Precipitation reactions
Definitions & Key takeaways
Precipitation reactions occur when two aqueous solutions are combined, causing a chemical reaction that produces a solid product. The ions present in the solutions combine to form an insoluble solid, which is referred to as the precipitate.
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 to nitrate which has the chemical formula P BN 32.
The other beaker contains an aqueous solution of potassium iodide which has the chemical formula K I. 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 P BN 032.
The subscript on lead is implied to be one. While the subscript on nitrate is two, we cross over these subscripts and say that our solution contains lead two plus cas and nitrate minus anions.
We can write these out as P B2 plus and N three minus. We can now do the same thing for potassium iodide.
Since that has the formula K I, we know that our solution has potassium one plus C A ions and iodide one minus anions. We can write these out as well K plus and I minus.
Now that we know the four ions in our solution, we can figure out the products of the reaction to do this. We can 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 kn three 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 P BI two or lead to iodide.
We can now write out our unbalanced chemical reaction, an aqueous solution of lead to nitrate plus an aqueous solution of potassium iodide yields potassium nitrate and lead to iodide.
We use 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.
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 ion silver one plus and lead 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.
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 potassium nitrate, 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 potassium iodide, 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 lead, two plus ions plus two nitrate ions plus two potassium ions plus two iodide anions yields two potassium ions, two nitrate ions and solid lead, two iodide.
Next, let's look at what's actually happening when the lead two plus ions combined with the iodide, an ions to form the solid precipitate of lead to iodide.
The other two ions, potassium and nitrate 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 potassium and nitrate.
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.
This leaves us with our net ionic equation which tells us the exact combination of ions that actually happened during our reaction.
Notice how the net ionic equation is already balanced. Now, let's look at this reaction slightly differently using particulate diagrams, we know from balancing charges and the balanced chemical equation that for every one lead two plus ion, we need to have two nitrate ions.
So if we look at our starting solution in the particulate diagram, we have two lead, two plus ions and four nitrate, one minus ions.
This makes sense because the ratio has to be 1 to 2. Looking at the particulate diagram for the starting solution of potassium iodide ions, we notice that we have equal numbers of potassium ions and iodide anions.
This makes sense because the ratio has to be 1 to 1. We can then look at the particulate diagram.
After we mix our two beakers together to form a precipitate in the precipitate that is forming, we see that there are three lead two plus C A ions in the diagram as well as six iodide, one minus an ions.
This ratio makes sense because of the 1 to 2 mole ratio between lead two plus and iodide. When our reaction is complete, we have lead to iodide in a three dimensional crystal structure.
If we look at the remaining ions in our beaker, we see that we have four left over potassium one plus ions as well as four nitrate one minus anions.
These remaining ions are the spectator ions for our reaction. Let's consider one more example of a precipitation reaction.
Suppose we mix a solution of barium chloride with a solution of potassium sulfate. Can we predict whether a precipitate will form?
Our first step is always to write down the chemical formulas for our reactants and products. Barium chloride has a chemical formula of BCL two and potassium sulfate has a chemical formula of K 24.
Since both of these are aqueous solutions, we can write them out in terms of their constituent ions by crossing over the subscripts to find charges, we find that barium chloride becomes barium two plus cat ions and chloride one minus an ions.
In solution or ba two plus and cl minus. Likewise, crossing over charges tells us that potassium sulfate becomes potassium plus cations and sulfate two minus anions in solution or K plus and so +42 minus.
Now that we know what ions are in our solution, we can figure out the products of the reaction. We start with our barium two plus cations.
Each barium two plus ion will combine with a single sulfate two minus anion resulting in barium sulfate which has the formula baso four.
Next, we look at our potassium one plus cations because the chloride anions have a charge of minus one. We can combine one potassium ion with one chloride ion to produce potassium chloride which has the chemical formula KCL.
We can now write down our unbalanced chemical reaction. An aqueous solution of potassium sulfate plus an aqueous solution of barium chloride yields barium sulfate and potassium chloride.
Now it's time to figure out whether either product barium sulfate or potassium chloride produces a precipitate in solution.
We once again need to use our solubility guidelines table. First, we can consider potassium chloride.
This is a salt containing an alkali metal ion potassium here. And so we know by rule two that this compound is soluble.
This makes the product potassium chloride aqueous and not a precipitate of the reaction. Next, we consider barium sulfate on the solubility guidelines table.
We can look at rule four rule four states that most sulfate salts are soluble, except specifically for the exception of barium sulfate.
So we know that barium sulfate is insoluble making it the precipitate for a reaction. In the chemical formula, we can add the subscript S to indicate that barium sulfate is a precipitate.
Next, we can balance our equation. There are two potassiums and two chlorides on the left side.
And so we need the same number to be on the right side. So we put a two as the coefficient of potassium chloride on the right side.
If we count up the number of barium and sulfate ions, we can see that there is one of each on the left and one of each on the right.
And so our equation is balanced. Now, we want to identify the spectator ions to do this.
We need to break up our chemical reaction equation into an overall ionic equation by writing each aqueous reactant in terms of its constituent ions.
When we write this out, we make sure to keep the coefficients intact. One barium cation plus two chloride, one minus anions plus two potassium ions plus one sulfate, two minus anions yields solid barium sulfate, two potassium ions and two chloride, one minus anions.
We can see that potassium and chloride ions appear on both sides of this equation in equal numbers and we can cancel them out.
Therefore, they are our spectator ions. The net ionic equation then becomes sulfate two minus anions plus barium two plus C a ions yields solid barium sulfate.
Notice that once again, our equation is already balanced. All right, as a quick recap, precipitation reactions occur when two aqueous solutions are combined, causing a chemical reaction that produces a solid product when determining how two aqueous solutions will react.
The first step is to use the chemical formulas of the reactants to determine the different ions in solution and their charges.
The easiest way to do this is by crossing over the subscripts of the chemical formulas to find charges. Next, combine the different ions in solution to produce two new ionic compounds, using the charges of the ions to determine the chemical formula of each product.
Once the products are known, the solubility guidelines table can be used to determine which if any of the products is a solid precipitate.
After finding the precipitate, the original chemical equation can be simplified by removing spectator ions which are ions that stay dissolved and don't participate in the precipitation reaction.
The resulting simplified reaction is the net ionic equation.
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