Why doesn't water play a role in reactions in aqueous reactions?

Question

$$\ce{Ba(NO3)2(aq) + Na2CO3(aq)}$$

In the above reaction both the substances are in water, however water plays no role in the reaction. Why is it that although water is present in the reaction, it does not play a reactant role? To clarify, why isn't the above written as:

$$\ce{Ba(NO3)2(aq) + Na2CO3(aq) + H2O}$$

Why does water act as a bystander and not a reactant?

Answer 1

Water is far from a bystander in aqueous solutions. It has a number of features that greatly facilitate chemical reactions, as well as being an important ligand (the aqua ligand) of transition metal complexes.

Here are some (unbalanced) examples:

• Water has a rather large dielectric constant and as such is very effective at shielding charges from one another, facilitating the charge separations inherent in the dissolution of ionic compounds. This is why you can easily dissolve sodium chloride in water, but not in hexane. The nature of this shielding is often articulated as several concentric 'solvation shells', where 'shells' of water molecules cluster around an ion with their dipoles approximately aligned towards or away from the ion.

Here's an image from Wikipedia qualitatively depicting the first solvation shell surrounding a sodium ion:

• More generally, the large dipole moment of water allows it to partake in hydrogen bonding, where water will weakly associate with other polar molecules or functional groups. This extends to water forming (short lived, highly fluxional) clusters with itself.

• Water is also happy to accept (forming hydronium, $\ce{H3O+}$) and donate (forming hydroxide, $\ce{OH-}$) protons in solution. So much so that it will actually do this with itself, in a process called autoionisation. This process makes water molecules significant in a wide range of acid/base and redox equilibria.

Answer 2

It does play a role--it's all hidden in the $_\ce{(aq)}$. Whenever an ionic compound is dropped in water, it dissociates into the ions. This dissociation is facilitated by the water, as it solvates the ions.

What happens is that a certain amount of energy is required to dissociate an ionic compound (lattice energy). If a pure compound was dissociated, there would be quite a bit of energy required to do this, and you would end up with something unstable (the molecule would probably just recombine.)

Now, water, being polar, solvates the ions. What happens is that water molecules aggregate around the ion, and the net charge gets effectively dispersed. This leads to a gain in stability (and a decrease in potential energy known as the "hydration energy"). Ionic compounds with enough hydration energy (specifically, when the absolute value of hydration energy>lattice energy) are said to be "soluble", and dissociate readily in water.

So the actual reaction is:

$\ce{Na2CO3 ->[\ce{H2O}] 2Na+_{(aq)} + CO3^{2-}_{(aq)}}$

Where the $\ce{_{(aq)}}$ signifies solvation. There will be a similar one for barium nitrate.

So, when one mentions $\ce{Ba(NO3)2_{(aq)} + Na2CO3_{(aq)}}$, it is actually $\ce{2Na+_{(aq)} + CO3^{2-}_{(aq)} + Ba^{2+}_{(aq)} + 2NO3^{-}_{(aq)}}$.

Now, what happens is that the barium and carbonate ions react to form insoluble barium carbonate. This precipitates out, and we get the following reaction:

$$\ce{\underbrace{Ba(NO3)2_{(aq)} + Na2CO3_{(aq)}}_{Mixture~of~ions} ->[\ce{H2O}] \underbrace{BaCO3}_{Insoluble~solid} v + \underbrace{NaNO3_{(aq)}}_{Mixture~of~ions}}$$

Water isn't reacting here, it is more of a catalyst. Without it, the reaction wouldn't occur (unless you melt the compounds--molten ionic compounds behave similar to their aqueous counterparts).