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 20Explanation / 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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