Question

A buffer consists of 0.72-M propanoic acid (ka = 1.4 × 10-5) and 0.84-M sodium propanoate....

A buffer consists of 0.72-M propanoic acid (ka = 1.4 × 10-5) and 0.84-M sodium propanoate.

a.) Calculate the pH of this buffer.

b.) Calculate the pH after the addition of 6.8 mL of 0.10–M HCL to 0.010 L of the buffer.

c.) Calculate the pH after the addition of 8.4 mL of 1.0–M HCL to 0.010 L of the buffer.
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Homework Answers

Answer #1

a) pH = pKa + log [sodium propanoate] / [propanoic acid]

Pka = - log [Ka] = - log [1.4 x 10-5] = 4.85

pH = 4.85 + log [0.84] / [0.72]

pH = 4.92

b) by adding 0.1 x 6.8 / 10 = 0.068 M HCl

[salt] = 0.84 - 0.068 = 0.772 M

[acid] = 0.72 + 0.068 = 0.788 M

pH = 4.85 + log [0.772] / [0.788]

pH = 4.84

c) by adding 1.0 x 8.4 / 10 = 0.84 M

[salt] = 0.84 - 0.84 = 0

[acid] = 0.72 + 0.84 = 1.56 M

only acid present in the solution

pH = 1/2 [pKa - log C]

pH = 1/2 [4.85 - log 1.56]

pH = 2.52

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