Chem. 31 – 11/22 Lecture - PowerPoint Presentation

1. Chem. 31 – 11/22 Lecture

2. AnnouncementsLabs Due TodayFormal Part AIC Part 1 11/22Today’s LectureChapter 8 – Advanced EquilibriumFailure of the ICE method with two equilibriaThe systematic method and its six stepsChapter 9 – Acid/Base EquilibriaThe “weak acid problem” (pH of weak acid in water)

3. 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 equilibriumEach additional reaction results in greater dissolutionTo properly solve problem we must consider 6 reactions not just 1Measured “[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 (vs. ~3.5 mL)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 (close to that observed)enhancements: (% over rxn 1 only)90%0%9%16%

4. The Systematic MethodThe Six StepsWrite out all relevant reactionsWrite a “Charge Balance Equation”Write “Mass Balance Equations”Write out all equilibrium equationsCheck 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 unknownsNote: 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

5. The Systematic MethodpH of 5.0 x 10-8 M HClDemonstrate Method on Board

6. The Systematic MethodConceptual Approach to Mass Balance EquationsWith 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+]AgCl(s)Ag+Ag+Ag+Ag+Cl-Cl-Cl-Cl-PhPhPhPh1,10-phenathrolineAg+Notes: with rxn 1) only, 2 Ag+s = 2 Cl-s; with rxn 2) also, 3 Cls = 2 Ags + 1 Ag(Ph)22nd Mass Balance Equation:[Ph]o = 0.010 M = [Ph]Total = [Ph] + 2[Ag(Ph)2+]Initially 4 Phs, then 2 Phs + one complex containing 2 Phs (so total # of Phs remains constant)

7. The Systematic Method2nd ExampleAn aqueous mixture of CdCl2 and NaSCN is madeInitial concentrations are [CdCl2] = 0.0080 M and [NaSCN] = 0.0040 MCd2+ reacts with SCN- to form CdSCN+ K = 95HSCN is a strong acidIgnore any other reactions (e.g. formation of CdOH+)Ignore activity considerationsGo through steps 1 through 5

8. The Systematic Method2nd ExampleA 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.

9. The Systematic MethodStong Acid/Strong Base ProblemsWhen do we need to use the systematic approach?when more than 1 coupled reaction occur (unless coupling is insignificant)examples: 4.0 x 10-3 M HCl. 7.2 x 10-3 M NaOHKey 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-]

10. The Systematic MethodGeneral CommentsEffects of secondary reactionse.g. MgCO3 dissolutionAdditional reactions increase solubilitySecondary 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)

11. Acid – Base Equilibria (Ch. 9)Weak Acid Problems:e.g. What is the pH and the concentration of major species in a 2.0 x 10-4 M HCO2H (formic acid, Ka = 1.80 x 10-4) solution ?Can use either systematic method or ICE method.Systematic method will give correct answers, but full solution results in cubic equationICE method works most of the timeUse of systematic method with assumptions allows determining when ICE method can be used

12. Acid – Base EquilibriaWeak Acid Problem – cont.:Systematic Approach (HCO2H = HA to make problem more general where HA = weak acid)Step 1 (Equations) HA ↔ H+ + A- H2O ↔ H+ + OH-Step 2: Charge Balance Equation: [H+] = [A-] + [OH-] 2 assumptions possible: ([A-] >> [OH-] – assumption used in ICE method or [A-] << [OH-])Step 3: Mass Balance Equation: [HA]o = 2.0 x 10-4 M = [HA] + [A-]Step 4: Kw = [H+][OH-] and Ka = [A-][H+]/[HA]Step 5: 4 equations (1 ea. steps 2 + 3, 2 equa. step 4), unk.: [HA], [A-] [H+], [OH-]

13. Acid – Base EquilibriaWeak Acid Problem – cont.:Assumption #1: [A-] >> [OH-] so [A-] = [H+]Discussion: this assumption means that we expect that there will be more H+ from formic acid than from water. This assumption makes sense when [HA]o is large and Ka is not that small (valid for [HA]o>10-6 M for formic acid)ICE approach (Gives same result as systematic method if assumption #1 is made)(Equations) HA ↔ H+ + A- Initital 2.0 x 10-4 0 0 Change - x +x +x Equil. 2.0 x 10-4 – x x x

14. Acid – Base EquilibriaWeak Acid Problem – Using ICE ApproachKa = [H+][A-]/[HA] = x2/(2.0 x 10-4 – x)x = 1.2 x 10-4 M (using quadratic equation)Note: sometimes (but not in this case), a 2nd assumption can be made that x << 2.0 x 10-4 to avoid needing to use the quadratic equation[H+] = [A-] = 1.2 x 10-4 M; pH = 3.92[HA] = 2.0 x 10-4 – 1.2 x 10-4 = 8 x 10-5 MNote: a = fraction of dissociation = [A-]/[HA]totala = 1.2 x 10-4 /2.0 x 10-4 = 0.60

15. Acid – Base EquilibriaWeak Acid Problem – cont.:When is Assumption #1 valid (in general)?When both [HA]o and Ka are high or so long as [H+] > 10-6 MMore precisely, when [HA]o > 10-6 M and Ka[HA]o > 10-12See chart (shows region where error < 1%)Failure also can give [H+] < 1.0 x 10-7 MAssupmption #1 WorksFails