| Question | Expected Answer | |----------|----------------| | Which ion precipitates first? | The one whose (K_sp) is smaller, or requires lower [precipitant] | | How to find that [precipitant]? | ( [X] = K_sp / [M^n+] ) or ( \sqrtK_sp/[M^n+] ) | | Can you separate completely? | Yes if (K_sp) differ by ≥ (10^4)–(10^6) | | What happens if you add too much precipitant? | The second ion also precipitates — loss of separation |
If you post a specific question from the POGIL (e.g., “Why does Pb²⁺ not precipitate until after Ag⁺ is gone?” or a table of (K_sp) values they gave), I can give you the exact reasoning and numeric answer.
Fractional Precipitation POGIL Answer Key
What is Fractional Precipitation?
Fractional precipitation is a technique used to separate two or more ions from a solution based on their different solubilities in water. This method is commonly used to purify substances and to separate ions that are present in small concentrations.
POGIL (Process of Guided Inquiry Learning) Activity
The POGIL activity on fractional precipitation is designed to help students understand the concept of solubility and how it can be used to separate ions. The activity involves a series of questions and exercises that guide students through the process of fractional precipitation.
Fractional Precipitation POGIL Answer Key
1. What is the solubility of BaSO4 in water?
Answer: 1.1 x 10^-5 M
2. What is the solubility of CaSO4 in water?
Answer: 2.5 x 10^-2 M
3. A solution contains 0.020 M Ba2+ and 0.020 M Ca2+. If SO42- is added to the solution, which ion will precipitate first? fractional precipitation pogil answer key best
Answer: Ba2+ will precipitate first because BaSO4 has a lower solubility than CaSO4.
4. What is the concentration of SO42- required to precipitate Ba2+?
Answer: [SO42-] = 1.1 x 10^-5 / 0.020 = 5.5 x 10^-4 M
5. What is the concentration of SO42- required to precipitate Ca2+?
Answer: [SO42-] = 2.5 x 10^-2 / 0.020 = 1.25 M
6. If 0.010 M SO42- is added to the solution, which ion will precipitate?
Answer: Ba2+ will precipitate because the concentration of SO42- is greater than 5.5 x 10^-4 M.
7. What is the molar ratio of Ba2+ to Ca2+ in the precipitate?
Answer: The molar ratio of Ba2+ to Ca2+ in the precipitate is 1:0, because only Ba2+ precipitates.
Conclusion
Fractional precipitation is an important technique used to separate ions based on their different solubilities. The POGIL activity on fractional precipitation helps students understand the concept of solubility and how it can be used to separate ions. By working through the questions and exercises, students gain a deeper understanding of the process of fractional precipitation and how it can be applied to real-world problems.
Tips and Variations
Fractional Precipitation POGIL (Process Oriented Guided Inquiry Learning) is a guided exercise designed to help chemistry students understand how to selectively remove specific ions from a mixture based on their varying solubilities. The "best" answer keys for this activity emphasize the relationship between the solubility product constant ( cap K sub s p end-sub ) and the reaction quotient ( cap Q sub s p end-sub ) to predict the order of precipitation. Core Concepts in Fractional Precipitation
Fractional precipitation is an analytical technique used to separate ions in a solution by adding a reagent that selectively causes one ion to precipitate while others remain dissolved. Chemistry Coach Selective Precipitation : The salt with the smaller cap K sub s p end-sub
value (least soluble) will typically precipitate first when a common ion is added gradually. Solubility Product ( cap K sub s p end-sub
: A constant that represents the equilibrium between a solid ionic compound and its dissolved ions. Reaction Quotient ( cap Q sub s p end-sub : Calculated using the same expression as cap K sub s p end-sub but with current concentrations. : The solution is unsaturated; no precipitate forms.
: The solution is supersaturated; precipitation occurs until Typical POGIL Model Walkthrough Most POGIL versions for this topic, such as those found on Course Hero
, use a standard experimental setup involving metal cations like cap Z n raised to the 2 plus power cap C u raised to the 2 plus power Step 1: Initial Concentration Analysis
The activity typically starts by asking for the initial concentrations of ions in the solution (e.g., cap Z n raised to the 2 plus power cap C u raised to the 2 plus power Step 2: Determining the First Precipitate
To find which ion precipitates first, you calculate the minimum concentration of the precipitating anion (e.g., cap C cap O sub 3 raised to the 2 minus power ) required to reach saturation for each salt.
open bracket cap A n i o n close bracket sub m i n end-sub equals the fraction with numerator cap K sub s p end-sub and denominator open bracket cap C a t i o n close bracket sub i n i t i a l end-sub end-fraction The cation that requires the concentration of the added anion to reach its cap K sub s p end-sub will precipitate first. Step 3: Assessing Separation Efficiency
A critical question in these keys is how much of the first ion remains in solution when the second ion just begins to precipitate.
required for the second ion's precipitation to solve for the remaining concentration of the first cation. Success Criterion
: Separation is generally considered "quantitative" if less than If you post a specific question from the POGIL (e
of the first ion remains when the second begins to precipitate. UCI Department of Chemistry Best Practices for Completing the POGIL 17.6: Fractional Precipitation - Chemistry LibreTexts
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
]
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⁺]).
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).
Key takeaway: The 1:2 stoichiometry dramatically changes the required cation concentration.
A solution contains 0.010 M Ba²⁺ and 0.010 M Ca²⁺. A student slowly adds solid Na₂CO₃ (which dissociates into 2Na⁺ + CO₃²⁻).
a) Which ion precipitates first?
b) What is the [CO₃²⁻] required to begin precipitation of the first ion?
c) When the second ion just begins to precipitate, what fraction of the first ion remains in solution?