Total ionic equations

[hotcommodity]

Hello, I'm having trouble writing out total ionic equations. In a total ionic equation, I know to write the ions separately as they exist in aqueous solution, and to leave solids as they are. But sometimes I come across examples where, even tho' one of the reactants or products is aqueous, it isn't broken up in ionic form. This is also giving me trouble writing net ionic equations, as I don't know what the spectator ions are. So my question is, what are the guidelines in writing total ionic equations? Any help would be appreciated, thanks...

[RyanJ]

Put simply spectator ions are ions that exist in a solution that do not interact with the other particles in any way. This means they can just be ignored because they don't preform any reacting

 

The guidelines are just ignore anything that doesn't react

[hotcommodity]

Put simply spectator ions are ions that exist in a solution that do not interact with the other particles in any way. This means they can just be ignored because they don't preform any reacting

 

The guidelines are just ignore anything that doesn't react

 

I get the idea of spectator ions, which do not appear in the net ionic equation. But I'm having trouble in trying to determine which aqueous solutions are written and examined in ionic form with respect to total ionic equations. I mean, does it have to be a strong acid or base to be written in ionic form? I just don't know what the guidelines are.

[RyanJ]

I get the idea of spectator ions, which do not appear in the net ionic equation. But I'm having trouble in trying to determine which aqueous solutions are written and examined in ionic form with respect to total ionic equations. I mean, does it have to be a strong acid or base to be written in ionic form? I just don't know what the guidelines are.

 

Not at all! Here's an example.

 

[ce]2KI + Pb(NO3)2 -> PbI2 + 2KNO3[/ce]

 

I guess we could write the net ionic equation as (note: not ignoring any spectators here):

 

[ce]2K^{+} + 2I^{-} + Pb^{2+} + 2NO3^{-} -> PbI2 + 2KNO3[/ce]

 

So we need to look at what is in the solution and what ions will do what in the solution. Once we know what they will do and how they will react we can predict the products. Does that help at all?

[hotcommodity]

Not at all! Here's an example.

 

[ce]2KI + Pb(NO3)2 -> PbI2 + 2KNO3[/ce]

 

I guess we could write the net ionic equation as:

 

[ce]2K^{+} + 2I^{-} + Pb^{+} + 2NO3^{-} -> PbI2 + 2KNO3[/ce]

 

So we need to look at what is in the solution and what ions will do what in the solution. Once we know what they will do and how they will react we can predict the products. Does that help at all?

 

Somewhat, I get what you're saying. I'll be right back, I'll find an example of one of the total ionic equations thats confusing me and edit this post...

[encipher]

I guess we could write the net ionic equation as:

 

[ce]2K^{+} + 2I^{-} + Pb^{+} + 2NO3^{-} -> PbI2 + 2KNO3[/ce]

 

 

That net ionic equation is incorrect.

 

The potassium Ion and the Nitrate Ion are spectator ions, and Lead has a charge of 2+, not +

 

Therefore the correct net ionic equation should be the following:

 

[ce]Pb_{(aq)}^{2+} + 2I_{(aq)}^{-} -> PbI2_{(s)}[/ce]

 

Generally, in most aqueous reactions like the one above. The group 1A alkali metals are spectators. I have indicated the state of each of the reactants and products to clarify, since the product will be the precipitate and is a solid due to its insolubility.

[RyanJ]

That net ionic equation is incorrect.

 

The potassium Ion and the Nitrate Ion are spectator ions, and Lead has a charge of 2+, not +

 

Therefore the correct net ionic equation should be the following:

 

[ce]Pb_{(aq)}^{2+} + 2I_{(aq)}^{-} -> PbI2_{(s)}[/ce]

 

Generally, in most aqueous reactions like the one above. The group 1A alkali metals are spectators. I have indicated the state of each of the reactants and products to clarify, since the product will be the precipitate and is a solid due to its insolubility.

 

I was writing out the whole thing as an example, thanks for pointing out the lead charge issue though. I missed that

[hotcommodity]

Alright, I found a few examples where an aqueous reactant and an aqueous product are not written in ionic form in a total ionic equation, which is what's confusing me.

 

example 1 (total ionic eq.):

 

H3PO4(aq) + 3[K^{+}(aq)+OH^{-}(aq)] -----> [3K^{+}(aq)+PO4^{3-}(aq)]+3H2O(l)

 

This is an example in my book, but I don't understand why H3PO4(aq) is not broken up into ions.

 

example 2(total ionic eq.):

2[H^{+}(aq)+Cl^{-}(aq)]+ CaCO3(s) ----> H2CO3(aq)+ [Ca^{2+}(aq)+2Cl^{-}(aq)]

 

Here I don't understand why the H2CO3(aq) is not broken up into ions.

[encipher]

H3PO4 (Phosphoric acid) Is a WEAK acid and does not dissociate extensively in water. Therefore, in an aqueous solution containing Phosphoric acid, there will exist more molecules of it, then there will exist ions. The rule regarding acids is similar to that of salts.

 

With salts it generally depends on solubility, with acids it depends on dissociation.

[hotcommodity]

H3PO4 (Phosphoric acid) Is a WEAK acid and does not dissociate extensively in water. Therefore, in an aqueous solution containing Phosphoric acid, there will exist more molecules of it, then there will exist ions. The rule regarding acids is similar to that of salts.

 

With salts it generally depends on solubility, with acids it depends on dissociation.

 

Ok, I think I understand now. You're saying if it's a strong acid, or a stong base, then it should be broken up into ions with respect to total ionic equations. And if the acid or base is weak, then it shouldn't be written as separate ions. I also understand why the H2CO3 wasn't broken up. I should have been looking at my solubility chart, and I would have seen that carbonates are generally insoluble. Please correct me if I'm wrong. Thanks for the replies.

[RyanJ]

Example 1:

 

[ce]H3PO4_{(aq)} + 3[K^{+}_{(aq)} + OH^{-}_{(aq)}] -> [3K^{+}_{(aq)} + PO4^{3-}_{(aq)}] + 3H2O_{(l)}[/ce]

 

Phosphoric acid is a weak acid therefore it has minimal dissociation in this context. So if we don't have much dissociation then there is little point writing it ionically.

 

In this case all that seems to be reacting is actually the dissociated hydrogen from the phosphoric acid (as an oxonium ion) and the hydroxyl group from the potassium hydroxide forming water, the rest are spectators too the reactants.

 

Example 2.

 

[ce]2[H^{+}_{(aq)} + Cl^{-}_{(aq)}] + CaCO3_{(s)} -> H2CO3_{(aq)} + [Ca^{2+}_{(aq)} + 2Cl^{-}_{(aq)}][/ce]

 

I'm not sure you could write calcium carbonate ionically here as it isn't in solution (its very insoluble in water, hot or cold).

 

Just a little hint, try using the chemistry LaTex tags, really helps when trying to present formulas and equations

[jdurg]

Example 1:

 

[ce]H3PO4_{(aq)} + 3[K^{+}_{(aq)} + OH^{-}_{(aq)}] -> [3K^{+}_{(aq)} + PO4^{3-}_{(aq)}] + 3H2O_{(l)}[/ce]

 

Phosphoric acid is a weak acid therefore it has minimal dissociation in this context. So if we don't have much dissociation then there is little point writing it ionically.

 

In this case all that seems to be reacting is actually the dissociated hydrogen from the phosphoric acid (as an oxonium ion) and the hydroxyl group from the potassium hydroxide forming water, the rest are spectators too the reactants.

 

Example 2.

 

[ce]2[H^{+}_{(aq)} + Cl^{-}_{(aq)}] + CaCO3_{(s)} -> H2CO3_{(aq)} + [Ca^{2+}_{(aq)} + 2Cl^{-}_{(aq)}][/ce]

 

I'm not sure you could write calcium carbonate ionically here as it isn't in solution (its very insoluble in water, hot or cold).

 

Just a little hint, try using the chemistry LaTex tags, really helps when trying to present formulas and equations

 

Correct. Calcium Carbonate would be written as a solid reactant and not as an ion.