I would like to know how do I begin to solve a problem like this?

That is finding the formula equation?

I would assume that it would be $$ \ce{NH3 + CH3CH2COOH -> NH3 + CH3CH2COOH + H2O} $$

I am really confused as to how I can balance this equation. Do I give the ammonia a hydrogen? Do I break up the acid into smaller units?

From there I would like to know the best way for me to switch the equation into an ionic equation and then finally a net ionic equation.

By the way the net ionic equation is $$ \ce{CaCO3 (s) + 2HC2H3O2+ (aq) -> CO2 (g) + H2O (l) + Ca^2+ (aq) + 2C2H3O2^- (aq) } $$

I am just clueless as to how they got there.

  • $\begingroup$ everyday-chemistry is more for questions that are things that one might run into outside of a classroom setting, like (for lack of a handy example) "why does my toothpaste have tin fluoride in it?" I think this is more of an acid-base question, so I will retag it as such. $\endgroup$ – jonsca Oct 19 '14 at 1:22

We know that ammonia is $\ce{NH3}$ and propanoic acid is $\ce{CH3CH2COOH}$.

Ammonia is a basic substance, and propanoic acid is... an acidic substance. More specifically, ammonia fits the definition of a Bronsted-Lowry base, so it will act as an $\ce{H+}$ accepter in solution.

Arrhenius acids are expected to increase the concentration of $\ce{H+}$ in solution, and propanoic acid is no exception. In water as propanoic acid dissociates...

$\ce{CH3CH2COOH ->[H2O] CH3CH2COO- + H+}$

This is where the $\ce{H+}$ comes (and is later bonded to the ammonia via)...

$\ce{NH3 + H+ -> NH4+}$

After those two steps occur, the products $\ce{NH4+ + CH3CH2COO-}$ are formed. Your ionic equation is as follows...

$\ce{NH3 + CH3CH2COOH -> NH4+ + CH3CH2COO-}$

There appear to be no spectator ions (ions that appear on both sides of the equation), so there is nothing to cancel out. As such, this is ALSO the net ionic equation. They both appear to be balanced.

So, yes, you had to do both. You were supposed to add an $\ce{H+}$ to the ammonia, but that happens if you dissociate the acid.

As for the equation with the calcium, carbonate ions sometimes create carbon dioxide and water (via carbonic acid readily decomposing, like in sodas).

  • $\begingroup$ Write a net ionic equation to represent the reaction of aqueous strontium hydroxide with nitric acid: So I know that nitric acid is $\ce{HNO3}$ and strontium hydroxide is $\ce{Sr(OH)2} $, thus $\ce{Sr(OH)2 + HNO3}$ What should be my second step after this? Place water and these two elements on the product side of this equation and start balancing? $\endgroup$ – Cetshwayo Oct 19 '14 at 0:51
  • $\begingroup$ You're right that there will be water in the products, but it's more about the mechanism than just saying "water exists". $\ce{Sr(OH)2}$ is going to dissociate into $\ce{Sr^{2+} + 2OH-}$, and the nitric acid will dissociate into $\ce{H+ + NO3-}$. The $\ce{H+ + OH-}$ will make water, and the other two are going to make $\ce{Sr(NO3)2}$. But that's soluble in water, so you can't add those two. Your ionic equation is $\ce{Sr^{2+} + 2OH- + H+ + NO3- -> Sr^{2+} + 2NO3- + H2O}$. Cancel out the spectator ions and that's your net ionic equation. And of course, balance. $\endgroup$ – Sparkery Oct 19 '14 at 0:54
  • $\begingroup$ So my last step should be to balance the equation? I thought that was supposed to be the first step? Also, is there any rhyme or reason to dissociation. Do they always split into free elements or polyatomic ions? $\endgroup$ – Cetshwayo Oct 19 '14 at 1:08
  • $\begingroup$ Dissociation comes down to energy and bonding. When the reaction $\ce{Na+ + NO3- -> NaNO3}$ occurs (or even $\ce{Cl-}$), the ionic bond that holds the reaction in place comes from the sodium's sole valence electron which "fills the gap" and forms an octet with the 7 valence electrons of chlorine. Ionic bonds are really strong, but they break apart in water because it takes less energy (the energy to break the bond and form one with water, overall, loses energy and increases entropy). The $\ce{NO3-}$ wouldn't "dissociate" because it's bonded so well. There's also solubility charts to memorize. $\endgroup$ – Sparkery Oct 19 '14 at 1:21

I am just clueless as to how they got there.

Well, you're not the only one. I really don't see how you get to equation 2 from equation 1.

To find out how to get the acid-base reaction equation, you first need to know which compounds react as bases (i.e. take up a proton) and acids (i.e. deliver a proton).

In your case, ammonia is the base (because the nitrogen has a lone pair that can be accessed by a proton) and propionic acid is the acid (nomen est omen). Then you know that the final equation has to be $$ \ce{NH3 + CH3CH2COOH <=>[][\ce{H2O}] NH4+ + CH3CH2COO-} $$

Also, how often does one need to balance these types of reactions or do they usually end up balanced or working themselves out as balanced?

Acids (and bases) can be multiprotic, which means that they can give off (or accept) more than just one proton. An example for a diprotic acid would be carbonic acid $\ce{H2CO3}$.

As a result of this, it is possible to write several (equilibrium) reactions. For example: $$ \ce{2NH3 + H2CO3 <=> NH3 + NH4+ + HCO3- <=> 2NH4+ + CO3^{2-}} $$ Where this equilibrium then lies depends on the equilibrium constants for each of these reactions.

The second equation can be expanded some more to show "intermediate" steps: $$ \ce{CaCO3 + 2 CH3COOH -> Ca^{2+} + H2CO3 + 2CH3COO- \\ <=> Ca^{2+} + 2CH3COO- + H2O + CO2 ^} $$ Here carbonic acid decomposes to give carbon dioxide which bubbles out of the solution, thereby making the reaction stay out of the equilibrium zone.

I hope this clarified it somewhat for you.

  • $\begingroup$ The second equation is the "net ionic equation". I just fixed it. Nonetheless, when it comes to "Acid Base Reactions" should the primary action I need to take to write (emphasis) one be to represent the reaction so that it shows that a hydrogen atom is being donated (positive proton) from the ionic acid? Also, how often does one need to balance these types of reactions or do they usually end up balanced or working themselves out as balanced? $\endgroup$ – Cetshwayo Oct 19 '14 at 0:40
  • $\begingroup$ @Cetshwayo I've edited my answer. $\endgroup$ – tschoppi Oct 19 '14 at 0:54

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