has a pH of 5.00 ($pK_{a}$)and $$[HAc] + [Ac^{-}] = 0.1 M$$

$$pH = pK_a + log \frac{[Conjugate Base]}{[Conjugate Acid]}$$
you can determine the ratio $$\frac {[Ac^{-}]} {[HAc]}$$.
This will give you the initial amount of $HAc$ and $Ac^{-}$.