You are instructed to create 50.0 mL of a 0.66 M Acetate buffer with a pH of 5.0

Question

You are instructed to create 50.0 mL of a 0.66 M Acetate buffer with a pH of 5.0. You have Acetic acid and the sodium salt NaC2H3O2, available. (Enter all numerical answers to three significant figures.)

Ka1 = 1.7105

1. What is the molarity needed for the acid component of the buffer?
2. What is the molarity needed for the base component of the buffer?
3. How many moles of acid are needed for the buffer?
4. How many moles of base are needed for the buffer?
5. How many grams of acid are needed for the buffer?
6. How many grams of base are needed for the buffer?

Ka1 = 1.7105

Explanation / Answer

We will take the molarity of both the buffer component = 0.66 M

1) What is the molarity needed for the acid component of the buffer

0.66 M

2) What is the molarity needed for the base component of the buffer

0.66M

3) pH = pKa + log [salt] / [Acid]

pKa = 4.77

5 = 4.77 + log [salt] / [acid]

0.23 = log [salt] / [acid]

Take antilog

1.698 = [salt] / [acid]

[salt] = 1.698 [acid]

[Salt] + [acid] = 0.66

[salt] = 0.66- [acid]

0.66-[acid] = 1.698[acid]

[acid] = 0.66 / 2.698 = 0.245 M

[Salt] = 0.415 M

Moles of acid = Molarity X volume in L = 0.245 X 0.05 = 0.0123

Moles of salt = Molarity X volume in L = 0.415 X 0.05 = 0.02075 = moles of base

Grams of acid = Moles X molecular weight = 0.0123 X 60 g / mole = 0.738 g

Grams of salt = Moles X molecular weight = 0.02075 X 82 = 1.7015 g

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