Buffer capacity refers to the amount of acid or base a buffer can absorb without

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 base components of the buffer. A 0.5 M buffer can "absorb" five times as much acid or base as a 0.1 M buffer for 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 in phosphate species, is: Concentration of HPO_4^2-: 0.304 M Molar quantity of HPO_4^2-; 18.2 mmol Concentration of H_3PO_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_4^3- and H_2PO_4^2- at equilibrium? What is the new HPO_4^2-/H_2PO_4^- ratio, and the new pH of the solution? The pKa of H_2PO_4^- is 664. Use the Henderson Hassel Bach 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.

Explanation / Answer

Q4.

(A) Moles of H+ or (HCl) added = Molarity X volume = 1M X 1.7mL = 1.7mmol

This 1.7mmol of HCl will react with HPO42- to form 1.7mmol of H2PO4-

Hence final molar quantity of H2PO4- = 11.8mmol+ 1.7mmol = 13.5mmol

final molar quantity of HPO42-  = 18.2mmol - 1.7mmol = 16.5mmol

(B) pH = pKa + log[HPO42- ] / [H2PO4- ] = 6.64 + log[16.5mmol] / [13.5mmol] = 6.64 + 0.087 = 6.73

(C) Moles of NaOH added = 1.00M X 3.7mL = 3.7mmol

3.7mmol of NaOH will react with H2PO4- to form 3.7 mmol of HPO42- .

New molar quantities

HPO42- = 18.2mmol + 3.7mmol = 21.9mmol

H2PO4- = 11.8 mmol - 3.7mmol = 8.1 mmol

pH = pKa + log[HPO42- ] / [H2PO4- ] = 6.64 + log[21.9mmol] / [8.1 mmol] =  6.64 + 0.433 = 7.073

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