Question

You need to prepare an acetate buffer of pH 5.80 from a 0.869 M acetic acid solution and a 2.06 M KOH solution. If you have 475 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a bufer of pH 5.80? The pKa of acetic acid is 4.76.

Answer 1

moles of acetic acid = 475 x 0.869 / 1000 = 0.413

CH3COOH +    KOH   -------------> CH3COOK + H2O

     0.413         2.06V                              0                0

0.413 - 2.06V     0                                2.06 V       2.06V

pH = pKa + log [CH3COOK / CH3COOH]

5.80 = 4.76 + log [2.06 V / 0.413-2.06 V]

[2.06 V / 0.413-2.06 V] = 10.96

V = 0.1837 L

volume of KOH = 184 mL