Question

You are asked to prepare 500. mL 0.150 M acetate buffer at pH 5.10 using only pure acetic acid ( MW = 60.05 g/mol, p K a = 4.76 ), 3.00 M NaOH , and water. How many grams of acetic acid will be needed to prepare the 500. mL buffer? Note that the given concentration of acetate refers to the concentration of all acetate species in solution.

Answer 1

Answer:

4.504g of acetic acid

Explanation:

The acetic acid in reaction with NaOH produce acetate ion, thus:

CH₃COOH + NaOH → CH₃COO⁻ + H₂O + Na⁺

That means the moles of acetate buffer comes, in the first, from the acetic acid

As you need 500mL (0,500L) of a 0.150M acetate buffer, moles are:

0.500L × (0.150mol / 1L) = 0.075 moles of acetate. That is:

0.075mol = [CH₃COO⁻] + [CH₃COOH]

Thus, grams of acetic acid you need to prepare the buffer are:

0.075 moles acetic acid × (60.05g / 1mol) = 4.504g of acetic acid