Question

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?

Answer 1

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.

This question is like the question in

http://socratic.org/questions/0-10-mol-of-solid-sodium-hydrogen-carbonate-and-0-20-mol-of-solid-sodium-carbona

1. Calculate the concentration ratios

The chemical equation for the equilibrium is

HCO₃⁻ + H₂O ⇌ CO₃²⁻ + H₃O⁺; `K_"a"` = 4.8 × 10⁻¹¹; `pK_"a"` = 10.32
HA + H₂O ⇌ A⁻ + H₃O⁺

The Henderson-Hasselbalch equation is

pH = `pK_"a" + log(([A⁻])/([HA]))`

10.80 = 10.32 + `log(([A⁻])/([HA]))`

`log(([A⁻])/([HA]))` = 10.80 – 10.32 = 0.48

`([A⁻])/([HA]) = 10^0.48` = 3.02 (3 significant figures)

This makes sense. pH > `pK_"a"`, so there should be more A⁻ than HA.

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

[HA] = `(1.00"mol/L")/4.02` = 0.2488 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 × `(0.2488"mol NaHCO₃")/(1"L") × (84.01"g NaHCO₃")/(1"mol NaHCO₃")` = 2.09 g NaHCO₃ (3 significant figures)

Mass of Na₂CO₃ = 0.1000 L × `(0.7512"mol Na₂CO₃")/(1"L") × (106.0"g Na₂CO₃")/(1"mol Na₂CO₃")` = 7.96 g Na₂CO₃ (3 significant figures)

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.