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Chem. 31 – 10/27 Lecture

Chem. 31 – 10/27 Lecture. Announcements. Water Hardness Lab Report due Today Today’s Lecture Advanced Equilibrium: (Ch. 7) the systematic method – a way to deal with multiple equilibria + numerous examples some consequences of application of the systematic method.

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Chem. 31 – 10/27 Lecture

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  1. Chem. 31 – 10/27 Lecture

  2. Announcements • Water Hardness Lab Report due Today • Today’s Lecture • Advanced Equilibrium: (Ch. 7) • the systematic method – a way to deal with multiple equilibria + numerous examples • some consequences of application of the systematic method

  3. 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

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

  5. 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 fully dissolved) • 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)

  6. 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

  7. The Systematic Method3rd Example • A student prepares a solution that contains 0.050 mol of AgNO3 and 0.0040 mol NH3 in water with a total volume of 1.00 L. The AgNO3 is totally soluble, NH3 is a weak base, and Ag+ reacts with NH3 to form Ag(NH3)2+. Assume the Ag+ does not react with water or OH-. Go through the first 5 steps of the systematic method.

  8. The Systematic MethodStong Acid/Strong Base Problems • When do we need to use the systematic approach? • examples: 4.0 x 10-3 M HCl. 7.2 x 10-3 M NaOH • Key point is the charge balance equation: • for strong acid HX, [H+] = [X-] + [OH-] • If [X-] >> [OH-], then [H+] = [X-] • for strong base NaOH, [H+] + [Na+] = [OH-]

  9. The Systematic MethodGeneral Comments Effects of secondary reactions e.g. MgCO3 dissolution Additional reactions increase solubility Secondary reactions also can affect pH (CO32- + H2O will produce OH- while Mg2+ + H2O will produce H+) Software is also available to solve these types of problems (but still need to know steps 1 → 5 to get problems solved)

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