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You are asked to prepare 500. mL of a 0.100 M acetate buffer at pH 5.00 using only pure acetic acid (MW=60.05 g/mol,...

You are asked to prepare 500. mL of a 0.100 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?

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Answer #1
Concepts and reason

An aqueous solution consists of a mixture of a weak acid and its conjugate base, or a weak base and its conjugate acid which is represented as a buffer solution. These buffer solutions can maintain pH at a nearly constant value, upon addition of small amount of strong acid or base.

Fundamentals

The pH can be calculated by the Henderson – Hasselbalch equation.

pH = pKa + log
(salt)
(acid]

pKa=-logKa
ka =dissociation constant
[salt]=concentration of salt
facid] =concentration of acid

The concentration can be calculated by,

moles
Molarity =
volume

(1)

CH,COOH(aq)+ NaOH(aq) +
CH,COONa (aq)+H,0(1)

Given:
pH = 5.00
pk = 4.76
[Salt] = x
[Acid]=y

substitute the values in the pH formula:
pH = pka + log,
[salt]
[acid]
5.0= 4.76+ log
log - 0.24
= 1.74
x = 1.74y

given that the total concentration of buffer = 0.100M
x + y = 0.100
1.74y + y = 0.100
2.74y = 0.100

y=0.0365M
x=1.74y
x= 1.740.0365M
x= 0.0635M

Moles of buffer = Molarity xvolume
= 0.100X0.500
Moles of buffer = 0.05

Molarity of sodium acetate=0.0635
Moles of sodium acetate = 0.0635 x 0.500
Moles of sodium acetate = 0.0318

Moles of sodium acetate = Moles of NaOH
Moles of sodium acetate = 0.0318
Moles of acetic acid = y
Moles of acetic acid = 0.03

Initially,
Moles of acetic acid = 0.0183+ moles of NaOH
= 0.0183+0.0318
Initial moles of acetic acid = 0.0501

Mass of acetic acid was calculated as follows

mass
Moles =
molecular mass
Mass = moles x molecular mass
moles of acetic acid = 0.0501

Molecular mass of acetic acid = 60.05g/mol
Mass of acetic acid = 0.0501x60.05
Mass of acetic acid = 3.008g

[Part 1]

Part 1

(2)

Given:
Molarity of NaOH = 3.00M
Calculation:

pH = pKa + log
(salt)
(acid]

Number of moles of sodium acetate required to attain the buffer with pH=5.00
and volume of 500 ml
was calculated as 0.0318

Moles of NaOH = moles of sodium acetate
Moles of sodium acetate=0.0318
Moles = molarity Volume
Moles of NaOH = 0.0318

3.00 V = 0.0318
V _ 0.0318
3.00
V = 10.60 mL
Volume of NaOH = 10.60mL

(3)

Swirl the flask till it dissolves completely.

Ans: Part 1

The mass of acetic acid = 3.008 g

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