You are asked to prepare 500. mL of a 0.300 M acetate buffer at pH 5.00 using on

Question

You are asked to prepare 500. mL of a 0.300 M acetate buffer at pH 5.00 using only pure acetic acid (MW=60.05 g/mol, pKa=4.76), 3.00 M NaOH, and water. Answer the following questions regarding the preparation of the buffer.

1. How many grams of acetic acid will you need to prepare the 500 mL buffer? Note that the given concentration of acetate refers to the concentration of all acetate species in solution.

2. What volume of 3.00 M NaOH must you add to the acetic acid to achieve a buffer with a pH of 5.00 at a final volume of 500 mL? (Ignore activity coefficients.)

Explanation / Answer

1) The total concentration of acetate = 0.3 M

[Acetic acid] + [acetate ion] = 0.3 M

[acetate] = 0.3 - [acetic acid] ....................(1)

Required pH = 5

We will use Hendersen Hassalbalch equation for buffer

pH = pKa + log [salt] / [acid]

5 = 4.76 + log [acetate ] / [acetic acid]

0.24 = log [acetate ] / [acetic acid]

Take antilog

1.737 = [acetate] / [acetic acid] ..........(2)

Put (1) in (2)

1.737 = 0.3 - [acetic acid] / [acetic acid]

1.737 [acetic acid] = 0.3 - [acetic acid]

2.737 [acetic acid] = 0.3

[acetic acid] = 0.3 / 2.737 = 0.1096 M

so concentration of acetate ion = 0.3 - 0.1096 = 0.1904 M

1) Moles of acetic acid required for 0.10.96 concentration will be

Moles = Concentration X volume in litres = 0.1096 X 0.5 = 0.0548

mass of acetic acid = Moles X molecular weight of acetic acid = 0.0548 X 60.05 = 3.29 grams

2) The concentration of NaOH required = 0.1904 M

Final volume = 500mL

Initial concentration = 3 M

so initial volume = Final volume X final concentration / Initial concentration = 500 X 0.1904 / 3 = 31.73 mL

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