Question

A beaker with 145 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 5.50 mL of a 0.430 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760

Answer 1

millimoles of buffer (salt + acid) = 145 x 0.1 = 14.5

pH = 5.00

pKa = 4.76

pH = pKa + log [salt / acid]

5.00 = 4.76 + log [salt / acid]

[salt/acid] = 1.74

salt + acid =14.5

1.74 acid + acid = 14.5

millimoles of [acid] = 5.29

millimoles of [salt] = 9.21

on addition of ’ C’ moles of acid to acidic buffer salt moles decreases and acid moles increases

so

millimoles of acid = 5.50 x 0.430 = 2.365

pH = pKa + log [Salt –C/acid + C]

      = 4.76 + log[9.21 - 2.365/5.29 + 2.365]

      = 4.71

pH change = 4.71 - 5.00

                   = - 0.29