Adding a Strong Acid to a Buffer?

A beaker with 200 mL 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 0.1 M. A student adds 5.30 mL of a 0.310 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.76.Adding a Strong Acid to a Buffer?
First, use the Henderson-Hasselbalch equation to find the ratio of base to acid in this buffer.

pH=pka+log(B/A)

5.00=4.76+log(B/A)

B/A = 1.74

The total molarity is 0.1M, with 1.74 times as much base as acid. So...

x+1.74x=0.1M.

x=0.036M. This is concentration of acid, and base is 1.74*0.036 = 0.064M.

Adding 0.00530L of 0.310mol/L HCl adds 0.00164 moles of H+ (0.00530*0.310). Since this is strong acid, it dissociates 100% and moles of HCl equals moles of H+.

H+ will react with the base, which is 0.200L of 0.064mol/L, or 0.0128 moles. We can see that H+ is limiting, but it will consume 0.00164 moles of the 0.0128 moles of base available. There are 0.0112 moles of base remaining, and an additional 0.00164 moles of acid formed. Now recalculate with the H-H equation, but remember that the volume has changed. (It is now 0.2053L)

(I got around 4.8, so the pH only changes by 0.2, but check and make sure; I tend to enter numbers in my calc too fast and mix them up, but this is the procedure for finding the pH of this buffer)