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You are asked to prepare 500. mL of a 0.250 M acetate buffer at pH 4.90...

You are asked to prepare 500. mL of a 0.250 M acetate buffer at pH 4.90 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 4.90 at a final volume of 500 mL? (Ignore activity coefficients.)

3. 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?

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Expert Solution

moles of acetic acid = 500 x 0.250 / 1000 = 0.125

mass of acetic acid = moles x molar mass

= 0.125 x 60.05

= 7.50 g

mass of acetic acid = 7.50 g

total moles of buffer = x + y

x = acetic acid

y = acetate

x + y = 0.125 ----------------------> (1)

pH = pKa + log (y / x)

4.90 = 4.76 + log (y/x)

y / x = 1.38

y = 1.38 x ------------------> (2)

1.38 x + x = 0.125

x = 0.0525

y = 0.0725

moles of acetate = 0.0725

moles of acetate = moles of NaOH

from the reaction :

CH3COOH + NaOH -------------------> CH3COONa + H2O

molarity = moles / volume

3.00 = 0.0725 / volume

volume = 0.0242 L

volume = 24.2 mL

volume of NaOH = 24.2 mL

volume of water needed = 500 - 24.2 = 475.8 mL

queen_honey_blossom answered 1 year ago

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