Question:
A solution contains 0.050 M Br⁻ and 0.050 M CrO₄²⁻. Solid Pb(NO₃)₂ is added slowly.
(K_sp) PbBr₂ = (6.6 \times 10^-6)
(K_sp) PbCrO₄ = (2.8 \times 10^-13)
Which precipitates first? At what [Pb²⁺] does the second begin to precipitate? What is [Br⁻] at that moment?
Solution (from a top-tier answer key):
For PbCrO₄ (1:1 salt):
[
[Pb^2+] = \frac2.8 \times 10^-130.050 = 5.6 \times 10^-12 M
] fractional precipitation pogil answer key best
For PbBr₂ (1:2 salt):
(K_sp = [Pb^2+][Br^-]^2 \Rightarrow [Pb^2+] = \frac6.6 \times 10^-6(0.050)^2 = \frac6.6 \times 10^-60.0025 = 2.64 \times 10^-3 M)
Order: PbCrO₄ precipitates first (much lower [Pb²⁺]). Question: A solution contains 0
Second precipitate (PbBr₂) begins at [Pb²⁺] = (2.64 \times 10^-3 M). At that [Pb²⁺], [CrO₄²⁻] remaining is: [ [CrO_4^2-] = \frac2.8 \times 10^-132.64 \times 10^-3 = 1.06 \times 10^-10 M ]
The [Br⁻] is still essentially 0.050 M (negligible precipitation of PbBr₂ has occurred yet). At what [Pb²⁺] does the second begin to precipitate
Key takeaway: The 1:2 stoichiometry dramatically changes the required cation concentration.
Understanding fractional precipitation isn't just about passing a test. Here’s where this knowledge applies in the real world:
The same logic applies if you have a solution containing two anions (e.g., CO₃²⁻ and SO₄²⁻) and add a cation like Ba²⁺. The "best" POGIL answer key will have you practice both scenarios. Always: