You need to prepare an acetate buffer of pH 5.24 from a 0.796 M acetic acid solution and a 2.65 M KOH solution. If you have 830 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pH 5.24 ? The pKa of acetic acid is 4.76. Be sure to use appropriate significant figures

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jufsanabriasa jufsanabriasa · Answer 1 · 2020-09-11T02:53:59+02:00

Answer:

187.3mL of 2.65M KOH you must add

Explanation:

The pH of the buffer of acetic acid is obtained as follows:

pH = pKa + log [Acetate] / [Acetic Acid]

Where [] could be moles of acetate and acetic acid

Replacing:

5.24 = 4.76 + log [Acetate] / [Acetic Acid]

0.48 = log [Acetate] / [Acetic Acid]

3.01995 = [Acetate] / [Acetic Acid](1)

As concentration of acetic acid is 0.796M and volume is 830mL, the moles of acetic buffer is:

0.830L * (0.796 moles / L) = 0.6607 moles

Thus:

0.6607 moles = [Acetate] + [Acetic Acid] (2)

Replacing (2) in (1):

3.01995 = 0.6607 moles - [Acetic Acid] / [Acetic Acid]

3.01995 [Acetic Acid] = 0.6607 moles - [Acetic Acid]

4.01995 [Acetic Acid] = 0.6607 moles

[Acetic Acid] = 0.16435 moles

[Acetate] = 0.6607 moles - 0.16435 moles = 0.49635 moles

The reaction of acetic acid with KOH to produce Acetate is:

Acetic acid + KOH → Acetate + Water

That means the moles of KOH you add, are the moles of acetate.

To add 0.49635 moles of 2.65M of KOH you need:

0.49635 moles * (1L / 2.65 moles) = 0.1873L =