Chemistry 30

3.4 Precipitation Reactions

Sometimes when two aqueous solutions are mixed together a solid is produced. This solid is called a precipitate, and the reaction is known as a precipitation reaction.

We can use our knowledge of solubility to predict whether a precipitate will form. For our discussion here, you will need to have the Solubility Table handy:

(Although there is a mathematical method is often used for predicting precipitates, we will keep our discussion simple here).

Before we can predict whether or not a precipitation reaction will occur, it is useful to review a category of chemical reactions called double displacement (or double replacement) reactions. In this type of chemical reaction, we will predict what happens when solutions of two ionic compounds are mixed. If any reaction occurs, the two positive cations will exchange places.

For example:

ZnBr_{2 (aq)} + 2 AgNO_{3 (aq)} → Zn(NO₃)₂ + 2 AgBr

In this reaction we see that zinc and silver have switched places.

When writing double replacement reactions it is very important that you write down the correct chemical formula for each product before balancing the entire equation.

KEY POINT: Keep in mind that our reactants, ZnBr₂ and AgNO₃, are present as solutions, meaning that

ZnBr_{2 (aq)} is really a short-hand form for writing Zn²⁺_(aq) + 2 Br^-_(aq)

AgNO_{3 (aq)} is really short-hand for Ag⁺_(aq) + NO₃^-_(aq)

It is important that you understand that the Zn²⁺ and Br^- ions are not joined together when in solution - they are separated from one another and present as separate ions rather than a single molecule/formula unit.

The same is true for AgNO_{3 (aq)} - the Ag⁺ and NO₃^- ions are present as separate ions, not bonded together.

It is possible that when these two solutions are mixed together every ion simply remains in solution, as separate ions. If that is the case, then no reaction actually occurs - now we just have one solution with four separate ions present.

But if we look at our solubility table, we see that Ag⁺ and Br^- forms an insoluble compound. These two ions will "find each other" in the solution and form an insoluble precipitate, a solid.

This occurs because the attraction between the Ag⁺ and Br^- ions is stronger than the attraction between the separate ions and the polar water molecules of the solvent. This is essentially what insoluble means.

On the other hand, we see in our solubility table that Zn²⁺ and NO₃^- form a soluble compound. Thus these two ions will remain in solution, not really forming a compound at all. The attraction between the polar water molecules and the Zn²⁺ and NO₃^- ions is stronger than the attraction between Zn²⁺ and NO₃^-. This is what soluble means.

We now need to return to our double displacement reaction and add the physical states to the two products:

ZnBr_{2 (aq)} + 2 AgNO_{3 (aq)} → Zn(NO₃)_{2 (aq)} + 2 AgBr_(s)

To get a clearer view of what really occurs during this reaction, we will write our solutions out in long form, rather than the short-cut form. Remember - anything that is (aq) is really present as separate ions. This, however, ONLY applies to aqueous states.

Thus we can rewrite our equation as:

Zn²⁺_(aq) + 2 Br^-_(aq) + 2 Ag⁺_(aq) + 2 NO₃^-_(aq) →

   Zn²⁺_(aq) + 2 NO₃^-_(aq)+ 2 AgBr_(s)

Notice that some ions are the same on both the reactant and product side of the equation, indicating that they did not undergo any change during the reaction. They were present, but did not precipitate.

Ions that are present in a reaction but do not participate are called spectator ions.

Sometimes we remove spectator ions from an equation in order to highlight what changes actually occurred during a reaction.

Reactions in which the spectator ions have been removed are called net ionic equations.

 

 

Therefore, for the overall equation:

Zn²⁺_(aq) + 2 Br^-_(aq) + 2 Ag⁺_(aq) + 2 NO₃^-_(aq) → Zn²⁺_(aq) + 2 NO₃^-_(aq)+ 2 AgBr_(s)

The net ionic equation is:

2 Ag⁺_(aq) + 2 Br^-_(aq) → 2 AgBr_(s)

(we usually arrange items so the positive cations are listed before the negative anions.)

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An important thing for chemistry students to always keep in mind is that while it is possible to predict the products of a double displacement reaction, that does not always mean that a reaction will actually take place.

What does that mean?

Let's look at what happens when we mix solutions of Mg(CH₃COO)₂ and (NH₄)₂SO₄.

Begin by writing a balanced equation and predicting the products of this double replacement reaction. Later we will predict the physical states of the products but we know that the reactants are both aqueous solutions.

Mg(CH₃COO)_{2 (aq)} + (NH₄)₂SO_{4 (aq)}→ MgSO₄ + 2 NH₄CH₃COO

Next, refer to a table of solubility to determine whether either, or both, of the two products will form insoluble precipitates.

We find that both MgSO₄ and NH₄CH₃COO are soluble compounds, meaning they will remain in solution and not form precipitates.

Thus, there is really NO REACTION occurring, since all participants simply remain in solution!

Lab Activity Precipitation Reactions

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Knowledge of precipitation reactions also allows us to use selective precipitation to separate mixtures of ions.

For example . . .

We have a solution that contains the following cations:

Ag⁺, Cu²⁺, and Mg²⁺

Of course, it isn't possible to have a beaker full of just cations - each has to be associated with a negative anion. For the sake a argument, let's say each of these cations is associated with nitrate ions, NO₃^-.

Nitrate ions are useful to use since compounds containing nitrate are ALWAYS soluble. So we really have a mixture of AgNO₃, Cu(NO₃)₂, and Mg(NO₃)₂. The nitrate ions will remain spectator ions, so we will not need to consider them further.

We wish to isolate each ion by causing one at a time to precipitate out of solution. Once a solid precipitate forms, we can filter the solid precipitate out, leaving the other ions still in solution.

We are given the following solutions to use:

Na₂S, NaCl, and NaOH.

As with nitrate, you should begin to recognize that sodium ions, Na⁺, also always form soluble compounds. So Na⁺ will remain as a spectator ion and we can ignore it. Thus, we are really given solutions of the following anions:

S^2-, Cl^-, and OH^-

We will be adding these one at a time to our cation mixture. We need to determine the order that we add these solutions so that we have ONE AND ONLY ONE precipitate occur with the first addition. This will remove one of the cations from the solution. We then continue until all cations are isolated from each other.

To help us analyze our situation, we'll prepare a chart, with the cations we need to separate along one axis and anions along the other axis:

.	Ag+	Cu2+	Mg2+
S2-	.	.	.
Cl-	.	.	.
OH-	.	.	.

Use a table of solubility to determine which combinations produce soluble compounds (sol) and which ones form compounds of low solubility. These are the ones that will precipitate out of solution (ppt - short-hand for precipitate).

.	Ag+	Cu2+	Mg2+
S2-	ppt	ppt	sol
Cl-	ppt	sol	sol
OH-	ppt	ppt	ppt

We need to determine the correct order for adding the negative ions. Be very careful with this next step - and be sure to read your chart in the right direction.

If we add S^2- first, we would get two cations to precipitate - Ag⁺ and Cu²⁺. This would not help us, as they still be mixed together, this time as solids.

Adding OH^- first would be even worse - all three cations would now be mixed together as solid precipitates.

Only by adding Cl^- first can we separate out one of the ions, Ag⁺. The others would remain in solution. The formula of our first precipitate would be AgCl_(s). It is important to realize that while both S^2- and OH^- also form a precipitate with Ag⁺, once it has been removed using Cl^- we no longer need to worry about it - it's all gone.

Once we've added Cl^-, we're removed Ag⁺ from solution. So if we now add S^2- we can again form a single precipitate - CuS_(s), thus removing Cu²⁺ from the mixture.

This will leave only Mg²⁺ still in solution. Sometimes this last cation is allowed to remain in solution. But if we chose we can add OH^- and produce our final precipitate, Mg(OH)_{2 (s)}

To summarize our solution:

1. First add NaCl to remove Ag⁺ , forming the precipitate AgCl_(s)
2. Second, add Na₂S to remove Cu²⁺, forming CuS_(s)
3. Last add NaOH to remove Mg²⁺, forming Mg(OH)_{2 (s)}

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Do the practice exercises, then complete Assignment 5 before continuing to the next section.