You are asked to prepare 500. mL of a 0.200 M acetate buffer at pH 5.00 using on

Question

You are asked to prepare 500. mL of a 0.200 M acetate buffer at pH 5.00 using only pure acetic acid (MW=60.05 g/mol, pKa=4.76), 3.00 M NaOH, and water. Answer the following questions regarding the preparation of the buffer.

1. How many grams of acetic acid will you need to prepare the 500 mL buffer? Note that the given concentration of acetate refers to the concentration of all acetate species in solution.

2. What volume of 3.00 M NaOH must you add to the acetic acid to achieve a buffer with a pH of 5.00 at a final volume of 500 mL? (Ignore activity coefficients.

Explanation / Answer

1)

Let us take the desired moles of the buffer solution

That means 0.2 M x 0.5L = 0.1 moles,
Hence required amount of acetic acid can be calculate as

(desired Moles x (Formula Weight) = 0.1 x 60.05 = 6.005 g Acetic Acid

2)

According to Handerson equation

pH = pKa + log (Base/Acid)

5.00 = 4.76 + log [b/a]
log [b/a] = 0.24, [b/a] = 1.74
[b] = [1.74] x [a] or
[a] = [0.575] x [b]. Substitute this into the below for [a.]

No. of moles of buffer solution (b+ a) = 0.2 x 0.5 = 0.1 moles.
b + 0.575 x b = 0.1
1.575b = 0.1
b = 0.06349 mols.

Hence required base (NaOH) was 0.06349 moles of 3.0 M
3.0M = 0.06349 mols/L

Required volume of NaOH (L) = 0.02116 L or 21.16 mL.

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