Precipitation and Solubility Limits

In this lesson, you have learned about the solubility rules for ionic compounds and how to apply them to determine if a particular compound is soluble. This approach is useful, but oversimplified in a couple important ways.

First, it only covers ionic compounds, and even then only a partial set of all the ionic compounds out there. You have also learned the principle of "like dissolves like" for molecular compounds, but this too is incomplete.

Second, it oversimplifies by creating two black-and-white categories of "soluble" and "not soluble." In reality, almost all compounds will dissolve in water - at least up to some very small extent - and very few compounds can be added to water without any limit, especially not ionic compounds.

Solubility and Saturation

In reality, most substances have a "solubility limit" - a maximum concentration of the substance that can be dissolved in a solvent. We will discuss this concept in more detail in the next lesson. For now, let's just say that almost every substance has a solubility limit. For compounds we call "insoluble" that limit is very low, sometimes almost undetectably low. For compounds we call "soluble" the limit is much higher, but it still exists. Even NaCl, a very soluble ionic compound, cannot be dissolved in water indefinitely. Eventually, if you keep spooning salt into water, it will simply settle to the bottom instead of dissolving.

Solubility limits can change depending on conditions. In particular, temperature can have a strong effect. For most substances, higher temperatures increase their solubility, both in water and other solvents.

Saturation

As you might imagine, a solution that is below its solubility limit is unsaturated. This means it has some solute in it, but more could be added.

It is sometimes possible for a solution to be supersaturated, meaning it is above its solubility limit. This sounds like an oxymoron; how could a solution be holding more solute than the maximum? There are two answers, each of which you will observe in the lab for this lesson.

First, supersaturated solutions can be created by manipulating temperature. As noted above, solubility generally increases with temperature. It is often possible to take a hot solvent, dissolve a lot of solute in it, then slowly and gently cool it down. If you are careful, the extra solute will remain dissolved. When that does happen, you have a very unstable situation. The extra solute, the solute that "should not" have dissolved, can crystallize out of solution quite easily. If you add just a little bit of crystal, then the excess solute will crystallize out of the solution, lowering the concentration until the normal solubility limit is reached. Sometimes, all you have to do is scratch the inside of the test tube or the container that the solution is in, and it will crystalize on the scratch. You will observe this phenomenon in lab this week.

Second, you can momentarily create a supersaturated solution by combining two solutions into one. For example, BaCl₂ and K₂SO₄ are both considered soluble ionic compounds. You could mix up two beakers, one with unsaturated BaCl₂ in it and one with unsaturated K₂SO₄. If you mix these beakers together, however, you are combining the barium and sulfate ions in solution, effectively making supersaturated BaSO₄ (you can use the solubility rules to observe that this compound is insoluble). In this case, the solution is much more unstable, and solid BaSO₄ will immediately be produced, crystallizing out in a reaction we call a precipitation.

Precipitation Reactions

Let's look at this process in a little more detail, starting at the beginning. First, we mixed up solutions of barium chloride and potassium sulfate, which we can represent with the equations below.

BaCl₂ (s) → Ba²⁺ (aq) + 2 Cl^- (aq)

K₂SO₄ (s) → 2 K⁺ (aq) + SO₄^2- (aq)

Next we mixed the solutions together. That forms a mixture we can visualize as shown below (note that there is no reaction arrow here; we're not showing a process, just visualizing all four of these ions being mixed up together in solution.

Ba²⁺ (aq) + 2 Cl^- (aq) + 2 K⁺ (aq) + SO₄^2- (aq)

The highlighted ions, barium and sulfate, form a compound which is not soluble according to our rules. That means they cannot coexist in solution, and will react, forming an ionic bond and "precipitating" from solution as a solid.

Ba²⁺ (aq) + 2 Cl^- (aq) + 2 K⁺ (aq) + SO₄^2- (aq) → BaSO₄ (s) + 2 Cl^- (aq) + 2 K⁺ (aq)

There are a few things you should observe about this reaction. First, note that nothing actually happens to the chloride or potassium ions (in blue); they are just along for the ride. For this reason, we call them spectator ions, and they are often omitted from the reaction.

Ba²⁺ (aq) + SO₄^2- (aq) → BaSO₄ (s)

Leaving them out, you can see that this reaction is essentially the reverse of a dissolution process. A precipitation is a kind of "un-dissolving."

If we don't want to leave out the potassium and chloride ions, but want to keep things a bit tidier, we can use something of a notational trick. Instead of writing out "Ba²⁺ (aq) + 2 Cl^- (aq)" we will simply write "BaCl₂ (aq)" and we will do the same with K₂SO₄ and with KCl in the products. This is a technique you will need to get used to - when you see an ionic compound with an "(aq)" next to it, it really means the separated ions, just written together for compactness. See how it makes the equation below more condensed while showing all the ions involved.

BaCl₂ (aq) + K₂SO₄ (aq) → BaSO₄ (s) + 2 KCl (aq)

Predicting Precipitations

Any time you mix aqueous solutions of two different ionic compounds, there is a possibility that a precipitation reaction might occur. By the end of this section, you should be able to look at any pair of ionic solutions and 1) predict whether or not a reaction will happen and 2) write a balanced equation for a reaction that does occur.

Let's consider solutions of ammonium chloride (NH₄Cl) and lithium carbonate (Li₂CO₃). We want to know: will mixing these solutions create a precipitation reaction? To answer, we go back to the solubility rules.

What you want to do is swap which cation is paired with which anion and see if an insoluble pair is produced. Our initial pairings are NH₄⁺ with Cl^- and Li⁺ with CO₃^2-, so our new pairings are NH₄⁺ with CO₃^2- and Li⁺ with Cl^-. Looking at the solubility rules, these new pairings are all still soluble, so no reaction would take place.

What if our compounds were KOH and MgBr₂? Switching the ions around, we pair K⁺ with Br^- and Mg²⁺ with OH^-. That second pairing is insoluble according to the rules, so this mixture will produce a precipitation reaction.

Next, we will work to write its balanced equation. Let's start by writing the four ions we are mixing together. Note: I am not concerning myself with the numbers of each ion right now. That part will come later when we balance the equation. For now, focus on the formulas.

Mg²⁺ (aq) + Br^- (aq) + K⁺ (aq) + OH^- (aq)

The insoluble compound formed by magnesium and hydroxide is Mg(OH)₂. So that will be our solid product, while the bromide and potassium ions will remain unchanged.

Mg²⁺ (aq) + Br^- (aq) + K⁺ (aq) + OH^- (aq) → Mg(OH)₂ (s) + Br^- (aq) + K⁺ (aq)

Now we will balance the ions in something of a chain reaction. The formula Mg(OH)₂ has two hydroxide ions, so we need two of those on the left. But hydroxide in this reaction only comes in the form of KOH, so if there are two hydroxide ions there must be two potassium ions. Furthermore, since magnesium bromide is MgBr₂, there must be two bromide ions on the left, and also two bromide ions on the right. This gives us the balanced equation below.

Mg²⁺ (aq) + 2 Br^- (aq) + 2 K⁺ (aq) + 2 OH^- (aq) → Mg(OH)₂ (s) + 2 Br^- (aq) + 2 K⁺ (aq)

This way of writing the reaction is a full ionic equation. It shows all four ions involved, even though only two react to form a solid. If we leave out the spectator ions (bromide and potassium in this case) we get the net ionic equation, which only shows the ions forming a bond.

Mg²⁺ (aq) + 2 OH^- (aq) → Mg(OH)₂ (s)

And like before, we can write what is called the molecular equation for the reaction. It has the advantage of being shorter than the full ionic, while being more complete than the net ionic. The downside is that it isn't very physically realistic; it implies that the aqueous ionic compounds are bonded together in a way that isn't really true. Nonetheless, it is the most common way that we right equations for precipitation reactions.

MgBr₂ (aq) + 2 KOH (aq) → Mg(OH)₂ (s) + 2 KBr (aq)

When writing equations for precipitation reactions, there are a couple key points to remember.

• Make sure you identify what type of equation is being asked for: full ionic, net ionic, or molecular. If the type is specified, assume molecular is being asked for.

• Make sure and represent the insoluble product as a solid, and everything else as an aqueous solution.

• Last and most importantly: the formulas you write have to come from balancing the charges of the cations and anions, not just by swapping ions. Let's illustrate with one last example, the reaction of Fe(NO₃)₃ with Na₂S.

Once you get some familiarity with precipitation reactions, it will become clear to you that at a certain level you are just swapping ions between compounds. We start with iron (III) nitrate and sodium sulfide, and end up with iron (III) sulfide and sodium nitrate. Consequently, you may be tempted to write something like the equation below.

Fe(NO₃)₃ (aq) + Na₂S (aq) → FeS (s) + Na₂(NO₃)₃ (aq)

This is not a correct equation. While it correctly predicts that iron and sulfide will form a precipitate here, the formulas of the products are wrong. Recall from CH 104 that ionic formulas come from the charges on the ions. In this case we have Fe³⁺ and S^2-; the correct formula for the compound they make is Fe₂S₃, not FeS. Similarly, the correct formula for sodium nitrate is NaNO₃, not Na₂(NO₃)₃. Putting in the correct formulas for the products - and balancing, of course - we get the correct equation for the reaction.

2 Fe(NO₃)₃ (aq) + 3 Na₂S (aq) → Fe₂S₃ (s) + 6 NaNO₃ (aq)

Your lab workbook has several practice problems you can complete dealing with precipitation reactions.