3/22/2016 12:00 AM A 6.5/10 Gradebook Print Calculator Periadic Table Question 1
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3/22/2016 12:00 AM A 6.5/10 Gradebook Print Calculator Periadic Table Question 19 of 22 Map sapling learning You are asked to prepare 500. mL of a 0.200 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 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 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 20 Previous Give Up & View Solution O Check Answer Next Exit HintExplanation / Answer
Answer for (1):
HA ç====è H+ + A- (HA = CH3COOH)
One need to use the Henderson-Hasselbalch equation to find the ratio of A- to HA.
pH = pKa + log [A-] / [HA]
pH = 4.90
pKa = 4.76
4.90 = 4.76 + log [A-] / [HA]
log [A-] / [HA] = 4.90 – 4.76 = 0.14
[A-] / [HA] = log 0.14 = 100.14 = 1.38
Therefore [A-] = 1.38 and [HA] = 1
We have to find decimal fractions (part/whole) from the ratio of [A-] and [HA].
Therefore [A-] = 1.38 / 1 + 1.38 = 1.38/2.38 = 0.58
[HA] = 1 / 2.38 = 0.42
Find the molarity (M) of each component in the buffer by simply multiplying the molarity of the buffer by the decimal fraction of each component.
Molarity of [A-] = 0.2 x 0.58 = 0.116 M/L
Molarity of [HA] = 0.2 x 0.42 = 0.084 M/L
Calculate the total moles of each component in the 500 ml buffer solution.
Moles = Molarity x Liters of buffer
Number of moles of HA required = 0.084 M x 0.5L = 0.042 Moles
Number of moles of NaOH required = 0.116 M x 0.5 L = 0.058 Moles
Since this buffer is prepared by the reaction of a weak acid (HAc) with a strong base (NaOH), one
must determine the total moles of the weak acid component needed because the conjugate
base is made in situ.
Total Number of moles of HA required = Number of moles of HA required + Number of moles of NaOH required = 0.042 + 0.058 = 0.1 Moles
Pure acetic acid molarity = 17.4 M
Number of moles of acetic acid required = 0.1 M/L
Volume of acetic acid required = 0.1 / 17.4 = 0.0057 L = 5.7 ml of 17.4 M pure acetic acid need to be used.
Number of grams of acetic acid required = 5.7 ml x 1.05 d = 5.98 g
Answer for (2):
Number of moles of NaOH required = 0.058 Moles
Volume of NaOH required = 0.058 / 3 = 0.0193 L = 19.3 ml of 3 M NaOH need to be used.
Initially acetic acid may be added to 250 ml of deionized water and then slowly treated with an aqueous solution of NaOH. The mixture is finally diluted to desired volume, 500 ml. This solution will have 0.2 M concentration and a pH of 4.90.
Answer for (3): Invert 2 times to get complete mixing of solutions.
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