NaHCO3 dissolved in water - how is pH calculated?
Background
A buffer is a solution that has the ability to keep pH within a certain and narrow range even if an excess of hydrogen ions (H+) is applied to the solution. In other words: buffers resist changes in pH when reasonable amount of either H+ or OH- is added to the buffer solution.
There are many examples of buffers: Blood in humans is a well buffered system buffered by several buffers. Seawater is a buffer and is buffered around pH=8.4, mainly because of sodium bicarbonate.
The purpose of this particular webpage is to show how pH can be calculated in solution made by dissolving sodium bicarbonate (NaHCO3) in water.
Example:
Consider a solution made of 0.1 M NaHCO3. The sodium bicarbonate will dissolve into Na+ and HCO3-. Furthermore, the HCO3- will partly split into fractions of CO2 and CO32-, respectively depending on pH. First of all we should start by finding out what will happen to NaHCO3 when it dissolves in water.
This is easier that it might seem. We simply use the fact that the charge balance is always zero in an aqueous solution. From the pdf document about equilibrium we know how to calculate the [CO2], [HCO3-] and [CO32-]. In fact it is not necessary to calculate the [CO2] but to help understand what is going on, it's included in the following calculations that can also be seen in an excel spreadsheet.
From the chapter about equilibriums we already now that:
| Compound / ion | Calculation |
| [CO2]* | TIC / (1 + Ka(CO2*) / [H+] + Ka(HCO3-).Ka(CO2*) / [H+]2) |
| [HCO3-] | TIC / (1 + [H+] / Ka(CO2*) + Ka(HCO3-) / [H+]) |
| [CO32-] | TIC / (1 + [H+]2 / (Ka(HCO3-).Ka(CO2*)) + [H+] / Ka(HCO3-)) |
In the case we are adding 0.1 M NaHCO3 we are adding 0.1 M Na+ and 0.1 M HCO3 that is distributed
among CO2*, HCO3- and CO32- depending on pH.
This pH or [H+] is easily calculated (not considering ionic strength).
Step 1: A charge balance is established
It must be that [Na+] - [HCO3-] - 2.[CO32-] = 0.
If not, the charge balance is not zero and
something is wrong. This equation can be fine-tuned to include [H+] and [OH-] so that the
equations is:
[Na+] - [HCO3-] - 2.[CO32-] + [H+] - [OH-] = 0
We do not know either the [HCO3-] or the [CO32-] as the [H+] is unknown.
However, by trying out with the different [H+]s in the equation:
[Na+] - TIC / (1 + [H+] / Ka(CO2*) + Ka(HCO3-) / [H+]) - 2.TIC / (1 + [H+]2 / (Ka(HCO3-).Ka(CO2*)) + [H+] / Ka(HCO3-)) + [H+] - 10-14/[H+] = 0
it is seen that when [H+] is 10-8.31 the equation is fulfilled. This leads us to the
conclusion that a 0.1 M NaHCO3 has a pH of 8.31. All calculations are shown in the excel spreadsheet. It is now up to the reader
to figure out why the pH of 0.5 M NaHCO3 solution is also 8.31 and why only very, very dilute
solutions of NaHCO3 has a pH closer to 7.
This page is also available as a pdf-document that can be accessed by clicking here.
Web resources
Dissolving salts
A model (for download) illustrating the dissolving of salts
pdf document