You are asked to prepare 500. mL of a 0.100 M acetate buffer at pH 5.10 using on

Question

You are asked to prepare 500. mL of a 0.100 M acetate buffer at pH 5.10 using only pure acetic acid (MW = 60.05 g/mol, pK_a = 4.76), 3.00 M NaOH, and water. Answer the following questions regarding the preparation of the buffer. 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. What volume of 3.00 M NaOH must you add to the acetic acid to achieve a buffer with a pH of 5.10 at a final volume of 500 mL? (Ignore activity coefficients.) Pour the beaker contents into a 500 mL volumetric flask. Rinse the beaker several times and add the rinses to the flask. Swirl the flask to mix the solution. Add water to the mark to make the volume 500 mL. Stopper the flask and invert how many times to ensure complete mixing? 1 2 20

Explanation / Answer

Q2.

V of NaOH required fo achieve 5.10 at V =500 mL

so...

initially, we have:

mol of acid = mass/MW = 3/60 = 0.05mol = 50 mmol of acid

for a pH = 5.10

we need

pH = pKa + log(A-/HA)

5.10 = 4.75 + log(A-/HA)

10^(5.1-4.75) = (A-/HA)

2.2387 = (A-/HA)

A- = formed when NaOH is added

HA = left after reaction between acid nad NaOH

so

initially

mmol of HA = 50

mmol of A- = 0

after adding, mmol of base = Mbase*Vbase = 3*Vbase

mmol of HA = 50 - 3*Vbase

mmol of A- = 0 + 3*Vbase

so

2.2387 = (A-/HA)

2.2387 *(50 - 3*Vbase) = 3*Vbase

2.2387 *(50) - 3*2.2387 *Vbase = 3*Vbase

111.935-6.7161Vbase = 3Vbase

(3+6.7161)*Vbase = 111.935

Vbase = 111.935/(3+6.7161) = 11.520 mL required

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