Answer:
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
