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Chem. 1B – 10/6 Lecture.

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Presentation on theme: "Chem. 1B – 10/6 Lecture."— Presentation transcript:

1 Chem. 1B – 10/6 Lecture

2 Announcements Drop Deadline = Friday Lab
Experiment 5 due next week Quiz (next week) on Titrations + Experiment 3 Mastering (Titrations – due today) Today’s Lecture Titrations (other topics; review) Solubility – using equilibrium Complex Ions (if time)

3 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 3

4 Chem 1B – Aqueous Chemistry Titrations (Chapter 16) – Review Questions
25 mL of M Ba(OH)2 is titrated with M HCl What volume of HCl is needed to reach the equivalence point? Is the slope of the pH (= y-axis) vs. VHCl positive or negative? What is the pH when 25 mL of HCl has been added? Phenolphthalein (pKa = 8.8) was selected as the indicator. It is clear at pH < 7.8 and pink at pH > What color change will be observed?

5 Chem 1B – Aqueous Chemistry Titrations (Chapter 16) – Review Questions
Looking at the titration below: What type of titration is it? Give the pKa for the weak acid being titrated.

6 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 6

7 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 7

8 Chem 1B – Aqueous Chemistry Solubility (Chapter 16)
Solubility in a common ion: For example CaF2 (Ksp = 1.46 x 10-10) has a solubility of 3.32 x 10-4 M (in water) Could we add F- at M (diluted concentration) to milk ([Ca2+] = 0.06 M) or to orange juice ([Ca2+] = M) without losing F- due to precipitation? (F- is prescribed to small children – usually added to juice – for teeth health in regions without water fluoridation) Can answer the above question by determining [F-] in equilibrium with given [Ca2+] and seeing if it is > or < M 8

9 Chem 1B – Aqueous Chemistry Solubility (Chapter 16)
Solubility in a common ion: Example 2: How much PbCl2 can dissolve into a solution containing 0.14 M NaCl? Example 3: What is the solubility of Mg(OH)2 in a pH = 10.0 buffer? 9

10 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: CaCO3 (Ksp = 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 CaCO3 in dilute HNO3? CaCO3(s) ↔ Ca2+(aq) + CO32-(aq) Ksp = 4.96 x 10-9 CO32-(aq) + H+(aq) ↔ HCO3-(aq) K = 1/Ka2 = 1.78 x 1010 net = CaCO3(s) + H+(aq) ↔ Ca2+(aq) + HCO3-(aq) K = 89 solubility in pH 4 buffer = 0.09 M 10

11 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 - Mg3(PO4)2 Mg(OH)2 - BaSO4 Hg2Br2 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 11

12 Chem 1B – Aqueous Chemistry Solubility (Chapter 16)
Another Example “Ocean Acidification” Effect of Carbon Dioxide Increase of CO2 in the atmosphere leads to an increase in H2CO3 (an acid) in oceans (even if effect is to make ocean less basic) This leads to both increasing (due to acid addition from H2CO3) and decreasing (due to common ion effect of added CO32-) solubility of CaCO3 This matters because ocean creatures have CaCO3 containing structures (e.g. shells) Which effect matters more? Ka1 = 4.3 x Ka2 = 5.6 x and Ksp = 4.96 x 10-9 Compare: CaCO3(s) + H2CO3(aq) ↔ Ca2+(aq) + 2HCO3- and Ca2+(aq) + H2CO3(aq) ↔ CaCO3(s) + 2H+(aq) 12

13 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: M Ca M F- – do we get CaF2 (s)? Using Ksp reaction (backwards of what is occurring), we can compare Q with Ksp If Q < Ksp, no precipitation occurs (solution stays clear) If Q > Ksp, either precipitation occurs or supersaturated solution [Example problem] 13

14 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 CaCO3. The collected liquid has [Ca2+] = 0.20 M and [Pb2+] = 0.10 M. A chemist wants to save the lead (in form Pb2+) but not the Ca2+. Can she selectively precipitate out >98% of the Pb2+ without precipitating Ca2+ by addition of SO42-? Ksp(CaSO4) = 7.10 x 10-5 and Ksp (PbSO4) = 1.82 x 10-8 14

15 Chem 1B – Aqueous Chemistry Complex Ion Formation (Chapter 16)
Complex Ions – Why do we study? Metals bound as complexes do not react the same as “free” metals In determining solubility, for example, only the “free” metal is in the equilibrium equation Example: AgCl(s) ↔ Ag+(aq) + Cl-(aq) or K = [Ag+][Cl-] – Ag complexed as Ag(NH3)2+ ≠ Ag+ Complexation also affects reactivity and uptake For example: spinach is high in Fe, but also in oxalate (C2O42-) which complexes Fe (Kf = 2 x 1020 for Fe3+) making uptake more difficult. Additionally, oxalate potentially can bind Fe from the body 15

16 Chem 1B – Aqueous Chemistry Complex Ion Formation (Chapter 16)
Complex Ions – Why do we study? Uses of complex ions in analytical chemistry (besides for increasing solubility) Water hardness titration – done in experiment 6 Separations – complexed metals are more organic soluble Reduction of oxidation (for example Fe3+ + H2O2 produces strong oxidants, unless Fe3+ is bound) 16


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