Chemistry question involving buffers?

A beaker with 175mL 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 6.80mL of a 0.430M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.Chemistry question involving buffers?
[CH3COOH] + [CH3COO-]= 0.100 M



pH = pKa + log [CH3COO-] / [CH3COOH]



5.00 = 4.760 + log [CH3COO-] / [CH3COOH]



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



solving this system

[CH3COO-] = 0.0635 M

[CH3COOH]= 0.0365 M



Moles acetate = 0.0635 M x 0.175 L=0.0111

moles acetic acid = 0.0365 M x 0.175 L= 0.00639



moles HCl = 0.00680 L x 0.430 M=0.00292



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

moles acetate = 0.0111 - 0.00292 =0.00818

moles acetic acid = 0.00639 + 0.00292 =0.00931



total volume = 175 + 6.80=181.8 mL = 0.1818 L



[acetate]= 0.00818 / 0.1818 = 0.0500 M

[acetic acid ]= 0.00931 / 0.1818 = 0.0512 M



pH = 4.760 + log 0.0500/ 0.0512 = 4.750