# 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?

Jun 2, 2018

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

#### Explanation:

The equation for the buffer equilibrium is

$\text{HCO"_3^"-" + "H"_2"O" ⇌ "CO"_3^"2-" + "H"_3"O"^"+"; "p} {K}_{\textrm{a}} = 10.32$

Let's rewrite this equation as

$\text{HA" + "H"_2"O" ⇌ "A"^"-" + "H"_3"O"^"+}$

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"]))

$9.0 = 10.32 + \log \left(\left(\left[\text{A"^"-"])/(["HA}\right]\right)\right)$

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

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

$\left(\left[\text{HA"])/(["A"^"-}\right]\right) = {10}^{1.32} = 20.9$

$\left[\text{HA"] = 20.9["A"^"-}\right]$

Also

$\left[\text{HA"] + ["A"^"-}\right] + = 0.25$

$20.9 \left[\text{A"^"-"] + ["A"^"-"] = 21.9["A"^"-}\right] = 0.25$

$\left[\text{A"^"-}\right] = \frac{0.25}{21.9} = 0.0114$

["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 ${\text{NaHCO}}_{3}$ and ${\text{Na"_2"CO}}_{3}$

${\text{Mass of NaHCO"_3 = 0.239 color(red)(cancel(color(black)("mol NaHCO"_3))) × ("84.01 g NaHCO"_3)/(1 color(red)(cancel(color(black)("mol NaHCO"_3)))) = "20 g NaHCO}}_{3}$

${\text{Mass of Na"_2"CO"_3 = 0.0114 color(red)(cancel(color(black)("mol Na"_2"CO"_3))) × ("105.99 g Na"_2"CO"_3)/(1 color(red)(cancel(color(black)("mol Na"_2"CO"_3)))) = "1.2 g Na"_2"CO}}_{3}$