Worked example: Predicting whether a precipitate forms by comparing Q and Kₛₚ | Khan Academy

• [Instructor] For this problem, our goal is to figure out whether or not a precipitate will form if we mix 0.20 liters of a 4.0 times 10 to the negative third Molar solution of lead two nitrate with 0.80 liters of an 8.0 times 10 to the negative third Molar solution of sodium sulfate.

The first step is to figure out the identity of the precipitate that might form. So, we're mixing an aqueous solution of lead two nitrate with an aqueous solution of sodium sulfate. In the lead two nitrate solution, there are lead two plus cations and nitrate anions. In the sodium sulfate aqueous solution, there are sodium cations and sulfate anions.

So we take the cation from one and the anion from the other. One possible product would be lead sulfate. So let's write on here, PbSO4, after we cross over our charges. We take the other cation and the other anion, and so the other product would be sodium nitrate. So we write it in here, NaNO3.

To balance the equation, we need a two in front of NaNO3. Since nitrates are soluble, sodium nitrate is an aqueous solution, and lead sulfate would be our possible precipitate. Now that we know our possible precipitate, let's go ahead and write a net ionic equation showing the formation of that precipitate.

So, lead two plus ions would come together with sulfate anions to form PbSO4. So PbSO4 is the possible precipitate. Since lead sulfate is our possible precipitate, we really only care about the concentration of lead two plus ions and sulfate anions in solution. We don't need to worry about sodium cations or nitrate anions because those are the spectator ions in our overall reaction.

Running the overall equation and the net ionic equation are really optional for a problem like this. But what we really need to do is identify the precipitate and then write out the dissolution equation. So PbSO4 would be our possible precipitate. And if it dissolves in water, we would form lead two plus cations in aqueous solution and sulfate anions in aqueous solution, right? aq over here.

The reason why it's important to write out the dissolution equation is because we can write a Ksp expression from it. So Ksp is equal to, it would be the concentration of lead two plus raised to the first power because we have a coefficient of one in the balanced equation, times the concentration of sulfate also raised to the first power.

Pure solids are left out of equilibrium constant expressions. Therefore, we’re not going to include lead sulfate. For lead two sulfate, Ksp is equal to 6.3 times 10 to the negative seventh at 25 degrees Celsius. The concentrations of lead two plus and sulfate in the Ksp expression are equilibrium concentrations.

For our problem, we're gonna calculate Qsp, which has the same form as Ksp; the difference is the concentrations can be at any moment in time. And we're gonna calculate Qsp at the moment our two solutions are mixed, and then we're going to compare Qsp to Ksp.

I've drawn out some diagrams to help us understand how Qsp compares to Ksp and what that means for the solution. However, these aren't perfect diagrams; they're just to help get the point across. If Qsp is less than Ksp, the solution is unsaturated, which means no precipitate would form.

For an unsaturated solution, you can dissolve more lead two sulfate in it. So lead two sulfate is a white solid; so, if we were to put a small amount of lead two sulfate in our unsaturated solution, it would dissolve, and it would continue to dissolve until Qsp is equal to Ksp.

And the system is at equilibrium. At equilibrium, the solid is turning into the ions at the same rate the ions are turning back into the solid. Since the rate of dissolution is equal to the rate of precipitation when the system is at equilibrium, the concentrations of lead two plus ions and sulfate ions are constant, and this represents a saturated solution.

And since the solution is saturated at equilibrium, if we tried to add some more lead two sulfate at the same temperature, we wouldn't be able to dissolve any more; we would just increase the pile of lead two sulfate on the bottom of the beaker. That concept helps us understand what happens when Qsp is greater than Ksp.

When Qsp is greater than Ksp, the solution is oversaturated. So it's exceeded the limit of what can dissolve, and therefore you can imagine some lead two plus ions combining with some sulfate ions to form a precipitate. Therefore, when Qsp is greater than Ksp, a precipitate will form. The precipitate will continue to form until Qsp is equal to Ksp, and the system reaches equilibrium.

Next, we need to go back to what we were given in our initial problem when we mixed our two solutions together. Remember, we only cared about the concentration of lead two plus ions and sulfate ions. So we're gonna calculate the concentration of those two ions at the moment in time when the two solutions are mixed.

First, let's calculate the concentration of lead two plus ions. The original solution of lead two nitrate had a concentration of 4.0 times 10 to the negative third Molar. So molarity is equal to moles over liters. We can plug in the concentration, and we can also plug in the volume of that solution, which was 0.20 liters, and solve for X.

X is equal to 8.0 times 10 to the negative fourth moles. That's how many moles of lead two nitrate there are, and that's also how many moles of lead two plus ions there are. Therefore, to find the concentration of lead two plus ions after the solutions are mixed, we plug in 8.0 times 10 to the negative fourth moles, and for the volume, we're adding these two solutions together, so the total volume of the solution is 0.20 plus 0.80.

So the concentration of lead two plus ions will be equal to 8.0 times 10 to the negative fourth Molar. We can do the same type of calculation to find the concentration of sulfate ions after the two solutions have been mixed. So we take the concentration of the original solution of sodium sulfate and plug that into the molarity equation, plug in the volume, and solve for X.

6.4 times 10 to the negative third moles is how many moles of sodium sulfate there are. That's also how many moles of sulfate ions there are, so we plug in that number into the concentration for sulfate, and once again, since we're adding the two solutions together, we divide that by the total volume to get a concentration of sulfate ions of 6.4 times 10 to the negative third Molar.

Now that we know the concentrations of lead two plus ions and sulfate ions after the two solutions have been mixed, we can plug those concentrations into our Qsp expression and solve for Qsp. So at this moment in time, Qsp is equal to 5.1 times 10 to the negative six. At 25 degrees Celsius, the Ksp value for lead two sulfate is equal to 6.3 times 10 to the negative seventh.

Qsp at this moment in time is 5.1 times 10 to the negative six. Therefore, Qsp is greater than Ksp. Since Qsp is greater than Ksp, we've exceeded the limit of what can dissolve, and therefore the solution is oversaturated. So yes, a precipitate will form, and the precipitate of lead two sulfate will continue to form until Qsp is equal to Ksp.