Example Problem


Problem: (from an actual death test) A solution is initially [Cl¯] = 7.1E-3 and [CrO4-2] = 1.0E-4. Silver ions (Ag+) are slowly added (without altering the volume). What percent of the chloride will remain in solution when Ag2CrO4 first appears? What mass of silver chloride will have been formed? Ksp of AgCl = 1.8 E-10, Ksp of Ag2CrO4 = 9.0 E-12


Solution: Here's the 411 on this problem. It's way easier than it actually appears. Tip number one when confronting a problem like this is not freaking out. When you look at problem like this, it helps a lot to run through a scenario of what actually happens as you add the silver ions. Basically, Ksp is a measure of how sticky a combination of ions are. The stickiers the ions, the more likely collisions between them will stick, the less total ions a solution can support before a precipitate (a solid) is formed. As you start adding silver ions, [Ag+] increases until you reach the threshold point, the Ksp, of the total number of ions the solution can handle before a precipitate is formed from the stickier combination of ions ( in other words, the less soluble one, or the one with the smaller Ksp).


Before we go on, let's write out some reactions.


  1. AgCl --> Ag+ + Cl¯
    Ksp = 1.8 E-10
  2. Ag2CrO4 --> CrO4-2 + 2Ag+
    Ksp = 9.0 E-12


Now let's figure out which anion (negative ion) in solution precipitates first with the silver. This is the one that requires a lower concentration of silver to start precipitation.


Ksp (AgCl) = 1.8 E-10 = [Ag+][Cl¯]
Since we know [Cl¯]:
Ksp (AgCl) = [Ag+][7.1 E-3]
[Ag+] = 1.8 E-10 / 7.1 E-3 = 2.5 E-8 M


Ksp(Ag2CrO4) = 9 E-12 = [Ag+]2[CrO4-2]
Since we know [CrO4-2]:
Ksp(Ag2CrO4) = [Ag+]2[1.0 E-4]
[Ag+]2 = 9 E-12 / 1.0 E-4 = 9 E-8
[Ag+] = 3 E-4 M


Okay, Cl¯ requires a lower concentration of Ag+ to precipitate, so it goes first.


To resume the scenario, when the less soluble pair starts to precipitate as AgCl, the Cl¯ starts dropping to the bottom of the beaker, and the conc. of Cl¯ starts decreases. This decrease in Cl¯ allows the conc. of Ag+ to start increasing again according to the Ksp equation for AgCl. The Ag+ conc. increases until we get a conc. of 3 E-4 M (see math). The question is simply what is the conc. of Cl¯ at this point? All ya gots ta do is plug and chug with the Ksp equation for AgCl:


Ksp(AgCl) = 1.8 E-10 = [Ag+][Cl¯] (use 3 E-4M for Ag+)
Ksp(AgCl) = [3 E-4][Cl¯]
[Cl¯] = 1.8 E-10 / 3 E-4 = 6 E-7 M


Now that we got our value for how much Cl¯ is in solution when the more soluble compound, the chromate compound, first appears, only a couple of simple calculations remain:


% Cl¯ remaining in solution = (6 E-7 / 7.1 E-3) * 100 = 0.0%
(One thing you gotta learn about Mr. McAfoos is that he loves using numbers that seem unbelievable in order to force his students to be confident in their math. What a guy! For example, look out for negative pH's, reactions that don't happen at all, and wacked out answers in problems that require a lot of unit conversions)


Gmm AgCl = 107.9 + 35.5 = 143.4 g/mol
Mass of silver chloride formed = (7.1E-3) M * 1L * 143.4 g/mol = 1.0 g


(Mr. McAfoos actually slipped a little in this problem by not specifying a volume...but he's entitled to be imperfect . . . right? Just don't let him hear about it)


Congratulations! Now move it on over to Type 3 problems.


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