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Chem. 31 – 10/22 Lecture

Chem. 31 – 10/22 Lecture. Announcements. HW 2.1 Additional Problem due today Water Hardness Lab Report due Monday Quiz Today Today’s Lecture Advanced Equilibrium: (Ch. 7) more on activity in equilibria real pH equation the systematic method – a way to deal with multiple equilibria.

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Chem. 31 – 10/22 Lecture

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  1. Chem. 31 – 10/22 Lecture

  2. Announcements • HW 2.1 Additional Problem due today • Water Hardness Lab Report due Monday • Quiz Today • Today’s Lecture • Advanced Equilibrium: (Ch. 7) • more on activity in equilibria • real pH equation • the systematic method – a way to deal with multiple equilibria

  3. Ionic Strength EffectsEffects on Equilibrium - Quantitative • Calculate expected [Mg2+] in equilibrium with solid MgCO3 for cases both with and without NaCl. • Go to Board

  4. Ionic Strength EffectsReal Equation for pH • pH = -logAH+ = -log(gH+[H+]) • Example Problem: Determine the pH of a solution containing 0.0050 M HCl and 0.020 M CaCl2. • Note: H2O  H+ + OH- also affected by ionic strength

  5. Second Part to Chapter 7The Systematic Method • Question: Why can’t we apply the ICE (initial, change, equilibrium) method to any type of equilibrium problem? • Answer: That method is best designed for cases where there is only one relevant equilibrium reaction. • Examples of failures: • Solubility of MgCO3 • pH of 5.0 x 10-8 M HCl solution (Show failure of Chem. 1B method) • Note: both problems can be solved using ICE method, but problem set up is more complicated

  6. The Systematic MethodSolubility of MgCO3 – Why did it fail? • MgCO3 Mg2+ + CO32- x x Equil. (in ICE) • So x = (Ksp)1/2 = 1.87 x 10-4 M (neglecting ionic strength effects) • Problem is both ions can react further: CO32- + H2O  HCO3- + OH- And HCO3- + H2O  H2CO3 + OH- Also, Mg2+ + OH-  MgOH+ And Mg2+ + CO32-  MgCO3 (aq) Finally, we also have H2O  H+ + OH- re-establishing equilibrium • Each additional reaction results in greater dissolution • To properly solve problem we must consider 6 reactions not just 1 • Measured “[CO32-]” from titration = [CO32-] + 0.5[OH-] + 0.5[HCO3-] + [MgCO3] + 0.5[MgOH+] • The “further” reactions makes [Mg2+] ≠ [CO32-], so ICE method fails (or needs modification by ICE tables for other reactions) • Actual solubility is greater than ICE method finds [Mg2+]total = solubility ~ 3.3 x 10-4 M (from systematic approach) Predicted HCl needed = 3.3 mL (close to that measured) These calculations didn’t include activity which would lead to a ~10% increase in solubility (~3.6 mL HCl needed). In 0.1 M NaCl, I get 6.1 mL HCl needed enhancements: (% over rxn 1 only) 90% 0% 9% 16%

  7. The Systematic MethodThe Six Steps • Write out all relevant reactions • Write a “Charge Balance Equation” • Write “Mass Balance Equations” • Write out all equilibrium equations • Check that the number of equations (in 2 to 4 above) = (or maybe >) the number of unknowns (undefined concentrations) • Solve for the desired unknown(s) by reducing the equations to one equation with one unknown. Then solve for remaining unknowns Note: the emphasis of teaching the systematic method is steps 1 to 5. Step 6 will be reserved for “easy” problems with 2 to max 3 unknowns

  8. The Systematic MethodpH of 5.0 x 10-8 M HCl • Demonstrate Method on Board

  9. The Systematic MethodConceptual Approach to Mass Balance Equations • With every source of related species, there should be one mass balance equation (or one set for ionic compounds) • Example: • Solubility of AgCl in water with 0.010 M 1,10-phenathroline (Ph) • Reactions: 1) AgCl(s)  Ag+ + Cl- 2) Ag+ + 2Ph  Ag(Ph)2+ • Mass Balance equations: • if only rxn 1) [Cl-] = [Ag+] • w/ rxn 2) [Cl-] = [Ag+] + [Ag(Ph)2+] 1,10-phenathroline Ag+ Ph Ph Ph Ph Ag+ Cl- Ag+ Cl- Ag+ Ag+ Cl- Cl- 2nd Mass Balance Equation: [Ph]o = 0.010 M = [Ph]Total = [Ph] + 2[Ag(Ph)2+] AgCl(s) Notes: with rxn 1) only, 2 Ag+s = 2 Cl-s; with rxn 2) also, 3 Cls = 2 Ags + 1 Ag(Ph)2 Initially 4 Phs, then 2 Phs + one complex containing 2 Phs (so total # of Phs remains constant)

  10. The Systematic Method2nd Example • An aqueous mixture of CdCl2 and NaSCN is made • Initial concentrations are [CdCl2] = 0.0080 M and [NaSCN] = 0.0040 M • Cd2+ reacts with SCN- to form CdSCN+ K = 95 • HSCN is a strong acid • Ignore any other reactions (e.g. formation of CdOH+) • Ignore activity considerations • Determine the concentrations of all species

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