Adding strong acid to buffer?

A beaker with 165mL 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 0.100M . A student adds 9.00mL of a 0.250M solution to the beaker. How much will the pH change? The of acetic acid is 4.760.Adding strong acid to buffer?
0.100 M * 165 mL = 16.5 mmoles CH3COOH and CH3COONa.



I'm going to assume that you're adding 9.00 mL 0.250 M strong acid to the solution.



9.00 mL * 0.250 M = 2.25 mmoles HCl



The strong acid will react with the conjugate base form to produce more weak acid.



16.5 mmoles CH3COONa - 2.25 mmoles HCl = 14.25 mmoles CH3COONa leftover.



16.5 mmoles CH3COOH + 2.25 mmoles CH3COOH created = 18.75 mmoles CH3COOH



Use the Henderson-Hasselbalch equation to find the new pH:



pH = pKa + log[A-/HA]



You normally would convert the new mmoles to concentration, but you do not have to because this is a ratio between the two.



pH = 4.760 + log(14.25/18.75) = 4.64



Since the solution before the addition of the acid is an even mixture of acid and conjugate base, the pH = pKa initially.



So the change in pH would be:



dpH = pHf - pHi = 4.64 - 4.76 = -0.12