Question

How to calculate the ratio of sodium carbonate and sodium bicarbonate in 'x' Molar sodium carbonate-sodium biocarbonate buffer?

For instance, I wish to prepare 0.25M carbonate-bicarbonate buffer of pH9.0. How do I determine the amount of sodium carbonate and sodium bicarbonate to prepare the buffer?

Answer 1

Use 20 g sodium hydrogen carbonate and 1.1 g sodium carbonate.

Explanation:

The equation for the buffer equilibrium is

`"HCO"_3^"-" + "H"_2"O" ⇌ "CO"_3^"2-" + "H"_3"O"^"+"; "p"K_text(a) = 10.32`

Let's rewrite this equation as

`"HA" + "H"_2"O" ⇌ "A"^"-" + "H"_3"O"^"+"`

The Henderson-Hasselbalch equation is

`"pH" = "p"K_text(a) + log((["A"^"-"])/(["HA"]))`

(a) Calculate the buffer ratio

`"pH" = "p"K_text(a) + log((["A"^"-"])/(["HA"]))`

`9.0 = 10.32 + log((["A"^"-"])/(["HA"]))`

`log((["A"^"-"])/(["HA"])) = "9.0 - 10.32 = -1.32"`

`(["A"^"-"])/(["HA"]) = 10^"-1.32"`

`(["HA"])/(["A"^"-"]) = 10^1.32 = 20.9`

`["HA"] = 20.9["A"^"-"]`

Also

`["HA"] + ["A"^"-"] + = 0.25`

`20.9["A"^"-"] + ["A"^"-"] = 21.9["A"^"-"]= 0.25`

`["A"^"-"] = 0.25/21.9 = 0.0114`

`["HA"] = "0.25 - 0.0114 = 0.239"`

So, you dissolve `"0.239 mol HA (NaHCO"_3)` and `"0.0114 mol A"^"-" ("Na"_2"CO"_3)` in enough water to make 1 L of solution.

(b) Calculate the masses of `"NaHCO"_3` and `"Na"_2"CO"_3`

`"Mass of NaHCO"_3 = 0.239 color(red)(cancel(color(black)("mol NaHCO"_3))) × ("84.01 g NaHCO"_3)/(1 color(red)(cancel(color(black)("mol NaHCO"_3)))) = "20 g NaHCO"_3`

`"Mass of Na"_2"CO"_3 = 0.0114 color(red)(cancel(color(black)("mol Na"_2"CO"_3))) × ("105.99 g Na"_2"CO"_3)/(1 color(red)(cancel(color(black)("mol Na"_2"CO"_3)))) = "1.2 g Na"_2"CO"_3`