[Wil"]

calcium carbonate
conside this equation
CaCO3(aq) + CO2(aq) + H2O(l)--> Ca2+(aq) + 2HCO3-(aq)

Since CO2 will become carbonic acid when dissovles in water, can I rewirte the equation like this?

CaCO3(aq) + H2CO3(aq)--> Ca2+(aq) + 2HCO3-(aq)

but if I write like this, the product should be different. when a carbonated compound reacts with acid, the product should be the mineral salt, water and carbon dioxide...



so what do you think?

[Borek]

Carbonic acid is a weak one, only partially dissociated. It is also very unstable, so it is safer to write CO₂ and H₂O.

In reality what you have in solution is a system with several reactions each with its own equilibrium constant. Both equations as you wrote them give some insight into what is going on, but they by no means describe the situation completely.

[shelanachium]

Most CO2 dissolved in water remains as such, very little if any exists as H2CO3. What little reaction there is, is best written:

CO2(aq) + H2O <=> H+(aq) + [HCO3]-(aq), with the equilibrium far to the left.

making the solution weakly acidic but with few or no H2CO3 molecules formed at all.

The amount of [CO3]2- formed is small indeed as for [HCO3]-(aq) <=> H+(aq) + [CO3]2-(aq) the equilibrium is even further to the left, as expected in the second ionisation of any acid - you are removing a positive charge from a species which already has a negative charge.

When CaCO3 dissolves in carbonated water, the reactions are:

CO2(aq) + H2O <=> H+(aq) + [HCO3]- then

[CO3]2- + H+(aq) <=> [HCO3]-(aq)

Giving as the net reaction [CO3]2- + H2O + CO2 <=> 2[HCO3]-

[cetirizine]