Adding a Strong Acid to a Buffer?

A beaker with 140 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 0.100 M. A student adds 4.10 mL of a 0.280 \it M \rm HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.Adding a Strong Acid to a Buffer?
5.00 = 4.760 + log Acetate / acetic acid

5.00 - 4.760=0.24

10^0.24 = 1.74 = acetate / acetic acid



1.74 acetic acid = acetate

acetate = 0.100 - acetic acid



1.74 acetic acid = 0.100 - acetic acid

acetic acid = 0.0365 M

acetate = 0.0635 M



moles acetate = 0.140 L x 0.0635 = 0.00889

moles acetic acid = 0.0365 x 0.140=0.00511



moles H+ added = 4.10 x 10^-3 L x 0.280 M= 0.00115



moles acetete = 0.00889 - 0.00115=0.00774

moles acetic acid = 0.00511 + 0.00115=0.00626



pH = 4.760 + log 0.00774/ 0.00626= 4.852