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Chem. 31 – 2/26 Lecture

Chem. 31 – 2/26 Lecture. Announcements I. Exam 1: Next Week on Wednesday To see last semester’s exams plus key, go to link on class website Typically about 25 to 30% multiple choice or short answer and the rest problems

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Chem. 31 – 2/26 Lecture

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  1. Chem. 31 – 2/26 Lecture

  2. Announcements I • Exam 1: • Next Week on Wednesday • To see last semester’s exams plus key, go to link on class website • Typically about 25 to 30% multiple choice or short answer and the rest problems • Homework, quiz and in lecture questions are pretty typical of what to expect on exam • Help Session: Monday – 12 to 1, Sequoia 446 (Quant lab) • Expected topics: Ch1, 3, 4, and Ch. 6 to complex ions • Review of Exam topics on Monday

  3. Announcements II • Necessary Changes to Office Hours • Monday (1 to 1:30 and 4 to 4:30) • Tuesday (now 10:00 to 11:00) • HW1.3 • Will be posting text solutions soon • AP1.3 due on Monday • Covering today: • Chapter 6 (Equilibria) • Solubility and Precipitation • Complex ion formation

  4. Solubility Product Problems- Solubility of Mg(OH)2 in Common Ion If we dissolve Mg(OH)2 in a common ion (OH- or Mg2+), from Le Châtelier’s principle, we know the solubility will be reduced Example 1) What is the solubility of Mg(OH)2 in a pH = 11.0 buffer? No ICE table needed because, from pH, we know [OH-]eq and buffer means dissolution of Mg(OH)2 doesn’t affect pH.

  5. Solubility Product Problems- Solubility of Mg(OH)2 at pH 11 – cont. [H+] = 10-pH = 10-11 M and [OH-] = Kw/[H+] = 10-3 M Ksp = [Mg2+][OH-]2 Moles Mg(OH)2 dissolved = moles Mg2+ [Mg2+] = Ksp/[OH-]2 = 7.1 x 10-12/(10-3)2 [Mg2+] = 7.1 x 10-6 M

  6. Solubility Product Problems- Solubility of Mg(OH)2 in Common Ion Example 2) Solubility of Mg(OH)2 in 5.0 x 10-3 M MgCl2.

  7. Solubility Product ProblemsPrecipitation Problems What occurs if we mix 50 mL of 0.020 M BaCl2 with 50 mL of 3.0 x 10-4 M (NH4)2SO4? Does any solid form from the mixing of ions? What are the concentrations of ions remaining?

  8. Precipitations Used for Separations Example: If we wanted to know the concentrations of Ca2+ and Mg2+ in a water sample. EDTA titration gives [Ca2+] + [Mg2+]. However, if we could selectively remove Ca2+ or Mg2+ (e.g. through titration) and re-titrate, we could determine the concentrations of each ion. Determine if it is possible to remove 99% of Mg2+ through precipitation as Mg(OH)2 without precipitating out any Ca(OH)2 if a tap water solution initially has 1.0 x 10-3 M Mg2+ and 1.0 x 10-3 M Ca2+.

  9. Complex Ions Example Reaction: Ag+ + 2NH3(aq) ↔ Ag(NH3)2+ Metal Ligand Complex Ion Why does reaction occur? Metal is a Lewis acid (electron pair acceptor) NH3 is a Lewis base (electron pair donator) Metal-ligand bonds are intermediate strength

  10. Complex Ions – Why Study? Crown ether (12-crown-4) • Useful in separations • Complexed metals become more organic soluble • Effects on metal solubility (e.g. addition of NH3 on AgCl solubility) • Complexometric titrations (e.g. water hardness titration) Na+ Crown ether added Diethyl ether Sodium conc. given by gray shading water

  11. Complex Ions Step-wise vs. full reactions: Example: addition of NH3 to Ag+ Reaction occurs in steps: 1) Ag+ + NH3(aq) ↔ AgNH3+ K1 (= β1) 2) AgNH3+ + NH3(aq) ↔ Ag(NH3)2+ K2 Net) Ag+ + 2NH3(aq) ↔ Ag(NH3)2+β2 = K1·K2

  12. Complex Ions Due to large exponents on ligand concentration, a small change in ligand concentration has a big effect on how metal exists Example: Al3+ + 3C2O42-↔ Al(C2O4)33-β3 = 4.0 x 1015 [C2O42-] [Al(C2O4)33-]/[Al3+] 10-4 M 4000 10-5 M 4 10-6 M 0.004

  13. Complex Ions –“U” Shaped Solubility Curves Many sparingly soluble salts release cations and anions that form complexes with each other Example: calcium oxalate (CaC2O4) CaC2O4(s) ↔ Ca2+ + C2O42- (Ksp = 1.3 x 10-8M) increased [C2O42-] decreases Ca2+ solubility for above reaction only, but ... Ca2+ + C2O42- ↔ CaC2O4(aq) K1 = 46 CaC2O4(aq) + C2O42- ↔ Ca(C2O4)22-K2 = 490 β2 = K1·K2 = 2.3 x 104 = [Ca(C2O4)22-]/([Ca2+][C2O42-]2)

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