You are asked to prepare 500. mL of a 0.250 M acetate buffer at pH 4.90 using on
ID: 1037555 • Letter: Y
Question
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. Number 7.51 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.) Number 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
1)
No of moles of acetic acid required = (0.250mol/1000ml)×500ml = 0.125
Molar mass of acetic acid = 60.05g/mol
mass of acetic acid required = 60.05g/mol × 0.125mol = 7.5063g
2)
Henderson - Hasselbalch equation is
pH = pKa + log([A-] /[HA])
[A-] /[HA] = 10pH - pKa
[A-] /[HA] = 104.90 - 4.76
[A-] / [HA] = 1.3804
[A-] = 1.3804[HA]
Total buffer concentration is 0.250M
Therefore,
[HA] + 1.3804[HA]= 0.250M
2.3804M[HA] = 0.250M
[HA] = 0.1050M
[A-] = 0. 250M - 0.1050M
[A-] = 0.1450M
Therefore,
[CH3COOH] required = 0.1050M
[CH3COO-] required = 0.1450M
CH3COOH + NaOH - - - - - > CH3COONa + H2O
Stoichiometrically, 1mole NaOH produce 1mole of CH3COO-
No of moles of CH3COO- required =(0.1450mol/1000ml)×500ml= 0.0725
No of moles of NaOH required = 0.0725M
Volume of 3.00M NaOH required = (1000ml/3.00mol)×0.0725mol =24.17ml
Therefore,
Volume of 3.00M NaOH should be added = 24.17ml
3)
20
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