Question

Buffer capacity refers to the amount of acid or base a buffer can absorb without a significant pH change. It is governed by the concentrations of the conjugate acid and bate components of the buffer. A 0.5 M buffer can "absorb" five time as much acid or lease as a 0.1 M buffer tor a given pH change. In this problem you begin with a buffer of known pH and concentration and calculate the new pH after a particular quantity of acid or base is added. You are given 60 ml. of 0.50 M phosphate buffer, pH = 6.83, to test. The starting composition of the buffer, both in terms of the concentration and the molar quantity of the two major phosphate species, is: Concentration of HPO^2-_4: 0.304 M Molar quantity of HPO^2-_4: 18.2 mmol Concentration of H_2PO^-_4: 0.196 M Molar quantity of H_2PO^-_4: 11.8 mmol a. You add 1.7 mL of 1.00 M HCl to the buffer. Calculate the molar quantity of H_3O^+ added as HCl, and the final molar quantity of HPO^2-_4 and H_2PO^-_4 at equilibrium? b. What is the new HPO^2-_4/H_2PO^-_4, ratio, and the new pH of the solution? The pKa of H_2PO^-_4 is 6.64. Use the Henderson-Hasselbach equation to calculate the new pH. c. Now take a fresh 60 ml. of the 0.50 M pH 6.83 buffer and add 3.7 mL of 1.00 M NaOH. Using steps similar to those above, calculate the new pH of the solution.

Answer 1

4. Buffer H2PO4-/HPO4^2-

a. molar quantity of H3O+ added = 1.0 M x 1.7 ml = 1.7 mmol

new molar quantity of HPO4^2- = 18.2 - 1.7 = 16.5 mmol

new molar quantity of H2PO4^- = 11.8 + 1.7 =13.5 mmol

b. New pH

using Hendersen-Hasselbalck equation,

pH = pKa + log(HPO4^2-/H2PO4-)

      = 6.64 + log(16.5/13.5)

      = 6.73

c. When,

3.7 ml of 1.0 M NaOH is added to buffer

molar quantity of OH- added = 1.0 M x 3.7 ml = 3.7 mmol

new molar quantity of HPO4^2- = 18.2 + 3.7 = 21.9 mmol

new molar quantity of H2PO4^- = 11.8 - 3.7 = 8.1 mmol

New pH

using Hendersen-Hasselbalck equation,

pH = pKa + log(HPO4^2-/H2PO4-)

      = 6.64 + log(21.9/8.1)

      = 7.072