How many milliliters of a 0.150 M NaOH solution must be added to a 25.0-mL sample of 0.150 M butanoic acid, CH3(CH2)2CO2H, in order to create a buffer with a pH of 4.82? Ka of butanoic acid = 1.5 x 10-5

12.5 mL of NaOH

Explanation:

0.150 M NaOH

CH3-CH2-CH2-COOH  0.150M

This is the data given. The reaction of both reactants make a neutralization and a salt

CH3-CH2-CH2-COOH  +  NaOH →  NaCH3-CH2-CH2-COO  +  H₂O

As both compounds produce a buffer, we can apply the Henderson Hasselbach equation:

pH = pKa + log ( mmol NaOH / Initial mmol of acid - mmol NaOH)

Final mmol of acid = mmol of NaOH

Final mmol of acid = Initial mmol of acid - mmol of NaOH we added.

Let's find the data to replace in the formula:

pKa butanoic acid = -  log Ka → - log 1.5×10⁻⁵  = 4.82

Initial mmol of acid = M . mL → 25 mL . 0.150M → 3.75 mmol

Notice that M = mmol/mL → mmol = M . mL.

We replace → 4.82 = 4.82 + log mmol NaOH / ( 3.75 mmol - mmol NaOH)

As pH = pKa, notice that log = 1 so:

mmol NaOH / ( 3.75 mmol - mmol NaOH) = 1

mmol NaOH = 3.75 mmol - mmol NaOH

2mmol NaOH = 3.75 → mmol NaOH = 3.75 / 2 = 1.875

M = mmol /mL → mL = mmol / M → 1.875 mmol / 0.150M = 12.5 mL

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