How many moles of sodium hydroxide would have to be added to 225 mL of a 0.377 M acetic acid solution, in order to prepare a buffer with a pH of 4.670?

0.0385 moles of NaOH

Explanation:

One of the way to make a buffer is to combine a weak base with a strong acid. For this case, we have:

CH₃COOH → CH₃COO⁻  +  H⁺

NaOH →  Na⁺  + OH⁻

As every weak acid, we have the equilibrium:

CH₃COOH  +  H₂O  ⇄  CH₃COO⁻  +  H⁺

For this buffer we should know that:

mmoles CH₃COO⁻ at the end = mmoles NaOH.    In conclusion:

mmoles CH₃COO⁻ at the end = mmmoles of CH₃COOH - mmoles of base, we added.

mmoles of  acetic → 0.377 M = mmoles / 225mL ⇒ 84.8 mmmoles

(0.377 M . 225 mL)

Now we can apply the Henderson Hasselbach:

pH = pKa + log (mmmoles NaOH / mmoles of acetic acid - mmoles NaOH)

pKa of acetic → 4.75

4.670 = 4.75 + log ( x / 84.8 - x)

-0.08 = log ( x / 84.8 - x)

10⁻⁰'⁰⁸ = x / 84.8 - x

0.831 (84.8 - x) = x

70.5 - 0.831x = x

70.5 = 1.831x

x = 70.5 mmmoles / 1.831 = 38.5 mmmoles of NaOH

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it is a solid

explanation: