Question

2. 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 forms of the buffer. A 0.5 M buffer will require five times as much acid or base as a 0.1 M buffer for 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. In the laboratory you will carry out some stepwise additions of acid or base and measure the resulting pH values.

Starting with 60 mL of 0.50 M phosphate buffer, pH=6.83, you add 1.7 mL of 1.00 M HCl. Using the Henderson-Hasselbalch equation with pK2 for phosphate of 6.64 calculate the following values to complete the ICE table.

What is the composition of the buffer to begin with, both in terms of the concentration and the molar quantity of the two major phosphate species? (Units required.)

What is the molar quantity of H3O+ added as HCl, and final molar quantity of HPO and H2PO4- at equilibrium?

What is the new HPO42-/H2PO4- ratio, and the new pH of the solution?(Note: you can use the molar ratio rather than the concentration ratio because both species are in the same volume.)

Now take another 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

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

Initial concentration

6.83 = 6.64 + log([HPO4^2-]/[H2PO4-])

[HPO4^2-] = 1.55[H2PO4-]

[H2PO4-] + [HPO4^2-] = 0.5 M x 60 ml = 30 mmol

[H2PO4-] + 1.55[H2PO4-] = 30 mmol

[H2PO4-] = 11.76 mmol

concentration of [H2PO4-] = 11.76 mmol/60 ml = 0.196 M

[HPO4^2-] = 30 - 11.76 = 18.24 mmol

concentration of [HPO4^2-] = 18.24 mmol/60 ml = 0.304 M

Amount of HCl added [H3O+] = 1.00 M x 1.7 ml = 1.7 mmol

New concentration of [H2PO4-] = (0.196 M x 60 ml + 1.7)mmol/61.7 ml = 0.218 M

New concentration of [HPO4^2-] = (0.304 M x 60 ml - 1.7)mmol/61.7 ml = 0.268 M

New [HPO4^2-]/[H2PO4-] = 0.268/0.218 = 1.23

New pH = 6.64 + log(1.23) = 6.73

When NaOH = 1.0 M x 3.7 ml = 3.7 mmol was added

New concentration of [H2PO4-] = (0.196 M x 60 ml - 3.7)mmol/63.7 ml = 0.126 M

New concentration of [HPO4^2-] = (0.304 M x 60 ml + 3.7)mmol/63.7 ml = 0.344 M

New [HPO4^2-]/[H2PO4-] = 0.344/0.126 = 2.73

New pH = 6.64 + log(2.73) = 7.266