Question

A beaker with 1.90×102 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 6.40 mL of a 0.450 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

Answer 1

millimoles of acid + conjugate base = 1.90 x 10^2 x 0.1 =19 ------------->1

pH = pKa + log [conjugate base / acid ]

5.00 = 4.74 + log [conjugate base / acid ]

1.82 = conjugate base / acid

conjugate base = 1.82 acid -------------> 2

from 1 and 2

1.82 acid + acid = 19

2.82 acid = 19

acid = 6.74

conjugate base = 12.26

millimoles of strong acid = C = 6.40 x 0.450 = 2.88

on additon of C millimoles of acid to buffer

pH = pKa + log [conjugate base - C / acid + C]

pH = 4.74 + log (12.26 - 2.88 / 6.74 + 2.88)

pH = 4.729

pH chnage = 4.729 - 5.000

= -0.271

pH chnage  = - 0.271