Why can't the product of an acid reacting with water react with water and produce $\ce{OH-}$? Take the reaction below as an example:

$$\ce{H2CO3(aq) + H2O(l) <=> HCO3-(aq) + H3O+(aq)}$$ $$\ce{HCO3-(aq) + H2O(l) <=> CO3^2-(aq) + H3O+(aq)}$$

Why can't the $\ce{CO3^2-}$ react with water (not $\ce{H3O+}$) and produce $\ce{OH-}$ ions? If it can then why is the solution acidic and not basic?

PS. I am a high school student and would really appreciate a straightforward answer.

  • 8
    $\begingroup$ It can and does. They all sit together in a nice equilibrium. $\endgroup$ Jan 16 at 10:34
  • $\begingroup$ The following hydrolysis reaction is a well known equilibrium : $\ce{CO3^{2-} + H2O <=> HCO3^- + OH-}$ But $\ce{CO3^{2-}}$ is not an acid. It is a base. $\endgroup$
    – Maurice
    Jan 16 at 10:46
  • 1
    $\begingroup$ Convenient reference for text/formula formatting: Notation basics / Formatting of math/chem expressions / upright vs italic // For more: Math SE MathJax tutorial. // Not to be applied in CH SE titles. $\endgroup$
    – Poutnik
    Jan 16 at 10:57
  • $\begingroup$ @Maurice I see. But if this reaction also happens in a container with water and H2CO3, why is the solution and acid and not a base? $\endgroup$ Jan 16 at 13:32
  • $\begingroup$ Imagine CO2 solution with anions OH-, HCO3- and CO3^2-. As the only present cation is H3O+ and there is requirement of electroneutrality, there is more H3O+ than OH- and the solution must be then acidic. $\endgroup$
    – Poutnik
    Jan 16 at 15:43

2 Answers 2


It definitely does. Let's simplify it to a monoprotic acid (like acetic acid, $\ce{CH3COOH}$), which I'll just represent as $\ce{HA}$ for simplicity. In a solution of the acid, you will have all of the following species floating around: $\ce{HA}$, $\ce{A-}$, $\ce{H2O}$, $\ce{H3O+}$, $\ce{OH-}$. Every conceivable reaction between these species will happen, such as:

$$\begin{align} \ce{HA + H2O &<=> A- + H3O+} \tag{1} \\ \ce{A- + H2O &<=> HA + OH-} \tag{2} \\ \ce{2 H2O &<=> H3O+ + OH-} \tag{3} \end{align}$$

To obtain a completely accurate value for the pH, and the equilibrium position, it is necessary to take all of these equations into account.*

But often we can get away with less work, and only bother with the first. Why? One qualitative way of looking at it is: when we add a weak acid $\ce{HA}$ to water, we know that very little of it will dissociate anyway; this is kind of by definition. So, there isn't much $\ce{A-}$ in the system anyway; and the second reaction can pretty much be ignored.

That's not to say that there is no $\ce{OH-}$ in the system. There is still $\ce{OH-}$ floating around. For example, in a pH 5 solution (i.e. $[\ce{H+}] = \pu{10^{-5} M}$), we have that $[\ce{OH-}] = \pu{10^{-9} M}$.* It's still there, it's just that there's very little of it.

A more formal explanation would involve solving it properly and verifying that the true solution is very close to the approximate solution obtained by ignoring eqns. (2) and (3).

On the other hand, if $\ce{HA}$ is a strong acid (like $\ce{HCl}$), then the reverse reaction $\ce{Cl- + H2O -> HCl + OH-}$ is completely negligible because $\ce{Cl-}$ is a pitifully weak base. (Strong acids have very weak conjugate bases.)

* In fact, these equations are not independent; note that (1) + (2) = (3), so there are only two 'pieces of information' to be obtained from these reactions, one corresponding to $K_\mathrm{a}$ and one corresponding to $K_\mathrm{w}$. But that's a story for another day.

  • 2
    $\begingroup$ Wow, this is the best explanation I have ever received! Although I am still wondering if the acid is so weak that we can ignore the second reaction (due to the small amount of A-) how can the H3O+ make a difference? (considering the fact that the amount of A- and H3O+ is the same?) $\endgroup$ Jan 16 at 17:46
  • $\begingroup$ In pure water, which is pH 7, there's only $\pu{10^{-7} M}$ of $\ce{H3O+}$. You really don't need a lot of extra $\ce{H3O+}$ to turn that into pH 4 or 5, so just the tiny bit that dissociates from the weak acid is enough. $\endgroup$ Jan 16 at 18:08
  • $\begingroup$ if this is the case, then how can we ignore the reaction that A- will have with water and the small amount of OH- that gets produced? $\endgroup$ Jan 16 at 18:15
  • $\begingroup$ @AuthenticMelody [H3O+]=[OH-]+[HCO3-] + 2[CO3^2-] -> [H3O+] >[OH-] -> pH <7. $\endgroup$
    – Poutnik
    Jan 16 at 18:24
  • $\begingroup$ @AuthenticMelody Because basicity (as judged by pH > 7) isn't caused by A-; it's caused by OH-. Reaction 1 starts from quite a lot of HA and generates a tiny bit of H3O+ and A-. Reaction 2 starts from a tiny bit of A-, and generates a vanishingly small amount of OH-. Of course, implicit in this is the assumption that HA is a weak acid and A- is a weak base. This is true, though: consider acetic acid, pKa = 4.76; that's a weak acid. The conjugate base, acetate, has a pKb of 14 - 4.76 = 9.24; that's also a weak base. So, reaction 2 pretty much just doesn't happen. $\endgroup$ Jan 16 at 18:32

I really like this question because it digs at wanting to understand what's going on beneath the equations you've been given to know why that equation works.

Orthocresl answered it really well, but I might be able to get a little more to what pH and pOH are and give a little more understanding to what's meant by acidic and basic, and what equilibrium is. I feel like that might be as much your question as whether or not H2O can react with CO3 2- (apologies, first time answering a question here, and not sure how to format this correctly).

tl;dr version up front:

Pure water has an equal amount of H+ and OH-, both of them exist in water. We consider this neutral, neither acidic or basic. Anything that moves the concentration to favor more H+ is considered an (Arrhenius) acid, anything that moves the concentration to favor more OH- is considered a base. Even so, there will still be both H+ and OH- in the solution. Individual water molecules will continue to accept protons (H3O+, usually shown as H+) and donate protons (OH-), but there will be more water molecules accepting a proton to form H3O+ than donating them to form OH- at lower pH.

non tl;dr version:

Very, very few chemical reactions are truly one way/irreversible unless the products leave the system entirely. Almost all will be occurring in both directions at the same time. But, one will be occurring more/faster. Let's say reaction A occurs twice as fast as the reverse reaction B (A: a + b -> c; B: c -> a + b). Since A occurs twice as fast a B, then there will be more products, c, than reactants, a and b. However, c will still be producing a and b. Eventually you'll get enough c that the reaction will reach a point that it will look like the reaction stopped. If reaction A occurs twice as fast as reaction B, this will be when there's twice as much products as reactants.

Basically, if a + b produce c twice per minute, and c produces a + b once per minute, you'll eventually get twice as much c as you have a and b, and the reaction will look like it stopped from a macro point of view, but will continue to occur in your solution.

Take what I said there from a conceptual view. It's useless for practical purposes. If you want to know the math, this video explains it really well (https://www.youtube.com/watch?v=2PM1yc_z4Bk&list=PL2ub1_oKCn7qmUZ80MJDPaRTdgJm8ZenX).

I don't see a good way to not discuss different definitions of acids even though I feel it's a little beyond this question. Arrhenius acid is anything that lowers the pH of pure water (base is the opposite). But that's not a good definition. A much better one is the Bronsted definition, anything that donates a proton acts as an acid, and anything that accepts a proton acts as a base (there's a 3rd definition, but it's difficult to understand at first). In this case, your carbonic acid is acting as an acid when it donates a proton to water to form H3O+, and water acts as a base. However, the reverse happens as well. H3O+ will donate a proton to HCO3-. In which case H3O+ is acting as an acid. H2O can also act as a Bronsted acid and donate a proton to form OH-. So yes, CO3 2- can and does act as a base and accept a proton from H2O, and you get OH- floating around.

In any aqueous solution, you'll end up with both H3O+ and OH-. This is usually measured with pH. pH is actually the result of this equation pH = -log(10)([H+]). That equation means that pH is the negative base 10 logarithm of the concentration of protons (hydronium ions) in the solution. However, because we know there's an equilibrium between [H+] and [OH-], we can actually measure this in terms of the concentration of OH- ions instead, pOH. pOH is remarkably, the negative logarithm of the concentration of hydroxide ions, pOH = -log(10)([OH-]).

So let's say you have a pH of 5, a slightly acidic solution. You'll have a pOH of 9. The pOH measures in the opposite direction, pOH below 7 is basic, above 7 is acidic. These two measure will equal 14. A pH of 1 will correspond to a pOH of 13. So even at a very high acidity of pH 1, you will still have H2O acting as an acid and donating protons to form OH-. It will just be happening very slowly, and so there won't be many of them.

  • $\begingroup$ Thank you for the thorough answer. It really helped me gain a better understanding of the topic. $\endgroup$ Jan 17 at 7:26

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.