You need to prepare 100.0 mL of a pH 4.00 buffer solution using 0.100 M benzoic acid (p K a = 4.20) and 0.220 M sodium benzoate. How many milliliters of each solution should be mixed to prepare this buffer?

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Chemistry

4 July, 23:38

You need to prepare 100.0 mL of a pH 4.00 buffer solution using 0.100 M benzoic acid (p K a = 4.20) and 0.220 M sodium benzoate. How many milliliters of each solution should be mixed to prepare this buffer?

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Answers (2)

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Melany Sutton · Answer 1

34.5 mL

Explanation:

Given, we ned to prepared buffer solution with pH = 4.00, volume = 100.0 mL.

[C6H5COOH] = 0.100 M, [C6H5COO-] = 0.120 M, pKa = 4.20

First we need to calculate the ratio of the conjugate base and acid

We know, Henderson Hasselbalch equation

pH = pKa + log [C6H5COO-] / [C6H5COOH]

4.00 = 4.20 + log [C6H5COO-] / [C6H5COOH]

log [C6H5COO-] / [C6H5COOH] = 4.00-4.20

= - 0.20

Antilog from both side

[C6H5COO-] / [C6H5COOH] = 0.631

Now we are given the molarity of each acid and its conjugate base and we need to calculate for volume

Sum of volume = 100 mL = 0.100 L

volume of benzoic acid = x

volume of benzoate = 0.10 - x,

so Volume of acid + volume of conjugate base = 0.100 L

[C6H5COO-] / [C6H5COOH] = 0.631

[C6H5COO-] = 0.631 * [C6H5COOH]

0.120 (0.1-x) = 0.631 * 0.100x

So, x = 0.065

So, volume of benzoic acid = x = 0.0655 L

= 65.5 mL

So, volume of sodium benzoate = 0.1 - x

= 0.1-0.0655

= 0.0345 L

= 34.5 mL

So we need to mix 65.5 mL of 0.100 M benzoic acid and 34.5 mL of 0.120 M sodium benzoate form 100.0 mL of a pH=4.00 buffer solution.

Carley Hurst · Answer 2

Volume benzoic acid = 77.74 mL

Volume sodium benzoate: = 22.26 mL

Explanation:

Step 1: Data given

Volume = 100.0 mL

pH = 4.00

Molarity benzoic acid = 0.100 M

pKa benzoic acid = 4.20

Molarity sodium benzoate = 0.220 M

Step 2

pH = pKa + log ([base]/[acid])

4 = 4.2 + log ([base]/[acid])

-0.2 = log ([base]/[acid])

10^-0.2 = [base]/[acid]

0.63 = [base]/[acid]

Step 3

start with 100 mL 0.1 M benzoic acid

0.100 L * 0.1M = 0.01 moles.

Since base/acid = 0.63, add 0.63*0.01 moles base = 0.0063 moles base.

And 0.0063 moles of a 0.22 M salt solution are x mL:

0.22 moles = 1000 mL

1 mole = 1000/0.22 mL

0.0063 moles = 1000*0.0063/0.22 mL = 28.63 mL.

So a mixture of 100 mL 0.1 M bencoic acid + 28.63 mL of 0.22 benzoate has pH = 4.00

Step 4: For a total volume of 100 mL we'll have:

Volume benzoic acid: 100*100 / (128.63) = 77.74 mL

Volume sodium benzoate: 28.63*100 / (128.63) = 22.26 mL