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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 ) (pKa=4.20) and 0.140 M 0.140 M sodium benzoate. How many milliliters of each solution should be mixed to prepare this buffer? benzoic acid: mL mL sodium benzoate: mL

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jufsanabriasa

Answer:

68.9mL of benzoic acid

31.1mL of sodium benzoate

Explanation:

Using Henderson-Hasselbalch formula:

pH = pka + log [Benzoate] / [Benzoic acid]

Where [Benzoate], [Benzoic acid] could be seen as moles of each compound.

4.00 = 4.20 + log [Benzoate] / [Benzoic acid]

0.631 = [Benzoate] / [Benzoic acid]

0.631 [Benzoic acid] = [Benzoate] (1)

0.1000L = [Benzoic acid] / 0.100M +  [Benzoate] / 0.140M (2)

Replacing (1) in (2):

0.1000L = [Benzoic acid] / 0.100M +  0.631 [Benzoic acid] / 0.140M

0.1000L = 14.5 [Benzoic acid]

6.893x10⁻³ = Moles of benzoic acid

Replacing this value in (1):

4.350x10⁻³ = Moles of sodium benzoate

Converting these moles in volume using molar concentration:

6.893x10⁻³mol ₓ (1L /0.1mol ) = 0.0689L ≡ 68.9mL of benzoic acid.

4.350x10⁻³mol ₓ (1L /0.14mol ) = 0.0311L ≡ 31.1mL of sodium benzoate.

I hope it helps!