Question

(30 Points) You need to produce a buffer solution that has a pH of 5.02. You already have a solution that contains 10. mmol (millimoles) of acetic acid. How many millimoles of acetate (the conjugate base of acetic acid) will you need to add to this solution? The pKa of acetic acid is 4.74. Express your answer numerically in millimoles.

Answer 1

pH = pKa + log [acetate]/ [acetic acid] ( Handerso-Hasselbalch equation)

5.35 = 4.74 + log acetate / 0.010 moles

5.35 - 4.74 =0.61 = log acetate / 0.010

10^0.61 =4.1 = acetate / 0.010

moles acetate = 0.041

millimoles acetate required = 41

Answer 2

By definition:

pH = pKa + log [acetate]/ [acetic acid]

so

5.02 = 4.74 + log [acetate] / 10 mmole

10mmole = 10/1000 = 0.01 mole

5.02 = 4.74 + log [acetate] / 0.01

5.02 - 4.74 = 0.28 = log [acetate] /0.01


10^0.28 = 1.90546 = [acetate] / 0.01
[acetate] = 0.019 mole

= 19 millimoles