Post on 17-Jan-2016
Chem. 1B – 10/8 Lecture
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
Announcements II
• Today’s Lecture– Titrations
• polyprotic acids and indicators
– Solubility and Precipitation: • Ksp and solubility
• common ion effect• pH effects on solubility• selective precipitation
Chem 1B – Aqueous ChemistryTitrations (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)
7Equiv. ptpH
Chem 1B – Aqueous ChemistryTitrations (Chapter 16)
• More complex titrations– polyprotic acid by a strong base (e.g. H2SO3 + OH-)
– This example has pKa1 = 1.81 and pKa1 = 6.97
– Titration involves 2 reactions:1)H2A + OH- ↔ HA- + H2O
2)HA- + OH- ↔ A2- + H2O
Veq1
Veq2
Veq2 = 2Veq1
V(NaOH)
Also has 2 buffer regions: 1) H2A + HA- present, 2) HA- + A2- present
buffer region 1)
buffer region 2)
Chem 1B – Aqueous ChemistryTitrations (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 orangeAcid form HInBase form In-
Called Methyl Orange, because at pH = pKa(HIn), equal amounts of each form
Chem 1B – Aqueous ChemistryTitrations (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– 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
Chem 1B – Aqueous ChemistryTitrations (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
7 indicator range
end point equivalence point
equivalence point
In this example, end point comes early
Chem 1B – Aqueous ChemistrySolubility (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(OH2) Ksp = 2.06 x 10-13
Chem 1B – Aqueous ChemistrySolubility (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
Chem 1B – Aqueous ChemistrySolubility (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 0.0002 M (diluted concentration) to milk ([Ca2+] = 0.06 M) or to orange juice ([Ca2+] = 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 [Ca2+] and seeing if it is > or < 0.0002 M
Chem 1B – Aqueous ChemistrySolubility (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
Chem 1B – Aqueous ChemistrySolubility (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
Chem 1B – Aqueous ChemistrySolubility (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 CO3
2-) solubility of CaCO3
– This matters because ocean creatures have CaCO3 containing structures (e.g. shells)
– Which effect matters more?
– Ka1 = 4.3 x 10-7 Ka2 = 5.6 x 10-11 and Ksp = 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 CaCO3 is a reactant and a product and compare K values (other products are H+ or HCO3
-)
Chem 1B – Aqueous ChemistrySolubility (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 Ca2+ + 0.0010 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]
Chem 1B – Aqueous ChemistrySolubility (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 SO4
2-?
Ksp(CaSO4) = 7.10 x 10-5 and Ksp (PbSO4) = 1.82 x 10-8