Chemistry help please?

A beaker with 175 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 6.90 mL of a 0.340 M HCL solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760Chemistry help please?
let x = [acetic acid]

let y = [acetate]



5.00 = 4.760 + log y/x

10^0.24 = y/x

1.74 = y/x

1.74 x = y



but we know that x + y = 0.100

y = 0.100-x

1.74 x = 0.100-x

2.74 x = 0.100

x = 0.0365 and y = 0.0635 M



moles acetic acid = 0.0365 x 0.175 L=0.00639

moles acetate = 0.0635 x 0.175 L=0.0111



moles H+ added = 6.90 x 10^-3 L x 0.340 M=0.00235

the reaction that occur is

CH3COO- + H+ = CH3COOH

moles acetate = 0.0111 - 0.00235 =0.00875

moles acetic acid = 0.00639 + 0.00235 =0.00874

total volume = 6.90 + 175 = 181.9 mL =%26gt; 0.1819 L

[acetic acid ] = 0.00874 / 0.1819 =0.0480 M

[acetate] = 0.00875/0.1819 =0.0481 M



pH = 4.760 + log 0.0481 / 0.0480=4.76