1. Chem. 1B – 10/8 Lecture

2. Announcements I Lab –Quiz 5 next Monday and Tuesday – Topics: titrations, solubility and experiments 3 and 4 –Experiment 4 – on Polyprotic acids Mastering Assignments – next (16b) due 10/13 Web site has: –Correct lecture notes from last time –Solutions to exams

3. Announcements II Today’s Lecture –Titrations polyprotic acids and indicators –Solubility and Precipitation: K sp and solubility common ion effect pH effects on solubility selective precipitation

4. 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 V(acid) 7 Equiv. pt pH

5. Chem 1B – Aqueous Chemistry Titrations (Chapter 16) More complex titrations –polyprotic acid by a strong base (e.g. H 2 SO 3 + OH - ) –This example has pK a1 = 1.81 and pK a1 = 6.97 –Titration involves 2 reactions: 1)H 2 A + OH - ↔ HA - + H 2 O 2)HA - + OH - ↔ A 2- + H 2 O V eq1 V eq2 V eq2 = 2V eq1 V(NaOH) Also has 2 buffer regions: 1) H 2 A + HA - present, 2) HA - + A 2- present buffer region 1) buffer region 2)

6. 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 HInBase form In - Called Methyl Orange, because at pH = pK a (HIn), equal amounts of each form

7. 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 pK a = 3.5 –At pH 4.5 –What type of titrations is it useful for? pH V(acid) 7 Yellow (pH > 4.5) Pink (pH < 2.5) indicator is only useful where it changes color

8. 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 (pK a = 6.7) for a weak acid – strong base titration pH 7indicator range end pointequivalence point In this example, end point comes early

9. 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) M p X q (s) ↔ pM q+ (aq) + qX p- (aq) Resulting Equilibrium Equation: K = K sp (for solubility product) = [M q+ ] p [X p- ] 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 ) K sp = 2.06 x 10 -13

10. Chem 1B – Aqueous Chemistry Solubility (Chapter 16) Is K sp 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 K sp value –The K sp values of Ag 2 CrO 4 and PbCl 2 are directly comparable due to same stoichiometries, but not between AgCl and the other two SaltK sp Solubility (M) AgCl1.77 x 10 -10 1.33 x 10 -5 Ag 2 CrO 4 1.12 x 10 -12 6.54 x 10 -5 PbCl 2 1.17 x 10 -5 1.43 x 10 -2

11. Chem 1B – Aqueous Chemistry Solubility (Chapter 16) Solubility in a common ion: For example CaF 2 (K sp = 1.46 x 10 -10 ) has a solubility of 3.32 x 10 -4 M (in water) Could we add F - at 0.0002 M (diluted concentration) to milk ([Ca 2+ ] = 0.06 M) or to orange juice ([Ca 2+ ] = 0.002 M) without losing F - due to precipitation? (F - is prescribed to infants – usually added to juice – for teeth health in regions without water fluoridation) Can answer the above question by determining [F - ] in equilibrium with given [Ca 2+ ] and seeing if it is > or < 0.0002 M

12. Chem 1B – Aqueous Chemistry Solubility (Chapter 16) Effect of pH on Solubility –Besides dissolving solids in water and in a common ion, addition of acids can affect solubility –Example: CaCO 3 (K sp = 4.96 x 10 -9 ) – a common mineral Molar solubility in water = (4.96 x 10 -9 ) 0.5 = 7.0 x 10 -5 M What if we dissolve CaCO 3 in dilute HNO 3 ? CaCO 3 (s) ↔ Ca 2+ (aq) + CO 3 2- (aq) K sp = 4.96 x 10 -9 CO 3 2- (aq) + H + (aq) ↔ HCO 3 - (aq) K = 1/K a2 = 1.78 x 10 10 net = CaCO 3 (s) + H + (aq) ↔ Ca 2+ (aq) + HCO 3 - (aq) K = 89 solubility in pH 4 buffer = 0.09 M

13. Chem 1B – Aqueous Chemistry Solubility (Chapter 16) Effect of pH on Solubility –Which of the following salts have solubility increase by addition of acids? –AgCl- Mg 3 (PO 4 ) 2 –Mg(OH) 2 - BaSO 4 –Hg 2 Br 2 RULE: –If conjugate base is a strong or weak base, acid addition increases solubility –If conjugate base is neutral (conjugate to strong acid), no effect on solubility

14. Chem 1B – Aqueous Chemistry Solubility (Chapter 16) Another Example –“Ocean Acidification” Effect of Carbon Dioxide –Increase of CO 2 in the atmosphere leads to an increase in H 2 CO 3 (an acid) in oceans (even if effect is to make ocean less basic) –This leads to both increasing (due to acid addition from H 2 CO 3 ) and decreasing (due to common ion effect of added CO 3 2- ) solubility of CaCO 3 –This matters because ocean creatures have CaCO 3 containing structures (e.g. shells) –Which effect matters more? –K a1 = 4.3 x 10 -7 K a2 = 5.6 x 10 -11 and K sp = 4.96 x 10 -9 –Can work in groups up to 5 members for 1 bonus pt (each group member) Hint: write reactions in which CaCO 3 is a reactant and a product and compare K values (other products are H + or HCO 3 - )

15. Chem 1B – Aqueous Chemistry Solubility (Chapter 16) Precipitation –If we mix an ion pair from a sparingly soluble salt together, how do we know if precipitation will occur? –Example: 0.040 M Ca 2+ + 0.0010 M F - – do we get CaF 2 (s)? –Using K sp reaction (backwards of what is occurring), we can compare Q with K sp –If Q < K sp, no precipitation occurs (solution stays clear) –If Q > K sp, either precipitation occurs or supersaturated solution [Example problem]

16. Chem 1B – Aqueous Chemistry Solubility (Chapter 16) Precipitation for selective ion removal –Example: An old battery plant had a leak of lead and sulfuric acid that was neutralized by addition of CaCO 3. The collected liquid has [Ca 2+ ] = 0.20 M and [Pb 2+ ] = 0.10 M. A chemist wants to save the lead (in form Pb 2+ ) but not the Ca 2+. Can she selectively precipitate out >98% of the Pb 2+ without precipitating Ca 2+ by addition of SO 4 2- ? K sp (CaSO 4 ) = 7.10 x 10 -5 and K sp (PbSO 4 ) = 1.82 x 10 -8