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Course: AP®︎/College Chemistry > Unit 7
Lesson 8: Solubility equilibriaWorked example: Predicting whether a precipitate forms by comparing Q and Kₛₚ
We can use the reaction quotient to predict whether a precipitate will form when two solutions containing dissolved ionic compounds are mixed. If Q < Kₛₚ, the newly mixed solution is undersaturated and no precipitate will form. If Q > Kₛₚ, the solution is oversaturated and a precipitate will form until Q = Kₛₚ. Created by Jay.
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How do we know which one will be a solid and which one will be aqueous (at1:33)(3 votes)- You can determine the solubility of an ionic pair using a solubility table. You can find one of these with a simple google search. They tell you how soluble an ionic compound is in water. The pairs range from soluble to insoluble. Soluble meaning they would appear as separate aqueous ions who dissolve in water and insoluble meaning they do not dissolve and would appear as solid precipitates.
Hope that helps.(4 votes)
Can we tell whether or not there will be a precipitate when Qsp=Ksp?(3 votes)- Remember what equilibrium means. It's when the forward for reverse reactions rates are equal. In this instance this means that the rate of lead sulfate dissociating is equal to the rate of solid lead sulfate reforming. So at equilibrium we will have solid lead sulfate (a precipitate).
Hope that helps.(1 vote)
Why we don't have to determine the limiting reactant (Pb(NO3)2 in this case) first, instead we solely use the concentration from the reactants to calculate Q?(2 votes)- We're not doing a stoichiometry problem here so we don't need to have a limiting reactant. We're calculating the reaction quotient and comparing it to the equilibrium constant. The reaction quotient just requires us to know the concentrations of the chemicals we care about. We essentially measure then if those concentrations are higher, lower, or are at equilibrium concentration values.
Hope that helps.(2 votes)
- How did you determine the Ksp for lead(II) sulfate? When I do a quick google search, the Ksp values I find are much different than 6.3x10^-7. The Ksp value for lead(II) sulfate that that I find in multiple tables is 2.53 x 10^-8...(2 votes)
- Well we determine equilibrium constants experimentally by letting a reaction reach equilibrium and simply measuring the concentrations of the reactants and products. I say simply, but in real life this is easier said than done, particularly with insoluble chemicals. One of the issues is that since insoluble chemicals dissolve barley at all the concentrations in solution are awfully minute and hard to measure.
Doing a quick search of chemical databases I've found a range of values at 25°C. The disagreement is more than likely due to the difficulty of measuring such small concentrations.
Wikipedia: 2.13 x 10^(-8)
A National Institute of Standards and Technology (NIST) database from 1980: 2.53 x 10^(-8)
The University of Rhode Island: 1.6 x 10^(-8)
The University of Massachusetts Amherst: 1.8 x 10^(-8)
Not sure where Jay got 6.3 x 10^(-7), but it's got to be an obscure source because I haven't found a value anywhere near that.
Hope that helps.(1 vote)
- Having started from the very beginning of this AP Chem course, I have not moved forward until achieving 100% on every quiz/test ... Yet the material in this particular example blew WAY over my head ... with so many steps that either we never covered or I have totally lost.
Should I continue forward and not worry about it, or are there other materials I need to go back and focus upon before moving ahead and getting further lost?(1 vote)- So KA is a good place to start learning. The material comes in small bites and is easily accessible. However it's also too light in most places. It's doesn't cover all the material that you would find in a class, it doesn't go into depth with a lot of the material it does cover, and it doesn't have a wide variety of problems and problem solving methods.
So essentially KA is good introductory material, but it shouldn't be thought of as the entirety of what you need to learn about on a certain topic. As annoying as it is, it's good to search out other materials outside of KA like a dedicated text book to get a more well-rounded education. Just my opinion from experience.
Hope that helps.(2 votes)
At the end of the video, he says that the precipitate will continue to form. How is that possible if the solution is supersaturated?(1 vote)
At3:07Is the 6.3 x 10^-7 at 25 degrees celsius given along with the problem or is it derived from values in the question?(1 vote)- The equlibrium constant usually given in a problem like this. The only exception is if the equlibrium concentrations were given and you were asked to calculate the equilibrium constant.
Hope that helps.(1 vote)
Not sure I understand why, at6:33, x = 8.0x10^-3 gives "how many moles of lead two nitrate there are and that's also how many moles of lead two plus ions there are". Am I just misunderstanding the statement or is Jay saying the total concentration of the molecule also = the concentration of the anion? If so, how does that work? I am struggling with learning about solubility so please give any tips if you've got them! :)(1 vote)- Lead(II) nitrate is a soluble salt in water and dissociates into lead(II) (Pb^(2+)) cations and nitrate (NO3^(1-)) anions. It follows the equation: Pb(NO3)2 → Pb^(2+) + 2NO^(1-). So we can see from the balanced chemical equation that every mole of lead(II) nitrate produces one mole of lead(II) cations in a 1:1 ratio. So the moles of lead(II) nitrate will equal the moles of lead(II).
Hope that helps.(1 vote)
- How can we find the mass of the precipitate?(1 vote)
1:33Video transcript
- [Instructor] For this problem,
our goal is to figure out whether or not a precipitate will form if we mix 0.20 liters over
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 they're 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. So one possible product
would be lead sulfate. So let's write on here,
PbSO4, after we cross over our charges and we take the
other cation and the other anion and so the other product
would be sodium nitrate. So we write it in here and a 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 are 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 we really need to do
is identify the precipitate and then write out the
disillusion 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 sulfates anions in aqueous solutions, 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 and
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 seven
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 differences 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
led two sulfate in it. So lead two sulfate as 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. So 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 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 solve for
X and 6.4 times 10 to 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.