Question

You have 225 mL of an 0.29 M acetic acid solution. What volume (V) of 2.10 M NaOH solution must you add in order to prepare an acetate buffer of pH = 4.68?

(The pKa of acetic acid is 4.76.)

Answer 1

pH = pKa + log(no. of moles of salt/no. of moles of aci)

initially no salt is formed so no.of moles of salt = no.of moles of NaOH added = 2.40 * V

no.of moles of acid = initial no. of moles of acid - no.of moles of salt = 0.29 * 0.225 - 2.20 * V

4.68 = 4.76 + log(2.10 * V/ 0.29 * 0.225 - 2.10 * V)
4.68 - 4.76 = log(2.10V / 0.06525 - 2.10V)
10^(-0.08) = 2.1V/0.06525 - 2.1V
0.8318 = 2.1V/0.06525 - 2.1V
0.8318(0.06525-2.1V) = 2.1V
0.05427 - 1.74678V = 2.1V
0.05427 = 3.84678V
V = 0.01411 L or simply 14.11 mL

Hope this helps