What mast 0/ CJ(NO;h; must added to [ OLo )LOM HF solution to begin erecipikation al Cafm? (You MJY AssUme there hn0 votume change on the additton ol calaum t trate | Cafiakt 4.010 u Hk7210 Describe hot You would scparte JMmorin [unethy Amenehydratint andeyridine Using Ion echanga cmonatogtjol. (Use Lhe follawing tablo for naeded valbes ) Aqueous-Equilibrium Constants TABLB Diesociallon Constants for F ABaraz 25C Hcnmn ommne Amecole R5: Anuyn Denatnutuen u 5410 ‘ Etuelinmt 2 67 * 10″ tdtalit 13710- Anentennc aee 1140m Melneee flan L7ro- nnn acer
The Correct Answer and Explanation is:
To solve this question, we need to approach it in two parts:
Part 1: How much Ca(NO₃)₂ must be added to 0.010 M HF to begin precipitation of CaF₂?
We are given:
- CaF2⇌Ca2++2F−CaF₂ ⇌ Ca²⁺ + 2F⁻CaF2⇌Ca2++2F−
- Ksp of CaF₂ = 4.0 × 10⁻¹¹
- Initial [HF] = 0.010 M
- Ka of HF = 6.8 × 10⁻⁴
Step 1: Calculate [F⁻] from HF
HF is a weak acid, so it only partially dissociates:HF⇌H++F−HF ⇌ H⁺ + F⁻HF⇌H++F−
Using the expression for Ka:Ka=[H+][F−][HF]Ka = \frac{[H⁺][F⁻]}{[HF]}Ka=[HF][H+][F−]
Assume initial [HF] = 0.010 M, let x be [H⁺] and [F⁻] at equilibrium:Ka=x20.010−xKa = \frac{x^2}{0.010 – x}Ka=0.010−xx2
Since Ka is small, assume 0.010−x≈0.0100.010 – x ≈ 0.0100.010−x≈0.010:6.8×10−4=x20.010⇒x2=6.8×10−6⇒x=[F−]≈2.61×10−3M6.8 × 10⁻⁴ = \frac{x^2}{0.010} \Rightarrow x^2 = 6.8 × 10⁻⁶ \Rightarrow x = [F⁻] ≈ 2.61 × 10⁻³ M6.8×10−4=0.010×2⇒x2=6.8×10−6⇒x=[F−]≈2.61×10−3M
Step 2: Use Ksp to find required [Ca²⁺]
CaF2⇌Ca2++2F−CaF₂ ⇌ Ca²⁺ + 2F⁻CaF2⇌Ca2++2F−Ksp=[Ca2+][F−]2Ksp = [Ca²⁺][F⁻]²Ksp=[Ca2+][F−]24.0×10−11=[Ca2+](2.61×10−3)2⇒[Ca2+]=4.0×10−116.81×10−6⇒[Ca2+]≈5.87×10−6M4.0 × 10⁻¹¹ = [Ca²⁺](2.61 × 10⁻³)^2 \Rightarrow [Ca²⁺] = \frac{4.0 × 10⁻¹¹}{6.81 × 10⁻⁶} \Rightarrow [Ca²⁺] ≈ 5.87 × 10⁻⁶ M4.0×10−11=[Ca2+](2.61×10−3)2⇒[Ca2+]=6.81×10−64.0×10−11⇒[Ca2+]≈5.87×10−6M
Answer: You must add Ca(NO₃)₂ to produce a [Ca²⁺] ≈ 5.87 × 10⁻⁶ M in the solution to begin precipitation of CaF₂.
Part 2: Separation of Aniline, Aminoacetophenone, and Pyridine using Ion Exchange Chromatography
Given: Their pKa values
- Aniline (weak base): pKa ≈ 4.6
- Aminoacetophenone (stronger base): pKa ≈ 9.7
- Pyridine (basic): pKa ≈ 5.2
Ion exchange chromatography separates compounds based on charge. We use cation exchange resin for separating basic compounds.
Step-by-step separation plan:
- Acidic mobile phase (pH ≈ 2):
- All bases are protonated and become positively charged.
- All will bind to the cation exchange resin.
- Gradual increase in pH:
- First eluted: The weakest base (Aniline, pKa 4.6) becomes neutral first and elutes.
- Second: Pyridine (pKa 5.2) elutes as it becomes neutral.
- Last: Aminoacetophenone (pKa 9.7) remains protonated longer and elutes at higher pH.
Conclusion: By adjusting the pH of the mobile phase from low to high, the three compounds can be separated based on their different protonation states and affinities for the cation exchange resin.
This approach uses equilibrium principles, acid-base dissociation constants, and solubility rules to solve both the precipitation and separation parts of the question.
