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.”