Fractional Precipitation Pogil Answer Key Best May 2026

Typical POGIL Question:
You have a solution containing Ba²⁺ (0.10 M) and Sr²⁺ (0.10 M). Which anion—SO₄²⁻ or C₂O₄²⁻ (oxalate)—would allow fractional precipitation?
(K_sp) BaSO₄ = (1.1 \times 10^-10), SrSO₄ = (3.2 \times 10^-7)
(K_sp) BaC₂O₄ = (1.6 \times 10^-6), SrC₂O₄ = (5.6 \times 10^-8)

Model Answer:

First, calculate for sulfate:

BaSO₄ precipitates first (lower required [SO₄²⁻]). The ratio of required concentrations is ~2900:1 — excellent separation.

Now for oxalate:

Here, SrC₂O₄ precipitates first (smaller required [C₂O₄²⁻]). But the required concentrations are very close (ratio only ~28:1). Complete separation would be difficult.

Best choice: Use sulfate. The larger difference in (K_sp) values favors better fractional precipitation.

We need to find how much $Cl^-$ is left when $[Ag^+] = 1.05 \times 10^-5\ M$. $$[Cl^-] = \fracK_sp(AgCl)[Ag^+]$$ $$[Cl^-] = \frac1.8 \times 10^-101.05 \times 10^-5$$ $$[Cl^-]_remaining = \mathbf1.71 \times 10^-5\ M$$

Scenario: A solution contains two anions, Chloride ($Cl^-$) and Chromate ($CrO_4^2-$). We wish to separate them by adding Silver Nitrate ($AgNO_3$) dropwise. fractional precipitation pogil answer key best

Solubility Product Constants ($K_sp$) at $25^\circ C$:

Key Definitions:

Below are the step-by-step solutions to the CTQs above. Use this to check your work.

Question:
As you continue adding AgNO₃, AgI continues to precipitate. At the moment just before AgCl begins to precipitate, what is the concentration of I⁻ remaining in solution? Typical POGIL Question: You have a solution containing

Model Answer:

AgCl begins to precipitate when [Ag⁺] reaches (1.8 \times 10^-8 M). At this [Ag⁺], the remaining [I⁻] is found from the (K_sp) of AgI:

[ [I^-] = \fracK_sp(\textAgI)[Ag^+] = \frac8.5 \times 10^-171.8 \times 10^-8 = 4.7 \times 10^-9 , M ]

Conclusion: By the time AgCl starts to precipitate, the [I⁻] has dropped from 0.010 M to (4.7 \times 10^-9 M). That’s a decrease by a factor of over 2 million. The separation is essentially complete. BaSO₄ precipitates first (lower required [SO₄²⁻])

Why this is the "best" key point:
This calculation demonstrates why fractional precipitation works. The first ion (I⁻) is reduced to a negligible level before the second ion (Cl⁻) begins to react.