Question

A beaker with 2.00×102 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 9.00 mL of a 0.330 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740. Express your answer numerically to two decimal places. Use a minus ( − ) sign if the pH has decreased.

Answer 1

acid + conjugate base   millimoles = 0.1 x 200 = 20 ------------------->1

pH = pKa + log (base / acid)

5.000 = 4.74 + log (base /acid)

base / acid = 1.82

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

from 1 and 2

acid = 7.1

base = 12.9

miilimoles of strong acid HCl = 0.330 x 9 = 2.97

now on additon of C moles of acid to buffer

new pH :  

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

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

pH = 4.740 + log (12.9 - 2.97 / 7.1 + 2.97 )

pH = 4.734

pH change = final - initial

                   = 4.734 - 5.000

                     = - 0.27

pH change    = - 0.27