Question

Prepare a 2.00 L of a 2.00 M buffer of pH 4.35. You can use citric acid (pka 3.09, 192.1 g/mol) and NaOH (10.0 M). Calculate how much of each reagent (and water) to make the buffer (show work).

The answers are:

Citric acid: 786 g, NaOH: 514 mL, water: 1.486 L

How do you get to these results?

Answer 1

Given : concentration of buffer = 2.00 M

This means concentration of citric acid and conjugate base = 2.00 M

[citric acid] + [conjugate base] = 2.00 M ....(1)

According to Henderson - Hasselbalch equation,

pH = pK_a + log([conjugate base] / [citric acid])

4.35 = 3.09 + log([conjugate base] / [citric acid])

log([conjugate base] / [citric acid]) = 4.35 - 3.09

log([conjugate base] / [citric acid]) = 1.26

[conjugate base] / [citric acid] = 10^1.26

[conjugate base] / [citric acid] = 18.2

[conjugate base] = [citric acid] * 18.2   ...(2)

Substituting the value of conjugate base from equation (1)

2.00 M - [citric acid] = [citric acid] * 18.2

2.00 M = (1 + 18.2) * [citric acid]

2.00 M = 19.2 * [citric acid]

[citric acid] = (2.00 M) / (19.2)

[citric acid] = 0.104 M

[conjugate base] = 2.00 M - [citric acid]

[conjugate base] = 2.00 M - 0.104 M

[conjugate base] = 1.896 M

moles citric acid present = (concentration citric acid) * (volume of buffer)

moles citric acid present = (0.104 M) * (2.00 L)

moles citric acid present = 0.208 mol

moles conjugate base present = (concentration conjugate base) * (volume of buffer)

moles conjugate base present = (1.896 M) * (2.00 L)

moles conjugate base present = 3.792 mol

moles NaOH added = moles conjugate base present

moles NaOH added = 3.792 mol

volume NaOH added = (moles NaOH added) / (concentration NaOH)

volume NaOH added = (3.792 mol) / (10.0 M)

volume NaOH added = 0.3792 L

volume NaOH added = 379.2 mL

initial moles citric acid added = (moles citric acid present) + (moles conjugate base present)

initial moles citric acid added = (0.208 mol) + (3.792 mol)

initial moles citric acid added = 4.00 mol

mass citric acid = (initial moles citric acid added) * (molar mass citric acid)

mass citric acid = (4.00 mol) * (192.1 g/mol)

mass citric acid = 768.4 g

volume of water added = (volume of buffer) - (volume NaOH added)

volume of water added = (2.00 L) - (0.3792 L)

volume of water added = 1.6208 L