Question

A beaker with 135 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 7.80 mL of a 0.300 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

Answer 1

let x = concentration acid
let y = concentration conjugate base

x + y = 0.100
5.00 = 4.740 + log y/x

we must solve this system

5.00 - 4.740 = log y/x
0.26 = log y/x
10^0.26 =1.819 = y/x

1.819 x = y

x + 1.819 x = 0.100
2.819 x = 0.100
x =0.0354 M = concentration acid
0.100 - 0.0354 =0.0645 M= concentration conjugate base

moles acid = 0.135 L x 0.0354 M= 0.00477
moles conjugate base = 0.0645 M x 0.135 L=0.00870

moles HCl = 7.80 x 10^-3 L x 0.300 M=0.00234

A- + H+ = HA

moles conjugate base = 0.00870 - 0.00234=0.00636
moles acid = 0.00477 + 0.00234=0.00711

total volume = 135 + 7.80 = 142.8 mL = 0.1428 L

concentration acid = 0.00711/ 0.1482 =0.04797 M
concentration conjugate base = 0.00636/ 0.1482 =0.04291 M

pH = 4.740 + log 0.04291/ 0.04797=3.69

change pH = 5.00 - 3.69=1.31