Fractional Precipitation Pogil Answer Key

Mastering Selective Separation: The Complete Guide to Fractional Precipitation (POGIL Answer Key & Concepts)

Mistake #2: Forgetting to account for side reactions.

POGIL Insight: In reality, adding (Cl^-) to (Pb^2+) forms soluble complexes like (PbCl_3^-) at high ([Cl^-]). This is why in qualitative analysis, we add cold HCl specifically for (Pb^2+)—heat and excess (Cl^-) redissolve (PbCl_2).

Question 1: Which precipitates first?

Answer: (AgCl) (Silver chloride) precipitates first.

Reasoning:

Calculate the ([Cl^-]) needed to start precipitating (AgCl): [ K_sp(AgCl) = [Ag^+][Cl^-] = 1.8 \times 10^-10 ] [ [Cl^-] = \frac1.8 \times 10^-100.01 = 1.8 \times 10^-8 \text M ]
Calculate the ([Cl^-]) needed to start precipitating (PbCl_2): [ K_sp(PbCl_2) = [Pb^2+][Cl^-]^2 = 1.7 \times 10^-5 ] [ [Cl^-] = \sqrt\frac1.7 \times 10^-50.01 = \sqrt1.7 \times 10^-3 \approx 0.041 \text M ]

The Key: Since (1.8 \times 10^-8 \text M) is much less than (0.041 \text M), (AgCl) reaches its (K_sp) first and precipitates.

Part 6: How to Use This Answer Key Responsibly (Teacher & Student Guide)

Part 1: The Core Concepts Behind the POGIL Activity

Before diving into specific answer keys, let's review the three pillars of fractional precipitation. fractional precipitation pogil answer key

Model 1: Comparing (K_sp) Values

Question: Two salts have (K_sp) values of (A = 4.0 \times 10^-5) and (B = 2.0 \times 10^-15). You add a common anion dropwise. Which precipitates first? Answer: Salt B, because it has the smaller (K_sp). Exception: The salts must have the same stoichiometry (e.g., both (MX) or both (MX_2)). If not, you must calculate the required ([Anion]).

Example Problem (Typical for POGIL)

Problem: A solution contains (0.10) M (Ag^+) and (0.10) M (Pb^2+). A solution of (Cl^-) is slowly added.
(K_sp(AgCl) = 1.8 \times 10^-10), (K_sp(PbCl_2) = 1.7 \times 10^-5). Calculate the ([Cl^-]) needed to start precipitating (AgCl):

Step 1 – Which precipitates first?
[ [Cl^-] \text to ppt Ag^+ = \fracK_sp(AgCl)[Ag^+] = \frac1.8\times 10^-100.10 = 1.8\times 10^-9 \text M ]
[ [Cl^-] \text to ppt Pb^2+ = \sqrt\fracK_sp(PbCl_2)[Pb^2+] = \sqrt\frac1.7\times 10^-50.10 = \sqrt1.7\times 10^-4 \approx 0.013 \text M ]
Since (1.8\times 10^-9 \text M < 0.013 \text M), AgCl precipitates first.

Step 2 – Can they be separated?
Find ([Cl^-]) when ([Ag^+] = 1.0\times 10^-5) M (complete precipitation):
[ [Cl^-] = \fracK_sp(AgCl)[Ag^+]\textfinal = \frac1.8\times 10^-101.0\times 10^-5 = 1.8\times 10^-5 \text M ]
At this ([Cl^-]), check if (PbCl_2) has started:
(Q = [Pb^2+][Cl^-]^2 = (0.10)(1.8\times 10^-5)^2 = 3.24\times 10^-11)
Compare to (Ksp(PbCl_2) = 1.7\times 10^-5).
(Q \ll K_sp), so (Pb^2+) is still in solution. Separation is possible. The Key: Since (1