How you would make 100.0 ml of a 1.00 mol/L buffer solution with a pH of 10.80 to be made using only sodium carbonate, sodium hydrogen carbonate and water? How much sodium carbonate and sodium hydrogen carbonate would you use?
You would add 7.96 g of sodium carbonate and 2.09 g of sodium hydrogen carbonate to a volumetric flask and then add enough water to make the volume up to 100.0 mL.
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1. Calculate the concentration ratios
The chemical equation for the equilibrium is
HCO₃⁻ + H₂O ⇌ CO₃²⁻ + H₃O⁺;
HA + H₂O ⇌ A⁻ + H₃O⁺
The Henderson-Hasselbalch equation is
10.80 = 10.32 +
This makes sense. pH >
2. Calculate the concentrations
[A⁻] = 3.02[HA]
Also, [A⁻] + [HA] = 1.00 mol/L
3.02[HA] + [HA] = 4.02[HA] = 1.00 mol/L
[A⁻] = 3.02[HA] = 3.02 × 0.2488 mol/L = 0.7512 mol/L
3. Calculate the masses of NaHCO₃ and of Na₂CO₃
Mass of NaHCO₃ = 0.1000 L ×
Mass of Na₂CO₃ = 0.1000 L ×
You would transfer 2.09 g of NaHCO₃ and 7.96 g of Na₂CO₃ to a 100 mL volumetric flask and make up to the mark with distilled water.