Problem Help! Henderson Hasselbalch equation.

Here's the problem I am given: A beaker with 120mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.1M. A student adds 6.60mL of a 0.300M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.76.



The main thing I'm having a problem with is figuring out how to find the moles of acid and base using the henderson hasselbalch equation. I found the ratio of conjugate base to conjugate acid (1.74) but that's as far as I've gotten. If someone could show me how to do that I would really appreciate it.



Thanks for your help!Problem Help! Henderson Hasselbalch equation.
Ok I would change thew molarity as it wree to mmoles of ...in other words the 120 mL X 0.1 M = 12 mmoles total for the two forms acid and base. Since the pH is 5 then the amount of each form can be determined by th following H/H 5.0 = 4.76 + log base /12 - base since the pH is merely a function of the ratio of mmoles since the two forms are in the same volume. so 5.0 - 4.76 = log base /12-base

so 1.74 = base /12-base

20.88 = 2.74 base base = 7.62 mmoles and acid = 12 - 7.62 = 4.38 moles



check



pH = 4.76 + log 7.62/4.38 = 4.76 + log 1.739 = 4.76 + 0.24 = 5







so now the pH after adding the HCl will be driven more acidic because the ratio uis going to change 6.6mL x 0.3 M = 1.98 mmole of H+ so adding this to the buffer will convert 1.98 moles of the base form ( acetate ) to the acid form so the pH will now be



pH = 4.76 + log[ 7.62-1.98]/4.38 + 1.98 mmoles = .76 + log 5.64/6.36

= 4.76 -0.052 = 4.708 it changes by - .29 pH units