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Misha Larkins [42]

3 years ago

13

Starting with 0.250L of a buffer solution containing 0.250 M benzoic acid (C 6H 5COOH) and 0.20 M sodium benzoate (C 6H 5COONa),

what will the pH of the solution be after the addition of 25.0 mL of 0.100M HCl? (K a (C 6H 5COOH) = 6.5 x 10 -5)

1 answer:

Zigmanuir [339]3 years ago

5 0

Answer:

pH = 4.05

Explanation:

The pH of the benzoic buffer can be determined using H-H equation:

pH = pKa + log [A⁻] / [HA]

<em>Where pKa is -logKa = 4.187</em>

pH = 4.187 + log [Sodium Benzoate] / [Benzoic Acid]

<em>Where [] can be understood as moles of each specie.</em>

Thus, to find pH of the buffer we need to calculate moles of benzoic acid and sodium benzoate.

<em>Initial moles:</em>

<em />

Initial moles of benzoic acid and sodium benzoate are:

Acid: 250mL = 0.250L ₓ (0.250 moles / L) = <em>0.0625 moles of benzoic acid</em>

Benzoate : 250mL = 0.20L ₓ (0.250 moles / L) = <em>0.050 moles of sodium benzoate</em>

<em>Moles after reaction:</em>

Now, 0.0250L×(0.100mol/L) = 0.0025 moles of HCl are added to the buffer reacting with sodium benzoate, C₆H₅COONa, producing more benzoic acid, as follows:

HCl + C₆H₅COONa → C₆H₅COOH + NaCl

That means after reaction moles of both species are:

Benzoic acid: 0.0625 mol + 0.0025mol (Moles produced) = 0.065 moles

Sodium Benzoate: 0.050mol - 0.0025mol (Moles that react) = 0.0475 moles

Replacing in H-H equation:

pH = 4.187 + log [0.0475] / [0.065]

pH = 4.05

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