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AP®︎/College Chemistry

Course: AP®︎/College Chemistry>Unit 7

Lesson 8: Solubility equilibria

Worked 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 (at )
• 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.
• Can we tell whether or not there will be a precipitate when Qsp=Ksp?
• 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?
• 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.
• 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.
• 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)
• 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...
(1 vote)
• 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)
• At Is 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, at , 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)