Precipitation reactions

Precipitation is when a chemical reaction occurs between two solutions, and the reaction produces a product that is a solid.

Let’s look at an example of a precipitation reaction.

Let’s say we have a beaker with a solution of lead nitrate, and we have another beaker containing a solution of potassium iodide.

If we pour the potassium iodide solution into the beaker containing the lead nitrate, we observe a cloud of yellow solid appearing in the beaker.

That yellow solid is our product, which is the precipitate of the reaction.

We would say that the solid “precipitates out” when it falls to the bottom of the beaker.

Let's write out the equation for this precipitation reaction.

One beaker contains an aqueous solution of Lead (II) nitrate, which has the chemical formula Pb(NO3)2.

The other beaker contains an aqueous solution of potassium iodide, which has the chemical formula KI.

To predict the products of this reaction, we need to know what ions were in solution.

We can find the ions using the technique of “crossing over” charges.

Taking lead nitrate, we look at the subscripts in its chemical formula, Pb(NO3)2.

The subscript on lead is implied to be one, while the subscript on nitrate is 2.

We “cross over” these subscripts, and say that our solution contains lead 2+ cations, and nitrate “minus” anions.

We can write these out as

Pb2+

NO3-

We can now do the same thing for potassium iodide.

Since that has the formula KI, we know that our solution has potassium “one plus” cations and iodide “one minus” anions.

We can write these out as well

K+ I-

Now that we know the four ions in our solution, we can figure out the products of the reaction.

To do this, we take the cation from one reactant, and combine it with the anion from the other reactant.

Let’s start with our potassium cations.

These have a charge of plus one, and so we can combine them with the anion from the other reactant which is a nitrate ion.

We cross over the absolute value of the charges to get KNO3, potassium nitrate, as our first product.

Now, we look at lead two plus.

Lead two plus will combine with the remaining anion which is the iodide ion.

We cross over the absolute value of the charges to get our other product which is PbI2, or lead II iodide.

We can now write out our unbalanced chemical reaction

Pb(NO3)2(aq) + KI(aq) → KNO3 + PbI2

We used the “aqueous” subscript to indicate that we know both reactants dissolve into their component ions in solution.

We know that both of our reactants are aqueous, but now we want to find out which of our products are aqueous.

We can guess that at least one of the products is not aqueous, since we observed a solid yellow precipitate.

But we don’t know whether the precipitate is potassium nitrate or lead iodide.

Now we need to figure out which one of these formed our solid yellow precipitate.

To do this, we need some more information: a chart called a “solubility guidelines table.”

The first rule is that nitrate salts are soluble.

In our case, one of our products was potassium nitrate, which has a nitrate ion and which is thus a nitrate salt.

So we can guess that the potassium nitrate is soluble, and we can add an “aqueous” subscript to KNO3.

The second rule is that ionic salts containing alkali metal ions, such as lithium “one plus”, sodium “one plus”, or potassium “one plus”, are also soluble.

In our case, potassium nitrate also contains potassium ion, further confirming that it is soluble in water.

The third rule is that chloride, bromide, and iodide salts are soluble---although there are some exceptions.

Salts of the ions Ag “one plus” and Pb “two plus” are two exceptions.

In our case, we have an iodide salt, but it contains lead, which happens to be one of the exceptions to rule three: lead iodide is therefore not soluble.

We can now guess that lead iodide is our precipitate, and we can indicate this by putting an “s” for solid as a subscript in the chemical equation.

Pb(NO3)2(aq) + KI(aq) → KNO3(aq) + PbI2(s)

Next, we need to balance our chemical reaction. We can see that we have two nitrate ions on the left side, and so we need the same number on the right side.

So we put a two in front of KNO3.

Pb(NO3)2(aq) + KI(aq) → 2KNO3(aq) + PbI2(s)

But now we have two potassium ions on the right side, and so we need the same number on the left side.

So we put a two in front of KI

Pb(NO3)2(aq) + 2KI(aq) → 2KNO3(aq) + PbI2(s)

We count and find that we have one lead ion, and two iodide ions, on both the right and left sides, and so our reaction is balanced.

Now, we want to take this chemical reaction equation, and turn it into an overall ionic equation.

This process consists of taking each aqueous compound in the equation, and writing it out in terms of its constituent ions.

When we write these out, we need to take care to keep the coefficients of each ion intact

Pb(NO3)2(aq) + 2KI(aq) → 2KNO3(aq) + PbI2(s)

Pb2+ + 2NO3- + 2K+ + 2I- → 2K+ + 2 NO3- + PbI2(s)

Next let's look at what's actually happening when the lead two plus ions combine with iodide anions to form the solid precipitate of lead II iodide.

The other ions K+ and NO3- were there when the reaction started, as ions in aqueous solution.

They are also there in aqueous solution when the reaction is over, and so overall they did not participate in the reaction themselves---they merely watched the reaction.

So we call these “spectator ions” because they are present for the reaction, but they don’t participate.

So our spectator ions for this precipitation reaction are K+ and NO3-

We go to our overall ionic equation, and cancel out the two potassium ions on the left with the two potassium ions on the right.

We can also cancel the two nitrate ions on the right side with the two nitrate ions on the left side.