Chem. 1B – 10/4 Lecture

Announcements I Exam #1 - Results Average = 64.5 Worse average than last year’s 1st exam (but better than Sect. 8) Much of this class depends on you having a good knowledge of Ch. 14 and 15 Students did worse than I expected on conceptual problems Score Range # Students* 90-96 7 80s 15 70s 28 60s 36 50s <50 20

Announcements II Exam #1 – Results – cont. Plot of last year – Exam 1 score vs. final class score Only 2 out of 17 students finished with C or above after an Exam 1 score < 50

Announcements III Exam 1 – cont. - go over problems with less than 50% getting it right (A version #1, 3, 10, 11, 17, 22, 23) Lab – Starting Experiment 3 (Acid-Base Titrations) Mastering (Buffers Assignment – due today) Today’s Lecture Titrations (weak acid-strong base, weak base-strong acid, other topics) Solubility – using equilibrium (if time)

Weak Acid – Strong Base Titration How does pH Change as NaOH is added? Reaction: HA + OH- ↔ A- + H2O K = 1/Kb Example – acetic acid: Ka = 1.76 x 10-5 4 regions to titrations (different calculations in each region): initial pH (last time) before equivalence point at equivalence point after equivalence point Show pH at 5 mL, 12.5 mL, and 15 mL 0.100 M NaOH 0.050 M HA, 25 mL 5

Chapter 10 – Acid Base Titrations Weak Acid – Strong Base Titrations What affects shape of curve? pKa values (low pKa or stronger weak acid gives sharper titration) pKa affects position of curve before and at equivalence point Note: at 6.25 mL (half of equivalence point), pH = pKa 50 mM example 6

Chem 1B – Aqueous Chemistry Titrations (Chapter 16) Weak Base – Strong Acid Titration How does pH Change as HCl is added? Reaction: B + H+ ↔ BH+ K = 1/Ka Example – ammonia: Ka (BH +) = 5.69 x 10-10 4 regions to titrations (different calculations in each region): initial pH before equivalence point at equivalence point after equivalence point 0.100 M HCl 0.050 M B, 25 mL 7

Chem 1B – Aqueous Chemistry Titrations (Chapter 16) Weak Base – Strong Acid Has similar 4 regions as weak acid – strong base initial (weak base) before equiv. point (buffer problem) equiv. point (weak acid) after equiv. point (excess strong acid) pH 7 V(acid) 8

Chem 1B – Aqueous Chemistry Titrations (Chapter 16) Qualitative Understanding Question: Based on the shape of this titration curve the flask/buret contains? Weak acid/strong base Strong base/strong acid Strong acid/strong base Weak base/strong acid pH Equiv. pt pH 7 V(acid) 9

Chem 1B – Aqueous Chemistry Titrations (Chapter 16) More complex titrations polyprotic acid by a strong base (e.g. H2SO3 + OH-) This example has pKa1 = 1.81 and pKa2 = 6.97 Titration involves 2 reactions: H2A + OH- ↔ HA- + H2O HA- + OH- ↔ A2- + H2O Veq2 = 2Veq1 Also has 2 buffer regions: 1) H2A + HA- present, 2) HA- + A2- present Veq2 buffer region 2) Veq1 buffer region 1) V(NaOH) 10

Chem 1B – Aqueous Chemistry Titrations (Chapter 16) Indicators One of the reasons to bother to learn the shape of the titration curves is to be able to select an indicator Indicators are colored compounds that exist in acidic and basic forms Example: methyl orange Acid form HIn Base form In- Called Methyl Orange, because at pH = pKa(HIn), equal amounts of each form 11

Chem 1B – Aqueous Chemistry Titrations (Chapter 16) Indicators – cont. Indicators change color over a narrow pH range (visible over 1 to 2 pH units) Methyl Orange pKa = 3.5 At pH < 2.5 main species = HIn (pink) vs. at pH > 4.5 (yellow) What type of titrations is it useful for? pH Yellow (pH > 4.5) 7 indicator is only useful where it changes color Pink (pH < 2.5) V(acid) 12

Chem 1B – Aqueous Chemistry Titrations (Chapter 16) Titration Errors The observed equivalence point (where the indicator changes color) is called the end point Titration errors occur when the end point volume is before or after the equivalence point Example: Use of bromothymol blue indicator (pKa = 6.7) for a weak acid – strong base titration pH equivalence point 7 indicator range In this example, end point comes early end point equivalence point 13

Chem 1B – Aqueous Chemistry Solubility (Chapter 16) A particular type of aqueous equilibrium reaction has to do with solubility of ionic compounds Generic Reaction (for ionic compound MX) MpXq(s) ↔ pMq+(aq) + qXp-(aq) Resulting Equilibrium Equation: K = Ksp (for solubility product) = [Mq+]p[Xp-]q Value: Now we can calculate the exact concentration of dissolved species rather than label compounds as only “soluble” or “insoluble” Example problem: What is the molar solubility (defined as moles of solid dissolved per L solution) of Mg(OH) 2 Ksp = 2.06 x 10-13 14

Chem 1B – Aqueous Chemistry Solubility (Chapter 16) Is Ksp the only thing that affects solubility? Not exactly, stoichiometry also matters Examples: Stoichiometries giving more moles of products will lead to greater solubility for a given Ksp value The Ksp values of Ag2CrO4 and PbCl2 are directly comparable due to same stoichiometries, but not between AgCl and the other two Salt Ksp Solubility (M) AgCl 1.77 x 10-10 1.33 x 10-5 Ag2CrO4 1.12 x 10-12 6.54 x 10-5 PbCl2 1.17 x 10-5 1.43 x 10-2 15