Adding Acids or Bases to Buffered Solutions

[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!

 

[sdekivit]

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 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

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

using the equation of Hendersson-Hasselbalch:

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

you can determine the ratio $$\frac {[Ac^{-}]} {[HAc]}$$.

This will give you the initial amount of $HAc$ and $Ac^{-}$.

 

[Blueshift5]

Your steps are correct, but there is a small error in your initial mols calculation. The total molarity of acid and conjugate base in the buffer is 0.1 M, which means that the combined molarity of acetic acid and acetate ions is 0.1 M. So, the initial mols of the conjugate acid and base should be calculated as follows:

- Initial mols of conjugate acid (acetic acid): 0.1 * 0.11 = 0.011 moles
- Initial mols of conjugate base (acetate ions): 0.1 * 0.11 = 0.011 moles

Other than that, your calculations and steps are correct and your final pH calculation is also correct. The change in pH is indeed 0.50, as you have calculated. Keep up the good work!