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Whilst I understand why salts can be acidic or basic, the carbonate reaction confuses me (carbonate being a basic ion).

$$\mathrm{{CO_3}^{2-} + H_2O \ce{<=>} {HCO_3}^- + OH^-} \\ \mathrm{{HCO_3}^{-} + H_2O \ce{<=>} {H_2CO_3} + OH^-} $$

The second reaction produces an acid plus a base, so why don't they neutralise immediately? Also, I can see why it is a basic salt by releasing $\ce{OH-}$ but doesn't it also release carbonic acid which would disassociate to give off $\ce{H+}$ ions which would just react with the OH and give no net change in pH then? So basically, I'm confused as to why the acid and base don't neutralize immediately and the carbonic acid just "stores" protons.

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    $\begingroup$ The first reaction also produces an acid plus a base. Then again, have you heard about the autoionization of water? $$\ce{H2O<=>H+ +OH-}$$ $\endgroup$ – Ivan Neretin Feb 3 at 8:14
  • $\begingroup$ The concept of equilibrium between reagents and products is important. $\endgroup$ – Buck Thorn Feb 3 at 11:05
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    $\begingroup$ It is more complicated than you indicated. Read about carbonic acid at Wikipedia. So it is really the various equilibria between dissolved $\ce{CO2, H2CO3, HCO3^-}$ and $\ce{CO3^{2-}}$ $\endgroup$ – MaxW Feb 3 at 19:09
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    $\begingroup$ I think it might be more clear if you wrote the second reaction as $\ce{HCO3- + H3O+ <=> H2CO3 + H2O}$. Does that help you make sense of it? Basically, as you've written it, the equilibrium lies far to the left. The right side products are only favored if there is a high concentration of protons and low concentration of hydroxide ions, ie at low pH. $\endgroup$ – Andrew Feb 4 at 1:21
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There are two concepts you need to know.

What is equilibrium?

Not all reactions occur until one of the reactants is consumed completely. Some of them occur until a certain point, where the system doesn't continue to evolve; rather, the concentrations of all the species remain constant in what is called a chemical equilibrium. This was first observed experimentally and then some theories started to appear in order to explain this.

The most important variable is the equilibrium constant, a quotient of the equilibrium concentration. If you think of an equilibrium as two opposite reactions, then this constant measures which one is prevalent.

If we have:

$$ \ce{aA + bB <=> cC + dD} $$

Then the equilibrium constant K is:

$$ K = \frac{ [C]^c \times [D]^d }{ [A]^a \times [B]^b } $$

This means that if you mix A and B, or C and D, or those two with more reactants added, no matter what the initial result is, the final concentration of everyone has to be such that at the end, K is the same.

So, if K is big, it means at any given time there is more of C and D than A and B. The direct reaction (to the right) is prevalent. If K is small, then the opposite reaction (to the left) is prevalent.

So yeah, it is possible to have at the same time the reactants and the products of a reaction simultaneously in the same place. How? That is concept number two.

Equilibrium is dynamic

I told you to picture an equilibrium as two opposite reactions. That is because in reality, those two reactions are happening continuously! Indeed, in your example:

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

A certain amount of carbonate molecules are reacting with water to give $\ce{HCO3^-}$ and $\ce{OH-}$, while the exact same amount of other, different $\ce{HCO3^-}$ molecules are reacting with $\ce{OH-}$ to give water and carbonate.

Now, equilibrium isn't instantaneous. What happens is that when you throw reactants together, the more favorable reaction happens at light speed while the opposite almost doesn't happen. As products start to accumulate, the speeds start to become more moderate until they are exactly the same and the overall reaction stops. The concentrations stop changing even though there are still molecules reacting all the time.

Conclusion

To simplify, you can think of the process you described as a two-step process. First, $\ce{CO3^2-}$ gives $\ce{HCO3^-}$ and $\ce{OH-}$ and achieves equilibrium (equation 1 that you wrote). Then, the $\ce{HCO3^-}$ reacts and reaches equilibrium, and as initial conditions we take the products of the first step.

  1. Carbonate in water $$ \ce{CO3^2- + H2O <=> HCO3^- + OH-}\qquad K_1 = 2.13 \times 10^{-4} $$

    Okay, so here we see that most $\ce{CO3^2-}$ remains as-is, producing only a small amount of $\ce{HCO3^-}$ and $\ce{OH-}$.

  2. Everybody reacts

    $\ce{HCO3^-}$ has many options. It can undergo the same reaction as before, reversed: $$ \ce{HCO3^- + OH- <=> CO3^2- + H2O}\qquad K_2 = 1/K_1 = 4700^\mathbf{*} $$ It can also further react with water: $$ \ce{HCO3^- + H2O <=> H2CO3 + OH-}\qquad K_3 = 2.22 \times 10^{-8}\\ \ce{HCO3^- + H2O <=> CO3^{2-} + H+}\qquad K_4 = 4.7 \times 10^{-11} $$

    Here we see that the only significant reaction would be to reverse the first one (although the others also happen, but to a minimal extent). But we said before that an equilibrium consists of two simultaneous opposite reactions. Thus, the concentration remains largely the same, unaffected by the others which have very small K.

The amount of $\ce{OH-}$ produced is bigger than what was originally on water, $$\ce{H2O <=> H+ + OH-} \quad K_w = 10^{-14}$$

due to the value of the equilibrium constants ($K_w = 10^{-14} \ll K_1 = 2.13 \times 10^{-4}$).

So $\ce{CO3^2-}$ is indeed a basic salt.


$\mathbf{^*}$ Easy to understand if you look at the generic expression for K.

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