You are asked to prepare 500. mL of a 0.150 M acetate buffer at pH 5.00 using on
ID: 847373 • Letter: Y
Question
You are asked to prepare 500. mL of a 0.150 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.)
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?
Explanation / Answer
for acidic buffers
pH = pKa + log [ conjugate base / acid ]
pH = pKa + log [ CH3COONa / CH3COOH]
5 = 4.76 + log [ CH3COONa / CH3COOH]
[CH3COONa] / [ CH3COOH] = 1.7378
given
total acetate = 0.15 M
[CH3C00H] + [CH3COONa] = 0.15
[CH3COOH] + 1.7378 [CH3COOH] = 0.15
2.7378 [CH3COOH] = 0.15
[CH3COOH] = 0.05479 M
[CH3COONa ] = 1.7378 x 0.05479
[CH3COONa] = 0.0952 M
1)
moles of CH3COOH = molarity x volume (L)
moles of CH3COOH = 0.05479 x 0.5
moles of CH3COOH = 0.027395
mass of CH3COOH = moles x molar mass
mass of CH3COOH = 0.027395 x 60
mass of CH3COOH = 1.6437 g
so 1.6437 grams of acetic acid should be added
2)
CH3COOH + NaOH ----> CH3COONa + H20
moles of CH3COONa = 0.0952 x 0.5
moles of CH3COONa = 0.0476
moles of NaOH added = moles of Ch3COONa formed = 0.0476
volume of NaOH added = moles / molarity
volume of NaOh = 0.0476 / 3 = 0.01586 L
volume of NaOH = 15.86 ml
so 15.86 ml of NaOh should be added
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