Chemistry Acid Base Buffer?

A beaker with 155 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 7.60 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.Chemistry Acid Base Buffer?
let x = [acetate]

let y = [acetic acid]



x + y = 0.100



5.00 = 4.760 + log x/y



5.00 - 4.760= log x/y



10^ 0.24 =1.74 = x/y



1.74 y = x



1.74 y + y = 0.100

2.74 y = 0.100

y=0.0365 M

x = 0.100 - 0.0365=0.0635 M



moles acetate = 0.0635 x 0.155 L=0.00984

moles acetic acid = 0.0365 x 0.155=0.00566



moles H+ added = 7.60 x 10^-3 L x 0.260=0.00198



CH3COO- + H+ = CH3COOH

moles acetate = 0.00984 - 0.00198=0.00786

moles acetic acid = 0.00566 + 0.00198=0.00764



total volume = 0.1626 L

[acetate]= 0.00786/ 0.1626=0.0483

[acetic acid]= 0.00764/ 0.1626=0.0470 M



pH = 4.760 + log 0.0483/ 0.0470=4.77



delta pH = 0.23