Determine the volumes of 0.10 M CH3COOH and 0.10 M CH3COONa required to prepare 10 mL of the following pH buffers: pH 4.7, pH 5.7. (Note: the pKa of CH3COOH

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01-08-2020
Chemistry

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Determine the volumes of 0.10 M CH3COOH and 0.10 M CH3COONa required to prepare 10 mL of the following pH buffers: pH 4.7, pH 5.7. (Note: the pKa of CH3COOH

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jufsanabriasa jufsanabriasa · Answer 1 · 2020-08-02T02:48:07+02:00

Answer:

pH 4.7: 5mL of 0.10 M CH3COOH and 5mL 0.10 M CH3COONa

pH 5.7: 0.91mL of 0.10 M CH3COOH and 9.09mL 0.10 M CH3COONa

Explanation:

pKa acetic acid, CH3COOH = 4.7

It is possible to determine pH of a buffer using H-H equation:

pH = pka + log [A⁻] / [HA]

For the acetic buffer,

pH = 4.7 + log [CH3COONa] / [CH3COOH]

As you want a pH 4.7 buffer:

4.7 = 4.7 + log [CH3COONa] / [CH3COOH]

1 = [CH3COONa] / [CH3COOH]

That means you need the same amount of both species of the buffer to make the pH 4.7 buffer. That is:

5mL of 0.10 M CH3COOH and 5mL 0.10 M CH3COONa

For pH 5.7:

5.7 = 4.7 + log [CH3COONa] / [CH3COOH]

1 = log [CH3COONa] / [CH3COOH]

10 = [CH3COONa] / [CH3COOH] (1)

That means you need 10 times [CH3COONa] over [CH3COOH]

And as you know:

10mL= [CH3COONa] + [CH3COOH] (2)

Replacing (1) in (2):

10 = 10mL + [CH3COOH] / [CH3COOH]

10[CH3COOH] = 10mL + [CH3COOH]

11[CH3COOH] = 10mL

[CH3COOH] = 0.91mL

And [CH3COONa] = 10mL - 0.91mL =

[CH3COONa] = 9.09mL

That is:

0.91mL of 0.10 M CH3COOH and 9.09mL 0.10 M CH3COONa

ConcepcionPetillo ConcepcionPetillo · Answer 2 · 2022-03-25T10:05:41+01:00

The volumes according to the pH are as follows:

(i) 5mL of 0.10 M CH₃COOH and 5mL 0.10 M CH₃COONa for pH 4.7

(ii) 0.91mL of 0.10 M CH₃COOH and 9.09mL 0.10 M CH₃COONa pH 5.7

Calculating the volume of chemicals needed:

Given that pKa of acetic acid, CH₃COOH = 4.7

The pH of a buffer using the H-H equation is given by:

pH = pKa + log [A⁻] / [HA]

For the acetic buffer,

pH = 4.7 + log [CH₃COONa] / [ CH₃COOH]

4.7 = 4.7 + log [CH₃COONa] / [ CH₃COOH]

0 = log [CH₃COONa] / [ CH₃COOH]

takin antilog on both sides of the equation we get:

1 = [CH₃COONa] / [CH₃COOH]

It implies that the same amount of both species is needed to make the pH 4.7 buffer.

So,

5mL of 0.10 M CH₃COOH and 5mL 0.10 M CH₃COONa makes a buffer of pH 4.7

Similarly:

5.7 = 4.7 + log [CH₃COONa] / [CH₃COOH]

1 = log [CH₃COONa] / [CH₃COOH]

takin antilog on both sides of the equation we get:

10 = [CH₃COONa] / [CH₃COOH]

10[CH₃COOH] = [CH₃COONa]

It implies that we need 10 times [CH₃COONa] as much of [CH₃COOH]

We have to prepare 10 mL of buffer, so:

10mL= [CH₃COONa] + [CH₃COOH]

10mL = 11[CH₃COOH]

[CH₃COOH] = 0.91mL

So, [CH₃COONa] = 10mL - 0.91mL

[CH₃COONa] = 9.09mL

Therefore,

0.91mL of 0.10 M CH3COOH and 9.09mL 0.10 M CH3COONa is required to make a buffer of pH 5.7

Learn more about buffer:

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