Fractional Precipitation Pogil Answer Key -

Question: Why is fractional precipitation sometimes impossible? Answer: If the (K_sp) values of the two salts are too close (within a factor of (10^2) or (10^3)), or if the second salt requires a lower anion concentration than the first, then one salt will not be completely removed before the other starts precipitating. This causes coprecipitation (both solids form together).

Answers with Calculations:

Conclusion: Hg₂²⁺ precipitates at a very low [Cl⁻] (1.14×10⁻⁸ M), Ag⁺ next at 1.8×10⁻⁸ M, and Pb²⁺ last at 0.0412 M.

Correction: Always calculate the required precipitant concentration. For (Ag_2S) (very small (K_sp)) vs. (CuS), the sulfide ion needed might be different due to stoichiometry.

In the world of analytical chemistry, separating metal ions from a complex solution often feels like untangling a knot of earphones. If you have a solution containing two different metal ions—say, Silver ((Ag^+)) and Lead ((Pb^2+))—how do you remove just one of them? fractional precipitation pogil answer key

The answer is Fractional Precipitation.

For students working through a POGIL (Process Oriented Guided Inquiry Learning) activity on this topic, the goal is to understand how to use solubility rules and common ions to separate ions step-by-step. This article serves as a comprehensive guide, providing the conceptual answer key to common POGIL questions, worked examples, and the "why" behind the chemistry.

Disclaimer: This is an educational guide designed to help students check their understanding and learn the underlying principles. Always complete the POGIL activity yourself first; rote copying of answers defeats the purpose of inquiry-based learning.

A typical POGIL on fractional precipitation presents you with: Answers with Calculations:

Below, we break down the common questions and provide the answer key with full explanations.

Given example:
A solution contains ( \textBa^2+ ) and ( \textSr^2+ ), each at 0.10 M. You add ( \textNa_2\textSO4 ) dropwise.
(Ksp(\textBaSO4) = 1.1 \times 10^-10)
(Ksp(\textSrSO_4) = 3.2 \times 10^-7)

Step 1 – Precipitation [SO₄²⁻] needed for each:
[ [\textSO4^2-]\textstart Ba = \fracK_sp(\textBaSO_4)[\textBa^2+] = \frac1.1 \times 10^-100.10 = 1.1 \times 10^-9 , M ]
[ [\textSO4^2-]\textstart Sr = \frac3.2 \times 10^-70.10 = 3.2 \times 10^-6 , M ]
Conclusion: BaSO₄ precipitates first (lower required [SO₄²⁻]).

Step 2 – When does Sr²⁺ begin to precipitate?
At the moment SrSO₄ just starts:
[ [\textSO4^2-] = 3.2 \times 10^-6 , M ]
At this [SO₄²⁻], what is remaining [Ba²⁺]?
[ [\textBa^2+]\textremaining = \frac1.1 \times 10^-103.2 \times 10^-6 = 3.4 \times 10^-5 , M ]
So, Ba²⁺ is reduced from 0.10 M to (3.4 \times 10^-5 M) before Sr²⁺ starts — that’s >99.97% removed. Conclusion: Hg₂²⁺ precipitates at a very low [Cl⁻]

Step 3 – Key POGIL-type question:
If you stop adding SO₄²⁻ as soon as Sr²⁺ just begins to precipitate, are the two ions separated effectively?
Answer: Yes — Ba²⁺ is mostly precipitated, Sr²⁺ remains in solution.

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