I'm not sure how to find the pH for the following solutions.

Question

Determine if each of the following results in the formation of a buffer solution. Justify your answer. Calculate the pH of the buffer solutions.

1. A solution containing 0.100 M acetic acid and 0.150 M sodium acetate
2. A solution containing 1.2 mM NH₃ and 0.40 M NH₄Cl
3. A solution containing 0.25 M NaOH and 0.25 M HCl.
4. A solution prepared by mixing 10.00 mL of 0.100 M NaHCO₃ and 10.00 mL of 0.200 M Na₂CO₃.
5. A solution prepared by mixing 25.00 mL of 0.50 M citric acid with 75.00 mL of 0.50 M NaOH.

Answer 1

First, what is a buffer? It is a weak acid or a weak base plus the conjugate base or acid, respectively.

Next, to find the pH of a buffer, one can use the Henderson Hasselbalch equation (see below).

1.) 0.100 M acetic acid (HAc) + 0.150 M sodium acetate (NaAc). This is a buffer since it contains a weak acid (HAc) and the conjugate base (NaAc, or simply Ac^-)

pH = pKa + log [Ac^-] / [HAc] .. Henderson Hasselbalch equation for an acidic buffer

pH = 4.75 + log (0.150 / 0.100) = 4.75 + log 1.5 (looked up pKa acetic acid and found it to be 4.75)

pH = 4.75 + 0.18

pH = 4.93

2.) 1.2 mM NH₃ + 0.40 M NH₄Cl. This is a buffer since it contains a weak base (NH₃) and the conjugate acid (NH₄Cl, or simply NH₄⁺).

pOH = pKb + log [NH4+] / [NH3] .. Henderson Hasselbalch equation for a basic buffer

pOH = 4.75 + log (0.40 / 1.2) = 4.75 + log 0.333 (looked up pKb for NH3 and found it to be 4.75)

pOH = 4.75 + (-048)

pOH = 4.27

pH = 14 - pOH

pH = 9.73

3.) 0.25 M NaOH + 0.25 M HCl. This is NOT a buffer since it contains a strong base (NaOH) and a strong acid (NaOH). Equal concentrations will result in complete neutralization and a pH = 7.0

4.) 10.00 ml 0.100 M NaHCO₃ + 10.00 ml 0.200 M Na₂CO₃. This is a buffer since it contains a weak acid (HCO₃^-) and the conjugate base (CO₃^2-).

moles HCO₃^- = 10.00 ml x 1 L / 1000 ml x 0.100 mol/L = 0.001 mols HCO₃^-

moles CO₃^2- = 20.00 ml x 1 L / 1000 ml x 0.200 mol/L = 0.004 mols CO₃^2-

pH = pKa + log [CO₃^2-25.] / [HCO₃^-]

pH = 10.32 + log (0.004 / 0.001)

pH = 10.32 + log 4

pH = 10.32 + 0.60

pH = 10.92

5.) 25.00 ml 0.50 M citric acid + 75.00 ml 0.50 M NaOH. This is a little tricky. Citric acid is a weak acid, but it is tri-protic, meaning it has THREE acidic hydrogens. These will be titrated by the NaOH, but since the moles of NaOH is equal to the moles of acidic hydrogens, ALL of the citric acid will be converted to tri-sodium citrate (the salt). Thus, this is NOT a buffer. If you had less NaOH, then a buffer would have been formed. As it is, we have the following situation:

moles citric acid = 25.00 ml x 1 L / 1000 ml x 0.50 mol/L = 0.0125 mols citric acid

moles NaOH = 75.00 ml x 1 L / 1000 ml x 0.50 mol/L = 0.0375 mols NaOH

citric acid + 3NaOH ==> trisodium citrate + 3H₂O

Result is formation of 0.0125 moles of trisodium citrate in 100.00 mls of solution = 0.125 M trisodium citrate

To find the pH of this solution, we must look at the hydrolysis of this compound

Na₃-citrate + 3H₂O ==> citric acid + 3NaOH

trisodium citrate is a base so we use the Kb for Na3-citrate

For citric acid Ka1=8.4x10^-4; Ka2=1.8x10^-5; Ka3=4.0x10^-6 Avg = 2.87x10^-4 (approximation)

Kb = 1x10^-14 / 2.87x10^-4 = ca. 3.48x10^-11 (approximation)

Kb = 3.48x10^-11 = [citric acid][OH-]³ / [Na₃-citrate]

3.48x10^-11 = (x)(x)³ / 0.125 - x (assume x small relative to 0.125 and ignore it in denominator)

3.48x10^-11 = x⁴ / 0.125

x⁴ = 2.78x10^-10

x = [OH-] = 4.08x10^-3 M

pOH = -log 4.08x10^-3

pOH = 2.39

pH = 14 - pOH

pH = 11.61