Will a Precipitate Form?

We’re going to solve two different problems on this page. The first is relatively straight-forward, the second is the same sort of problem, but has an added issue that we’ll need to address.

A simple Problem 

 Let’s start by solving this problem:

A solution is made that is 0.0035 M \(Pb(NO_3)_2\) and 0.070 M \(NaCl\). Will a precipitate form? The \(K_{sp} ~of~ PbCl_2\) = \(1.7 \cdot 10^{-5}\).

There are several things we need to think about before we can start the problem:

• Since we are given the \(K_{sp}\) of \(PbCl_2\) then the relevant reaction must be the dissolving of \(PbCl_2\) even if that seems to come from out of nowhere.
• All nitrates are soluble (check your solubility table). That means that we don’t need to care about the \(NO_3^{-1}\) ions, on the \(Pb^{+2}\) ions, which have a concentration of 0.0035 M.
• All sodium compounds are soluble. So, like the nitrates, we can ignore the \(Na^{+1}\) ions and the \([Cl^{-1}]\) = 0.070 M.

Here is our relevant reaction and the current ICE table:

\(PbCl_{2~(s)} \rightleftarrows Pb^{+2}_{(aq)} + 2~Cl^{-1}_{(aq)}\)

It is tempting to think that this reaction must run backwards, since no reactant is present, but this is a solubility problem. The reactant is a solid, is not included in the math and therefore is NOT relevant.

What we actually need to do is to figure out what the question actually is. “Will a precipitate form? means “will a solid appear in the container?” For that to happen, the ions in solution need to stick together and form the solid compound. In other words:

“Will a precipitate form?” = “Does the reaction run LEFT?”

That question, of course, is really asking: Is \(K_{sp} \lt Q_{sp}\)?

And that is simple math.

\(K_{sp}\) > or = or \lt \(Q_{sp}\)

\(1.7 \cdot 10^{-5}\) ? \([Pb^{+2}] \cdot [Cl^{-1}]^2\)

\(1.7 \cdot 10^{-5}\) ? \([0.0035] \cdot [0.070]^2\)

\(1.7 \cdot 10^{-5} \lt 1.72 \cdot 10^{-5}\)

Since \(K \lt Q\), the reaction will run to the left and the answer is YES (a precipitate will form).

A More Complex Problem

Now, let's try a more complex problem:

A solution is made by mixing 100 mL of a 0.0035 M \(Pb(NO_3)_2\) solution and 200.0 mL of a 0.070 M \(NaCl\) solution. Will a precipitate form? The \(K_{sp} of PbCl_2 is 1.7 \cdot 10^{-5}\).

In the last problem, we were given the concentrations of the ions in the solution which had already been made. In this problem we are given volumes and concentrations of the two solutions that are being mixed. Since each compound will end up in a larger volume than it had been before, the concentrations of each must be determined before we can do the check for precipitation.

Here is the work for the lead II nitrate. We’ll do a FULL version of the math and then find the quicker and easier way.

The full version:

\(100 ~mL ~Pb(NO_3)_2 \cdot \frac{1 L}{1000 mL} \cdot \frac{0.0035~moles}{1 L~solution}= 3.5 \cdot 10^{-4} mols~PbCl_2\)

Then, calculating the new molarity of the \(Pb(NO_3)_2\) in the mixed solution with its volume of 300 mL (0.300 L) requires this math:

\(M~=\frac{moles}{L}=\frac{3.5 \cdot 10^{-4} mols~PbCl_2}{0.300~L}= 0.00117 M\)

There is a simpler and faster way to get the same value. We need to notice that the \(Pb(NO_3)_2\) started as 100 mL and ended up in 300 mL. We can, therefore, do the following math:

\(0.0035 \frac{mols}{L}\cdot \frac{100~mL}{300 ~mL} = 0.00117~M\)

or even more simply:

\(0.0035 \frac{mols}{L}\cdot \frac{1}{3} = 0.00117~M\)

The NaCl goes from 200 mL to 300 mL, so we can calculate that molarity:

\(0.070~M~NaCl \cdot \frac{2}{3}= 0.0467~M~NaCl\)

Finally, we can address the actual question, which was “Will a precipitate form?” which we know means: is \(K_{sp} \lt Q_{sp}\)?

So, with our newly determined concentrations, we can solve the problem: 

\(K_{sp} \gt ~or~ = ~or~ \lt Q_{sp}\)

\(1.7 \cdot 10^{-5}\) ? \([Pb^{+2}] \cdot [Cl^{-1}]^2\) 

\(1.7 \cdot 10^{-5}\) ? \([0.00117] \cdot [0.0467]^2\) 

\(1.7 \cdot 10^{-5} \gt 2.55 \cdot 10^{-6}\)

So, in this case, the answer is NO (a precipitate will not form).

Finding Solublity in g/100 mL

Let’s solve this problem:

The \(K_{sp}\) of \(PbCl_2\) is \(1.7 \cdot 10^{-5}\), what is the solubility of the compound in \(\frac{g}{100~mL}\)?

This problem exists because we don’t have devices that measure moles directly. Therefore, knowing the molar solubility is not all that helpful. In the end, this problem is about turning molar solubility into a usable value.

We start this problem by finding the molar solubility of the compound. Given that we solved this problem on this page, we’ll jump right to the answer for that portion:

Molar solubility = “x” = 0.016 M.

The first thing we need to do is fix the unit moles, by converting it to grams. That’s a relatively simple problem using molar mass:

\(0.016 \frac{moles~PbCl_2}{L} \cdot \frac{278.10~g~PbCl_2}{1~mole~PbCl_2}=4.45 \frac{g}{L}\)

This is now a useful and workable amount. It tells us that we can dissolve, at most, 4.45 g of \(PbCl_2\) in a liter of solution.

For some reason, that is beyond the reasoning of this author, it was decided that solubilities should be recorded in terms of the grams that will dissolve in 100 mL, rather than in 1 L.

Solving this is easy...IF you think. If you try to set up a conversion process, you will get it wrong. I promise.

Here’s the simple logic. 100 mL is \(\frac{1}{10}\) of a liter. Therefore you can just divide the value we had before by 10.

So, the solubility of \(PbCl_2\) is 0.445 \(\frac{g}{100~mL}\).

Finding Molar Solublity

Let’s solve this problem:

The \(K_{sp}\) of \(PbCl_2\) is \(1.7 \cdot 10^{-5}\), what is the molar solubility of the compound?

This problem is the SAME as finding the concentrations in a saturated solution, with the exception that the answer is ALWAYS “x”. Be sure to read all the way to the bottom of this page to make sure that you understand the implications of that statement!

Our reaction is:

\(PbCl_{2~(s)} \rightleftarrows Pb^{+2}_{(aq)} + 2~Cl^{-1}_{(aq)}\)

We can create an ICE table for this reaction, which would look like this:

We can now do the following math:

\(K_{sp}=[Pb^{+2}] \cdot [Cl^{-1}]^2~~~~~1.7 \cdot 10^{-5} = [x] \cdot [2x]^2\)

Simplified, that is:

\(1.7 \cdot 10^{-5} = 4x^3\)

which gives us x=0.016 M

Molar solubility is “x” therefore the answer is 0.016M.

What's interesting here is that it does NOT matter what the formula for the ionic compound is, the answer is always "x". 

A simple ionic compound like \(AgCl\) would simplify to:

\(K_{sp} = x^2\)

while a compound like \(Pb_3(PO_4)_2\) would work out to be:

\(K_{sp} = 108x^5\)

However, in both cases, the left column of the ICE table would be "some - x" and therefore the molar solubility is just "x".

Finding the Concentrations - Common Ion

Let’s solve this problem:

The \(K_{sp}\) of \(PbCl_2\) is \(1.7 \cdot 10^{-5}\). What is the concentration of chloride ion 

in a saturated solution that already has \([Pb^{+2}]\ = 0.25~M\)?

There are several things that are implied, but not stated, that we need to recognize:

The reaction involves \(PbCl_2\) breaking up into ions (this comes from the definition of \(K_{sp}\)).

The reaction will run forward until equilibrium is reached. (Again, this is the definition of saturated.)

So, our reaction is:

\(PbCl_{2~(s)} \rightleftarrows Pb^{+2}_{(aq)} + 2~Cl^{-1}_{(aq)}\)

We can create an ICE table for this reaction, which would look like this:

We can now do the following math:

\(K_{sp}=[Pb^{+2}] \cdot [Cl^{-1}]^2~~~~~1.7 \cdot 10^{-5} = [0.25 + x] \cdot [2x]^2\)

This will be a “little x” problem, so we can assume that 0.25 + x = 0.25. That simplifies the math to:

\(1.7 \cdot 10^{-5} = (0.25)\cdot 4x^2\)

which gives us x=0.0041 M

Since \([Cl^{-1} = 2x\) we know that \([Cl^{-1}]=0.0082~M\).

This is an example of the common ion effect. Simply put, the presence of lead ions in the solution did not allow as much of the lead II chloride to dissolve as would have in pure water.

Another, perhaps more useful, way to think about the common ion effect is to relate it to Le Chatelier’s Principle. For the reaction:

\(PbCl_{2~(s)} \rightleftarrows Pb^{+2}_{(aq)} + 2~Cl^{-1}_{(aq)}\)

additional amounts of \(Pb^{+2}\) ions will drive the reaction to the left, reducing the amount of chloride ion in solution.

Finding Concentrations in a Saturated Solution

Let’s solve this problem:

The \(K_{sp}\) of \(PbCl_2\) is \(1.7 \cdot 10^{-5}\), what is the concentration of chloride ion in a saturated solution?

There are several things that are implied, but not stated, that we need to recognize:

• The reaction involves \(PbCl_2\) breaking up into ions (this comes from the definition of \(K_{sp}\)).
• The solution is made by adding solid \(PbCl_2\) to pure water. This is implied by the term saturated solution. If the situation was anything OTHER than solid added to water, we would be told.
• The reaction will run forward until equilibrium is reached. (Again, this is the definition of saturated.)

So, our reaction is:

\(PbCl_{2~(s)} \rightleftarrows Pb^{+2}_{(aq)} + 2~Cl^{-1}_{(aq)}\)

We can create an ICE table for this reaction, which would look like this:

Although writing “some” in the midst of an ICE table seems odd, we can get away with it, since the solid is not part of the equilibrium expression.

We can now do the following math:

\(K_{sp}=[Pb^{+2}] \cdot [Cl^{-1}]^2~~~or~~~1.7 \cdot 10^{-5} = [x] \cdot [2x]^2\)

simplified, that is:

\(1.7 \cdot 10^{-5} = 4x^3\)

which gives us x = 0.016 M

Since \([Cl^{-1}] = 2x\) we know that \([Cl^{-1}]=0.032~M\).

Solubility

Understanding solubility requires us to understand how things (especially ionic compounds) dissolve, reversible reactions, and the mathematics of equilibrium - especially “little x” problems.

Let’s look at solubility with respect to one particular ionic compound - lead II chloride (\(PbCl_2\)). All of the ideas that are developed here, will then be applicable to other compounds. We can write the process of lead II chloride dissolving as a simple reaction:

\(PbCl_{2~(s)} \rightleftharpoons PbCl_{2~(aq)}\)

Although that reaction is correct, it isn’t very useful. It is much more interesting to write it as an ionic equation:

\(PbCl_{2~(s)} \rightleftharpoons Pb^{+2}_{(aq)} + 2~Cl^{-1}_{(aq)}\)

Remembering that solids are not included in the equilibrium expression, we can write the following:

\(K=[Pb^{+2}] \cdot [Cl^{-1}]^2\)

Since this equilibrium expression has an “odd” look (nothing on the bottom), we give it a special name and add a subscript to the K:

\(K_{sp}=[Pb^{+2}] \cdot [Cl^{-1}]^2\)

where \(K_{sp}\) stands for the Solubility Product.

There are several important ideas embedded in this:

1. \(K_{sp}\) is always for an ionic solid dissolved in water breaking up into its constituent ions
2. Any problem that mentions \(K_{sp}\) does not need to give the reaction, because it is implied.

There is also some vocabulary that flows from this:

• Saturated solutions are solutions for which \(Q_{sp} = K_{sp}\). In other words, a saturated solution is at equilibrium. Remember that this does NOT mean that the process of dissolving has stopped. It means that the rate at which ions are entering solution is matched by the rate at which ions are coming together and dropping out of solution. Therefore the concentrations of the ions will not be changing, nor will the amount of solid present. In addition, no solid must be visibly present. Saturation is based on the concentrations of dissolved ions only.
• Unsaturated solutions are those that can hold more solute than is currently dissolved. That is \(K_{sp} > Q_{sp}\). If solid is present, then the rate of solvation is faster than the rate of precipitation and the concentrations will be rising. If no solid is present, then the solution will remain as it is. 
• Supersaturated solutions are those where “too much” solute is dissolved. That is, solutions where \(K_{sp} < Q_{sp}\). Generally, we would expect such a situation to result in ions in solution coming together and precipitating, thus decreasing the concentrations of the ions and increasing the amount of solid present in the container. In rare occasions, when the crystalline structure of the solid is complex, it is possible for a solution to temporarily remain in a state of supersaturation until random chance allows the beginning of precipitation or until crystalization is seeded.
• Molar Solublity is the amount of a solid that will dissolve in water in units of \(\frac{moles~solid}{liters}\)

The only way to really understand all of this is, of course, to look at some problems. We’ll look at five different problems: 

1. Finding the concentrations of ions in a saturated solution 
2. Finding the concentrations of ions in a saturated solution in which one ion was already present (this is called the common ion effect) 
3. Finding the molar solubility of an ionic compound 
4. Finding the solubility of an ionic compound in \(\frac{g}{100~mL}\) 
5. Determining whether a precipitate will form when solutions are mixed

Finding the Density of a Solid

Density

Remember that density is a measure of the amount of stuff packed into a space. Different liquids and different solutions will have different densities and determining the density of a solution can help identify the solution.

Mathematically, we use the formula \(d=\frac{mass}{volume}\). This means that to find the density of something, we need to know the mass and volume. There are several ways to do this for a solid.

Finding the Density of a Regular Solid

Finding the density of a regular solid is very straightforward. Mass is measured on a balance and then volume is found by measuring the the solid and doing simple geometric math.

Cubes and Rectangular Prisms

Cylinders

Finding the Density of Irregular Solids

An irregular solid is one whose shape is not a simple geometric figure. This could be anything ranging from a slightly irregular block (where the corners aren’t quite square) to a random unsightly chunk.

Again, mass is easy. We simply put the solid on a balance.

Volume is now the challenge. In such a case, we find the volume using Archimedes Principle. The idea that a solid will displace a volume of water equal to its own volume.

If you aren’t familiar with the Archimedes Principle, the VERY short version of the story is that Archimedes was charged by the king to determine if the man who had made his new crown had cheated him by mixing the gold he had been given with another metal. The only way to identify pure gold at the time was to measure the density. However, the crown, being “crown-shaped” did not have an obvious volume. While pondering how to determine the volume of the crown, Archimedes went to the public baths. When he lowered himself into the tub, a large amount of water splashed over the side and Archimedes realized he could measure the volume of the crown by measuring the water displaced. He was so excited that he jumped up and ran through the town (naked) crying Eureka (I’ve found it).

Using the Archimedes Principle is simple. We find a graduated cylinder that has a wider diameter than the object and fill it part way with water. Carefully measuring the volume. Then we place the object into the cylinder so that it is completely submerged. As a result of this, the water will rise up the cylinder. We can then read the new volume. The volume of the object is simple the difference between the two measured volumes:

\(volume~of~object = volume~with~object - volume~without~object\)

So the density can be found this way:

\(density = \frac{mass}{(volume~with~object - volume~without~object)}\)