Adding a strong acid to 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 0.100 M. A student adds 6.20 mL of a 0.260 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.



please help??? I'm lost....Adding a strong acid to buffer?
5.00 = 4.760 + log [CH3COO-] / [CH3COOH]



10^0.24 = 1.74 = [CH3COO-] / [CH3COOH]



[CH3COOH] + [CH3COO-] = 0.100



[CH3COOH] = 0.100 - [CH3COO-]



1.74 = [CH3COO-] / 0.100 - [CH3COO-]



0.174 - 1.74 [CH3COO-] = [CH3COO-]



[CH3COO-] = 0.0635 M

[CH3COOH]= 0.100 - 0.0635 =0.0365 M



moles acetate = 0.200 L x 0.0635 =0.0127

moles acetic acid = 0.200 x 0.0365 =0.00730



moles HCl = 0.00620 L x 0.260 M=0.00161



CH3COO- + HCl %26gt;%26gt; CH3COOH

moles acetate = 0.0127 - 0.00161 =0.0111

moles acetic acid = 0.00730 + 0.00161 =0.00891

total volume =0.206 L

cocnentration acetate = 0.0111 / 0.206 =0.0538 M

concentration acetic acid = 0.00891 / 0.206 =0.0433 M



pH = 4.760 + log 0.0538 / 0.0433 =4.85