You are asked to prepare 500. mL of a 0.200 M acetate buffer at pH 5.10 using only pure acetic acid (MW=60.05 g/mol, pKa=4.76),

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3 years ago

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You are asked to prepare 500. mL of a 0.200 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.
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.)
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?

Chemistry

1 answer:

Alenkasestr [34] · Answer 1 · 2022-06-14T03:56:37+03:00

Answer:

1. 6.005 g

2. 22.9 mL

3. Until the mixtures becomes homogeneous.

Explanation:

A buffer is a solution where a weak acid is in equilibrium with its conjugate base (its anion) or a weak base is in equilibrium with its conjugate base (its cation). The buffer remains the pH almost unaltered because it shifts the equilibrium if an acid or base is added.

1. The pH of a buffer can be calculated by the Henderson-Hasselbalch equation:

pH = pKa + log[A⁻]/[HA]

Where [A⁻] is the concentration of the conjugate base (the anion) of the acid, and HA is the acid concentration.

5.10 = 4.76 + log[A⁻]/[HA]

log[A⁻]/[HA] = 5.10 - 4.76

log[A⁻]/[HA] = 0.34

[A⁻]/[HA] = $10^{0.34}$

[A⁻]/[HA] = 2.1878

Because the volume is the same, we can replace the concentration by the number of moles (n):

nA⁻/nHA = 2.1878

nA⁻ = 2.1878*nHA

The total number of moles of the substances in the buffer is: 0.200 mol/L * 0.5 L = 0.1 mol

nA⁻ + nHA = 0.1

2.1878*nHA + n HA = 0.1

3.1878nHA = 0.1

nHA = 0.0314 mol

nA⁻ = 0.0686 mol

The total number of moles of acetic acid needed is 0.1 mol (both substances may be from it):

m = MW*mol

m = 60.05*0.1 = 6.005 g

2. NaOH must react with acetic acid to form the anion, so for a 1:1 reaction, it will be needed 0.0686 mol of NaOH:

V = mol/concentration

V = 0.0686/3

V = 0.0229 L = 22.9 mL

3. The buffer must be a homogeneous solution, it means that it can't be noticed phases in the buffer, so the flask must be inverted until all the buffer is diluted in water, and it will be noticed when the solution becomes homogenous.