How do you do this chem problem?? A beaker with 140 mL of an acetic acid buffer with a pH of 5.00 is sittin on?

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 6.90 mL of a 0.410 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.How do you do this chem problem?? A beaker with 140 mL of an acetic acid buffer with a pH of 5.00 is sittin on?
let x = [acetate] and let y = [acetic acid]

5.00 = 4.760 + log x/y

10^0.24 =1.74 = x/y



1.74 y = x

x + y = 0.100 M



1.74 y + y = 0.100

2.74 y = 0.100

y = 0.0365 M

x = 0.0635 M



moles acetate = 0.0635 x 0.140 L=0.00889

moles acetic acid = 0.0365 x 0.140 L=0.00511



moles HCl = 6.90 x 10^-3 L x 0.410=0.00283



CH3COO- + H+ = CH3COOH

moles acetate = 0.00889 - 0.00283 =0.00606

moles acetic acid = 0.00511 + 0.00283 =0.00794



total volume = 140 + 6.90=146.9 mL = 0.1469 L



[acetate] = 0.00606/ 0.1469=0.0413 M

[acetic acid ]= 0.00794/ 0.1469=0.0541 M



pH = 4.760 + log 0.0413 / 0.0541= 4.64