Question

You have 525 mL of an 0.15 M acetic acid solution. What volume (V) of 2.00 M NaOH solution must you add in order to prepare an acetate buffer of pH = 5.32? (The pKa of acetic acid is 4.76.)

Answer 1

millimoles of acetic acid CH3COOH = 525 x 0.15 = 78.75

millimoles of base NaOH = V x 2.00 = 2V

CH3COOH + NaOH ----------------> CH3COONa + H2O

    78.75             2V                                 0                  0

78.75 - 2V           0                                  2V               2V

in the raction salt and acid remains so it forms acidic buffer.

pH = pKa + log [salt / acid]

5.32 = 4.76 + log (2V / 78.75 - 2V)

3.63 = (2V / 78.75 - 2V)

285.86 - 7.26 V = 2V

V = 30.87 ml

volume of NaOH = (V) = 30.87 mL