10
$\begingroup$

I was reading about acids and bases today and finally decided to question the statement that "baking soda is a base."

Let's start with the dissolving of baking soda, $\ce{NaHCO3}$. The equation to represent its dissolution in water is $\ce{NaHCO3 + H2O <=> Na+ + H3O+ + CO3^{-2}}$.

In this "reaction," baking soda is a proton donor, creating hydronium ions and a conjugate base, $\ce{CO3^{-2}}$. Doesn't this make baking soda an acid?

If what I have written here is correct, then baking soda will create a base when dissolved, but baking soda itself is not the base. Is this accurate?

$\endgroup$
5
  • 6
    $\begingroup$ It's "amphoteric", meaning it can act as an acid or a base. It's usually called a base because it acts as a weak base in aqueous solutions giving a slightly basic pH. But, if you add a strong base to a sodium bicarbonate solution, the bicarbonate will give up a proton (meaning it's acting as an acid) to the base. $\endgroup$
    – airhuff
    Apr 17, 2017 at 20:39
  • $\begingroup$ @airhuff But water is not considered a strong base, right? Then is my above equation still accurate? Or do we tend to ignore dissociations when we are interested in acid-base reactions? Dissociation is sort of a reaction, no? $\endgroup$ Apr 17, 2017 at 20:41
  • 2
    $\begingroup$ Really it's: $\ce{NaHCO3 + H2O <--> H2CO3 + OH− + Na+}$. Bicarbonate is a stronger base than water. $\endgroup$
    – airhuff
    Apr 17, 2017 at 20:48
  • 2
    $\begingroup$ And $\ce{H2CO3}$ (carbonic acid) decomposes to $\ce{CO2}$ and water. Note that tap water can have a pH of 8.5 or more, so you may not see CO2 bubbling out on addition of Na bicarbonate. Tap water can also be acidic, it just depends on where you live and your water system. (I don't know if you're trying this or if you were already well aware of that, just FYI.) Hopefully someone here has time to lay this all out clearly for you in a good answer ;) $\endgroup$
    – airhuff
    Apr 17, 2017 at 21:00
  • 2
    $\begingroup$ And yes, this dissociation absolutely is a chemical reaction, of the acid-base type. $\endgroup$
    – airhuff
    Apr 17, 2017 at 21:03

2 Answers 2

9
$\begingroup$

The statement you read that "baking soda is a base" comes from the fact that a solution of sodium bicarbonate (baking soda) and water has a pH of around $8.3$.

However, sodium bicarbonate is amphoteric with respect to Brønsted–Lowry acid/base theory, which means that it can act as either an acid or a base. More specifically, the bicarbonate ion is amphiprotic, meaning that it can either donate a proton, acting as an acid, or accept a proton, making it act as a base. The reason that a sodium bicarbonate solution is slightly basic is that it's ability to take a proton from a water molecule is greater than its ability to donate a proton to a water molecule. In other words, the chemical reaction causing the solution to be basic is:

$$\ce{HCO3- + H2O <--> H2CO3 + OH-}$$

The other equilibrium reaction at play, the reaction "trying" to make the system more acidic, is:

$$\ce{HCO3- + H2O <--> CO3^2- + H3O+}$$

Again, in an otherwise pure aqueous system, it is the first reaction that dominates, and thus giving the observed slightly basic solution.

Note that the $\ce{H2CO3}$ produced in the first reaction is also part of yet another equilibrium process:

$$\ce{H2CO3 <--> CO2 ^ + H2O}$$

Where the $\ce{CO2}$ is liberated as a gas. This is the cause of the well-known "fizzing" when baking soda is added to vinegar (acetic acid).

$\endgroup$
0
5
$\begingroup$

Sodium bicarbonate can indeed act as an acid. You just need a strong enough base with which it can react.

Take 0.1 mol sodium hydroxide and add it to a 100 mL distilled water sample, then add 0.1 mol sodium bicarbonate. The pH goes down in the second step as the solutes are combined to make the almost-but-not-quite strong base sodium carbonate. Compare this reaction:

$\ce{NaHCO3 + NaOH -> Na2CO3 + H2O}$

with this one where we see the same proton transfer process:

$\ce{HCl + NaOH -> NaCl + H2O}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.