Question

2. How would you make 100 mL of a carbonic acid buffer at 0.5 M and pH = 6.0 using 1.0 M NaHCO3 and either 1.0 M NaOH or 1.0 M HCl and water? (Show your work.)

Answer 1

Plan: use 1.0 M NaHCO3 as the source of both HCO3– (A–) and H2CO3 (HA) by protonating part of the bicarbonate with 1.0 M HCl. Use Henderson–Hasselbalch with pKa (H2CO3 ⇌ H+ + HCO3–) = 6.35.

1) Use HH to get the ratio [A–]/[HA]:
pH = pKa + log([A–]/[HA])
6.00 = 6.35 + log([A–]/[HA])
log([A–]/[HA]) = −0.35
[A–]/[HA] = 10^(−0.35) ≈ 0.4467

2) Total buffer concentration = [A–] + [HA] = 0.50 M.
Let r = 0.4467 = [A–]/[HA]. Then [A–] = 0.50 * r/(1 + r)
[A–] = 0.50 * 0.4467 / 1.4467 ≈ 0.1544 M
[HA] = 0.50 − 0.1544 = 0.3456 M

3) Convert to moles for 100 mL (0.100 L):
nA– = 0.1544 M × 0.100 L = 0.01544 mol
nHA = 0.3456 M × 0.100 L = 0.03456 mol
total bicarbonate needed initially = nA– + nHA = 0.05000 mol

4) Volumes of stock solutions (1.0 M):
Volume NaHCO3 required = 0.05000 mol / 1.0 M = 0.05000 L = 50.0 mL
Volume 1.0 M HCl required to protonate 0.03456 mol HCO3– = 0.03456 L = 34.6 mL

Procedure (practical):
- Measure 50.0 mL of 1.0 M NaHCO3 into a beaker and stir.
- Slowly add about 34.6 mL of 1.0 M HCl with stirring (add slowly—CO2 may evolve); do this in a fume hood or vented area.
- After reaction, transfer and dilute to 100.0 mL with deionized water and mix.
- Check pH and adjust slightly if necessary (small corrections with dilute HCl or NaOH).

Notes:
- pKa taken as 6.35; small differences in the chosen pKa or CO2 loss may require small pH adjustments experimentally.
- NaOH is not appropriate here because the target pH (6.0) is below the bicarbonate pKa, so acid (HCl) is needed to make the carbonic acid form.