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You are asked to prepare 500. mL of a 0.250 M acetate buffer at pH 5.10 using on

ID: 550766 • Letter: Y

Question

You are asked to prepare 500. mL of a 0.250 M acetate buffer at pH 5.10 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. or ace ane revers to he' concentraal acedaie sipeaies in given concentration Number 2. 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.) Number mL 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? O 2 O 20

Explanation / Answer

Q1

pH = pKa + log(acetate/acetic acid)

pH = 4.76 + log(A-/HA)

if pH = 5.10 , then

5.10 = 4.76 + log(A-/HA)

A-/HA = 10^(5.10-4.76)

A-/HA = 2.1877

A- + HA = M*V = 0.25*500 = 125

then.

A- = 2.1877*HA

2.1877*HA + HA = 125

HA = 125 / (1+2.1877) = 39.21 mmol

A- = 2.1877*39.21 = 85.77 mmol

then,

mol of acetic acid = mmol*MW = (39.21)(60) = 2352.6 mg = 2.352 g of acetic acid

this is in equilbirium... if the answer refers to total acetic aicd, then

A- +HA = 39.21 +85.77 = 125 mmol

mass = mol*MW = 0.125*60 = 7.5 g of of acetic acid initially

Q2

then, we need to neutralize 85.77 mmol of HA to form 85.77 mmol of A-

ratio is 1:1

mmol of A- = 85.77

mmol of base = 85.77

V = mmol/M = 85.77 3

V = 28.59 mL of base required

Q3

Only once,

if we do this more times then we will need to get rid of excess V = 500 mL solution mark

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