# How can I predict whether a reaction will occur? [duplicate]

I'd like to know how can I predict if two substances will react. I already know that there's a reaction if one of the products is a weak electrolyte or a precipitate but that leads me to the question:

If we have two substances such as salts and they react with one another forming two other salts, when is the reaction double replacement and when there is no reaction?

I'll try giving two examples from the problems I came across:

Which one of the following substances will react with one another in a solution –

• $\ce{(NH4)2CO3, CuCl2, K2SO4, Ba(NO3)2, HCl}$
• $\ce{Na2CO3, Ba(NO3)2, Na2SO4, AgNO3, CaCl2}$
• – Berry Holmes May 12 '17 at 11:53
• well thank you for that,so the way i understand it is that whenever a weaker acid is formed,there is a reaction. And it does apply to salts as well: whenever a salt thats from a weak acid is formed, there will be a reaction? – moondemon May 12 '17 at 11:58
• It would work if the constituent base of the salt obtained as the product and the constituent base of the reactant salt is the same. – Berry Holmes May 12 '17 at 12:01
• could you perhaps illustrate that with an example using the salts i mentioned in the post. I apologize about my inability to understand these things and i thank you for that wonderful edit! – moondemon May 12 '17 at 12:24

For the first example, I definitely wouldn't want to react $$\ce{CuCl2}$$ with $$\ce{HCl}$$, because then I would end up getting the same products again. I'm primarily driven by the motive to create something new, albeit the reaction must be a feasible one. Many factors influence feasibility, and you've mentioned the formation of a precipitate is one of them.

### Reactions involving neutralization:

$$\ce{HCl}$$ would react with $$\ce{(NH4)2CO3}$$, $$\ce{K2SO4}$$ and $$\ce{Ba(NO3)2}$$, as the reactions give out $$\ce{H2CO3}$$, $$\ce{H2SO4}$$ and $$\ce{HNO3}$$ respectively which are weaker acids.

### Reactions involving precipitation:

I see a very tempting reaction here:

$$\ce{K2SO4 + Ba(NO3)2 -> 2KNO3 + BaSO4\downarrow}$$

$$\ce{BaSO4}$$ is a very good precipitate, and I highly emphasize this fact. Just as a fun fact: It can be partially soluble if you try to dissolve it in boiling concentrated $$\ce{HCl}$$ (or boiling concentrated $$\ce{H2SO4}$$).

On similar grounds, $$\ce{Ba(NO3)2}$$ would react with $$\ce{(NH4)2CO3}$$ to give out the precipitate of $$\ce{BaCO3}$$. How do I know this? Partly because of the fact that while testing for the presence of $$\ce{CO3^2-}$$ ions in a laboratory we use the lime water test. Instead of using $$\ce{Ca(OH)2}$$, baryta water or $$\ce{Ba(OH)2}$$ is also used at times yielding strikingly similar results.

Also, I know this partly because $$\ce{CO3^2-}$$ ion is notorious for forming precipitates. It forms a precipitate with a large number of ions, some exceptions (they're the only exceptions I know) being $$\ce{Na+, K+, Rb+, Cs+}$$ and $$\ce{NH4+}$$.

However, what happens to the reactions like these?

$$\ce{(NH4)2CO3 + K2SO4 <=> K2CO3 + (NH4)2SO4}$$

Neither of the product is a precipitate. Hey that's easy? $$\ce{K2SO4}$$ is made up of $$\ce{KOH}$$ and $$\ce{H2SO4}$$. And look the base is the same for $$\ce{K2CO3}$$, that means I've made carbonic acid from sulfuric acid and thus the reaction takes place. Right? Wrong.

In your haste you forgot to take note that I could've used the same logic if I were to deal with $$\ce{(NH4)2CO3}$$ and $$\ce{(NH4)2SO4}$$. Wait! This reverses my previous reasoning! Yes it does. Or maybe we get a little confused.

Perhaps these cations are spectators, shall I strip them off? Yeah, let's give it a try too. We're now (sadly) left with:

$$\ce{CO3^2- + SO4^2- <=> CO3^2- + SO4^2-}$$

That's bizarre. What's happening? Nothing. And that's what it is in its true form, the ions basically are just combining with each other, there are four ions, two cations and two anions, four possible combinations. Does that make it a reaction? Perhaps not. It's just an equilibrium, there's no going forward or backward. When we say there's a reaction taking place, we are implying that the equilibrium sways to the the product side, which apparently isn't the case here.