Question

# You need to prepare 100.0 mL of a pH=4.00 buffer solution using 0.100 M benzoic acid...

You need to prepare 100.0 mL of a pH=4.00 buffer solution using 0.100 M benzoic acid (pKa = 4.20) and 0.180 M sodium benzoate. How much of each solution should be mixed to prepare this buffer?

total volume = 100 mL = 0.1 L

Let us assume , volume of  sodium benzoate = X

volume of benzoic acid = 0.1 -X

Hence,

[sodium benzoate ] = molarity x volume in Litres = 0.18 X mol

[benzoic acid ] = molarity x volume in Litres = 0.1 (0.1-X) mol = 0.01 - 0.1 X

pH = pKa + log [sodium benzoate] /[benzoic acid ]

4 = 4.2 + log [ 0.18 X /0.01 - 0.1 X ]

0.18 X /0.01 - 0.1 X = 0.63

0.18 X = 0.0063 - 0.063 X

0.243 X = 0.0063

X = 0.026 L

Therefore,

volume of  sodium benzoate = 0.026 L = 26 mL

volume of benzoic acid = 0.1 -X = 0.1- 0.026 = 0.074 L = 74 mL

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