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 $Pb(N{O}_3{)}_2$ and 0.070 M $NaCl$. Will a precipitate form? The $K_{sp}ofPbC{l}_2$ = $1.7\cdot {10}^{-5}$.

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

• Since we are given the $K_{sp}$ of $PbC{l}_2$ then the relevant reaction must be the dissolving of $PbC{l}_2$ 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 $N{O}_3^{-1}$ ions, on the $P{b}^{+2}$ ions, which have a concentration of 0.0035 M.
• All sodium compounds are soluble. So, like the nitrates, we can ignore the $N{a}^{+1}$ ions and the $[C{l}^{-1}]$ = 0.070 M.

Here is our relevant reaction and the current ICE table:

$PbC{l}_{2(s)}\rightleftarrows P{b}_{(aq)}^{+2}+2C{l}_{(aq)}^{-1}$

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 $K_{sp}<Q_{sp}$?

And that is simple math.

$K_{sp}$ > or = or \lt $Q_{sp}$

$1.7\cdot {10}^{-5}$ ? $[P{b}^{+2}]\cdot [C{l}^{-1}{]}^2$

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

$1.7\cdot {10}^{-5}<1.72\cdot {10}^{-5}$

Since $K<Q$, 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 $Pb(N{O}_3{)}_2$ solution and 200.0 mL of a 0.070 M $NaCl$ solution. Will a precipitate form? The $K_{sp}ofPbC{l}_{2}is1.7\cdot {10}^{-5}$.

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:

$100mLPb(N{O}_3{)}_2\cdot \frac{1L}{1000mL}\cdot \frac{0.0035moles}{1Lsolution}=3.5\cdot {10}^{-4}molsPbC{l}_2$

Then, calculating the new molarity of the $Pb(N{O}_3{)}_2$ in the mixed solution with its volume of 300 mL (0.300 L) requires this math:

$M=\frac{moles}{L}=\frac{3.5\cdot {10}^{-4}molsPbC{l}_2}{0.300L}=0.00117M$

There is a simpler and faster way to get the same value. We need to notice that the $Pb(N{O}_3{)}_2$ started as 100 mL and ended up in 300 mL. We can, therefore, do the following math:

$0.0035\frac{mols}{L}\cdot \frac{100mL}{300mL}=0.00117M$

or even more simply:

$0.0035\frac{mols}{L}\cdot \frac{1}{3}=0.00117M$

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

$0.070MNaCl\cdot \frac{2}{3}=0.0467MNaCl$

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

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

$K_{sp}>or=or<Q_{sp}$

$1.7\cdot {10}^{-5}$ ? $[P{b}^{+2}]\cdot [C{l}^{-1}{]}^2$ 

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

$1.7\cdot {10}^{-5}>2.55\cdot {10}^{-6}$

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