Announcements Labs Due Wednesday GC Todays Lecture Chapter 8 Advanced Equilibrium The systematic method and its six steps more practice Generalities about using the systematic method ID: 1039111
Download Presentation The PPT/PDF document "Chem. 31 – 11/27 Lecture" is the property of its rightful owner. Permission is granted to download and print the materials on this web site for personal, non-commercial use only, and to display it on your personal computer provided you do not modify the materials and that you retain all copyright notices contained in the materials. By downloading content from our website, you accept the terms of this agreement.
1. Chem. 31 – 11/27 Lecture
2. AnnouncementsLabs Due WednesdayGCToday’s LectureChapter 8 – Advanced EquilibriumThe systematic method and its six steps – more practiceGeneralities about using the systematic methodChapter 9 – Acid/Base EquilibriaThe “weak acid problem” (pH of weak acid in water)The weak base problem
3. 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
4. The Systematic Method3rd 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.
5. 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-]
6. 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)
7. 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
8. 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-]
9. 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
10. 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
11. 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
12. Acid – Base EquilibriaWeak Base Problem:As with weak acid problem, ICE approach can generally be used (except when [OH-] from base is not much more than [OH-] from water) Note: when using ICE method, must have correct reactionExample: Determine pH of 0.010 M NH3 solution (Ka(NH4+) = 5.7 x 10-10, so Kb = Kw/Ka = 1.75 x 10-5)ReactionNH3 + H2O NH4+ + OH-Go over on board
13. Acid – Base EquilibriaWeak Acid/Weak Base Questions:A solution is prepared by dissolving 0.10 moles of NH4NO3 into water to make 1.00 L of solution. Show how to set up this problem for determining the pH using the ICE method.A student is solving a weak base problem for a weak base initially at 1.00 x 10-4 M using the ICE method and calculates that [OH-] = 2.4 x 10-8 M. Was the ICE method appropriate?The pH of an unknown weak acid prepared to a concentration 0.0100 M is measured and found to be 3.77. Calculate a and Ka.