Chemistry 102(01) spring 2009
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Transcript of Chemistry 102(01) spring 2009
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Instructor: Dr. Upali Siriwardanee-mail: [email protected]: CTH 311 Phone 257-4941Office Hours: M,W, 8:00-9:00 & 11:00-12:00 a.m.;
Tu,Th,F 9:00 - 10:00 a.m. Test Dates: March 25, April 26, and May 18; Comprehensive Final Exam:
May 20,2009 9:30-10:45 am, CTH 328.March 30, 2009 (Test 1): Chapter 13 April 27, 2009 (Test 2): Chapters 14 & 15 May 18, 2009 (Test 3): Chapters 16, 17 & 18
Comprehensive Final Exam: May 20,2009 :Chapters 13, 14, 15, 16, 17 and 18
Chemistry 102(01) spring 2009
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Chapter 17. Aditional Aqueous Equilibria
17.1 Buffer Solutions 17.2 Acid-Base Titrations17.3 Acid Rain17.4 Solubility Equilibria and the Solubility
Product Constant, Ksp 17.5 Factors Affecting Solubility / Precipitation:
Will It Occur?
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Reaction of a basic anion with water is an ordinary Brønsted-Lowry acid-base reaction.
CH3COO-(aq) + H2O(l) CH3COOH(aq) + OH-(aq)
This type of reaction is given a special name.HydrolysisThe reaction of an anion with water to produce
the conjugate acid and OH-.The reaction of a cation with water to produce the
conjugate base and H3O+.
Hydrolysis
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How do you calculate pH of a salt solution?
• Find out the pH, acidic or basic?• If acidic it should be a salt of weak base• If basic it should be a salt of weak acid• if acidic calculate Ka from Ka= Kw/Kb
• if basic calculate Kb from Kb= Kw/Ka
• Do a calculation similar to pH of a weak acid or base
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What is the pH of 0.5 M NH4Cl salt solution? (NH 3; Kb = 1.8 x 10-5)Find out the pH, acidic
• if acidic calculate Ka from Ka= Kw/Kb
• Ka= Kw/Kb = 1 x 10-14 /1.8 x 10-5)• Ka= 5.56. X 10-10
• Do a calculation similar to pH of a weak acid
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Continued• NH4
+ + H2O <==> H 3+O + NH3
• [NH4+] [H3
+O ] [NH3 ]• Ini. Con. 0.5 M 0.0 M 0.00 M• Eq. Con. 0.5 - x x x• [H 3
+O ] [NH3 ] • Ka(NH4
+) = -------------------- =• [NH 4
+] • x2
• ---------------- ; appro.:0.5 - x . 0.5• (0.5 - x) •
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• x2 • Ka(NH4
+) = ----------- = 5.56 x 10 -10
• 0. 5• x2
= 5.56 x 10 -10 x 0.5 = 2.78 x 10 -
10
• x= sqrt 2.78 x 10 -10 = 1.66 x 10-5
• [H+ ] = x = 1.66 x 10-5 M• pH = -log [H+ ] = - log 1.66 x 10-5
• pH = 4.77• pH of 0.5 M NH4Cl solution is 4.77 (acidic)
Continued
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This is an example of Le Châtelier’s principle.Common ion effectThe shift in equilibrium caused by the addition of
an ion formed from the solute.Common ionAn ion that is produced by more than one solute
in an equilibrium system.Adding the salt of a weak acid to a solution of
weak acid is an example of this.
Common Ion Effect
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Common Ion Effect
Weak acid and salt solutionsE.g. HC2H3O2 and NaC2H3O2
Weak base and salt solutionsE.g. NH3 and NH4Cl. H2O + C2H3O2
- <==> OH- + HC2H3O2 (common ion)
H2O + NH4+ <==> H3
+O + NH3 (common ion)
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Solutions that resist pH change when small amounts of acid or base are added.
Two typesMixture of weak acid and its saltMixture of weak base and its salt
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
Add OH- Add H3O+
shift to right shift to left
Based on the common ion effect.
Buffers
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The pH of a buffer does not depend on the absolute amount of the conjugate acid-base pair. It is based on the ratio of the two.
Henderson-Hasselbalch equation.Easily derived from the Ka or Kb expression.
Starting with an acid
pH = pKa + log Starting with a base
pH = 14 - ( pKb + log )[HA][A-]
[A-][HA]
Buffers
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Henderson-Hasselbalch EquationHA(aq) + H2O(l) H3O+(aq) + A-(aq)
[H3O+] [A-] Ka = ----------------
[HA]
[H3O+] = Ka ([HA]/[A-])
pH = pKa + log([A-]/[HA])
when the [A-] = [HA]pH = pKa
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Calcualtion of pH of BuffersHenderson Hesselbach Equation [ACID]pH = pKa - log --------- [BASE] [BASE]pH = pKa + log --------- [ACID]
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Control of blood pHOxygen is transported primarily by hemoglobin in
the red blood cells.
CO2 is transported both in plasma and the red blood cells.
CO2 (aq) + H2O H2CO3 (aq)
H+(aq) + HCO3-(aq)
The bicarbonatebuffer is essential
for controllingblood pH
Buffers and blood
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Buffer Capacity• Refers to the ability of the buffer to retard
changes in pH when small amounts of acid or base are added
• The ratio of [A-]/[HA] determines the pH of the buffer whereas the magnitude of [A-] and [HA] determine the buffer capacity
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Adding an Acid or Baseto a Buffer
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Buffer Systems
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Titrations ofAcids and Bases
• Titration• Analyte• Titrant
analyte + titrant => products
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Acid-base indicators are highly colored weak acids or bases.
HIn In- + H+
color 1 color 2
They may have more than one color transition. Example. Thymol blue
Red - Yellow - Blue
One of the forms may be colorless - phenolphthalein (colorless to pink)
Indicators
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Acid-Base Indicator
HIn + H2O H3O+ + In-
acid base color color
[H3O+][In-]Ka =
[HIn]They may have more than one color transition.
Example. Thymol blue Red - Yellow - Blue
Weak acid that changes color with changes in pH
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What is an Indicator?
• Indicator is an weak acid with different Ka, colors to the acid and its conjugate base. E.g. phenolphthalein
• HIn <===> H+ + In-
• colorless pink• Acidic colorless• Basic pink
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Selection of an indicator for a titration
a) strong acid/strong base b) weak acid/strong base c) strong acid/weak base d) weak acid/weak base Calculate the pH of the solution at he
equivalence point or end point
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pH and Color of Indicators
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Red Cabbage as IndicatorC O H
O
N N N
CH3
CH3
(aq)
C O-
O
N N N
CH3
CH3
(aq) + H3O+ (aq)
yellow
red
+ H2O (l)
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• Acid-base indicators are weak acids that undergo a color change at a known pH.
phenolphthalein
pH
Indicator examples
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Titration Apparatus
Buret delivering base to a flask containing an acid. The pink color in the flask is due to the phenolphthalein indicator.
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Endpoint vs. Equivalence Point
Endpointpoint where there is a physical change,
such as color change, with the indicator
Equivalence Point# moles titrant = # moles analyte
#molestitrant=(V M)titrant
#molesanalyte=(V M)analyte
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Acid-Base Indicator Behavior
acid color shows when
[In-] 1 £ [HIn] 10
[H3O+][In-] 1 = [H3O+] = Ka [HIn] 10
base color shows when[In-] 1 ³ [HIn] 10
[H3O+][In-] = 10 [H3O+] = Ka [HIn]
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Indicator pH Range
acid color shows when
pH + 1 = pKa
and base color shows whenpH - 1 = pKa
Color change range ispKa = pH 1 or pH = pKa 1
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Titration curves
Acid-base titration curveA plot of the pH against the amount of acid or
base added during a titration.Plots of this type are useful for visualizing a
titration.It also can be used to show where an indicator undergoes its color change.
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Indicator and Titration Curve 0.1000 M HCl vs 0.1000 M NaOH
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EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HCl with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH.
at 0.00 mL of NaOH added, initial point
[H3O+] = CHCl = 0.1000 M
pH = 1.0000
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EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HCl with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH.
at 15.00 mL of NaOH added
Va Ma > Vb Mb thus
(Va Ma) - (Vb Mb)[H3O+] =
(Va + Vb) ((35.00mL) (0.1000M)) - ((15.00mL) (0.1000M))= (35.00 + 15.00)mL
= 4.000 10-2 M pH = 1.3979
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EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HCl with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH.
at 35.00 mL of NaOH added
Va Ma = Vb Mb , equivalence point
at equivalence point of a strong acid - strong base titration
pH º 7.0000
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EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HCl with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH.
at 50.00 mL of NaOH added
Vb Mb > Va Ma , post equvalence point
(Vb Mb) - (Va Ma)[OH-] =
(Va + Vb) ((50.00mL) (0.1000M)) - ((35.00mL) (0.1000M))=
(35.00 + 50.00)mL= 1.765 10-2 MpOH = 1.7533
pH = 14.00 - 1.7533 = 12.25
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0
5
10
15
0 10 20 30 40 50
Volume of Base Added
pH equivalence point x
Titration of Strong Acid with Strong Base
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Titration of Weak Acid with Strong Base
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Effect of Acid Strength on Titration Curve
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Titration of Weak Base with Strong Acid
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Titration of Diprotic Weak
Acid with Strong Base
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pH range of Indicators
• litmus (5.0-8.0)• bromothymole blue (6.0-7.6)
• methyl red (4.8-6.0)• thymol blue (8.0-9.6) • phenolphthalein (8.2-10.0)• thymolphthalein (9.4-10.6)
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Acid Rain
acid rain is defined as rain with a pH < 5.6
pH = 5.6 for rain in equilibrium with atmospheric carbon dioxide
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Sulfuric Acid from Sulfur burning
SO2
S + O2 => SO2
SO3
2 SO2 + O2 => 2 SO3
Sulfuric AcidSO3 + H2O => H2SO4
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Nitric Acid2 NO2(g) + H2O(l) => HNO3(aq) + HNO2(aq)
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How Acid Precipitation Forms
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Acid Precipitation in U.S.
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Solubility Productsolubility-product - the product of the
solubilitiessolubility-product constant => Ksp
constant that is equal to the solubilities of the ions produced when a substance dissolves
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Solubility Product ConstantIn General:
AxBy xA+y + yB-x
[A+y]x [B-x]y
K = [AxBy]
[AxBy] K = Ksp = [A+y]x [B-x]y
Ksp = [A+y]x [B-x]y
For silver sulfateAg2SO4 2 Ag+ + SO4-2
Ksp = [Ag+]2[SO4-2]
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Solubility Product
Constant Values
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Dissolving Slightly Soluble Salts Using Acids
Insoluble salts containing anions of Bronsted-Lowry bases can be dissolved in solutions of low pH
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Calcium CarbonateDissolved in Acid
Limestone Dissolving in Ground WaterCaCO3(S) + H2O + CO2 => Ca+2
(aq) + 2 HCO3-(aq)
Stalactite and Stalagmite FormationCa+2
(aq) + 2 HCO3-(aq) => CaCO3(S) + H2O + CO2
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The Common Ion Effectcommon ion• second source which is completely
dissociated• In the presence of a second source of the
ion, there will be less dissolved than in its absence
common ion effect• a salt will be less soluble if one of its
constitutent ions is already present in the solution
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EXAMPLE: What is the molar solubility of AgCl in pure water and in 1.0 M NaCl?
AgCl Ag+ + Cl-
Ksp = [Ag+][Cl-] = 1.82 10-10M2
let x = molar solubility = [Ag+] = [Cl-](x)(x) = Ksp = [Ag+][Cl-] = 1.82 10-10M2
x = 1.35 10-5M
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EXAMPLE: What is the molar solubility of AgCl in pure water and in 1.0 M NaCl?
AgCl Ag+ + Cl-
Ksp = [Ag+][Cl-] = 1.82 10-10M2
let x = molar solubility = [Ag+][Cl-] = 1.0 M
Ksp = [Ag+][Cl-] = (x)(1.0M) = 1.82 10-10M2
x = 1.82 10-10M
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Formation of Complexes
ligand - Lewis basecomplexes - product of Lewis acid-base
reaction
Ag+(aq) + 2 NH3(aq) [Ag(NH3)2(aq)]+
Ag+(aq) + Cl-(aq) AgCl(s)
AgCl(s) + 2 NH3(aq) [Ag(NH3)2(aq)]+ + Cl-(aq)
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Sodium Thiosulfate Dissolves Silver Bromide
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Formation Constants
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Amphoterism
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Reactant Quotient, Q
Reactant Quotient, Q• ion product of the initial concentration• same form as solubility product constant
Q < Ksp - no precipitate forms an unsaturate solutionQ > Ksp - precipitate may form to restore
condition of saturated solutionQ = Ksp - no precipitate forms, saturated solution
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Will Precipitation Occur?
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Kidney Stones
Kidney stones are normally composed of:calcium oxalate
calcium phosphatemagnesium ammonium phsphate
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Calcium Oxalate
Precipitate formed from calcium ions from food rich in calcium, dairy products, and oxalate ions from fruits and vegetables
Ca+2 + C2O4-2 CaC2O4
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PRECIPITATION REACTIONS
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Analysis of Silver Group
These salts formed are insoluble, they do dissolve to some SLIGHT extent.
AgCl(s) Ag+(aq) + Cl-(aq)When equilibrium has been
established, no more AgCl dissolves and the solution is SATURATED.
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED, expt. shows that [Ag+] = 1.67 x 10-5 M.
(SOLUBILITY) of AgCl.What is [Cl-]?
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Analysis of Silver Group
AgCl(s) Ag+(aq) + Cl-(aq)Saturated solution has
[Ag+] = [Cl-] = 1.67 x 10-5 MUse this to calculate Ksp
Ksp = [Ag+] [Cl-]
= (1.67 x 10-5)(1.67 x 10-5) = 2.79 x 10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
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Because this is the product of “solubilities”, we call it
Ksp = solubility product constant See Table in the Text
Ksp = solubility product
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Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-
(aq) Ksp = 1.9 x 10-5
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Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) + 2 I-(aq)
Calculate Ksp if solubility = 0.00130 M
Solution1.Solubility = [Pb2+]
= 1.30 x 10-3 M [I-] = 2 x [Pb2+] = 2.60 x 10-3 M2.Ksp = [Pb2+] [I-]2
= [Pb2+] {2 • [Pb2+]}2
= 4 [Pb2+]3 = 4 (solubility)3
Ksp = 4 (1.30 x 10-3)3 = 8.8 x 10-9
Solubility of Lead(II) Iodide
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Precipitating an Insoluble Salt
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 1.1 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0.010 M, what [Cl-] is req’d to
just begin the precipitation of Hg2Cl2?
(maximum [Cl-] in 0.010 M Hg22+ without
forming Hg2Cl2?)
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Precipitating an Insoluble SaltHg2Cl2(s) Hg2
2+(aq) + 2 Cl-(aq)
Ksp = 1.1 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs.
Precip. begins when product of
ion concs. EXCEEDS the Ksp.
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Precipitating an Insoluble Salt
Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)
Ksp = 1.1 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0.010 M,
If this conc. of Cl- is just exceeded, Hg2Cl2
begins to precipitate.
[Cl ] = Ksp0.010
= 1.1 x 10-8 M
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Precipitating an Insoluble SaltHg2Cl2(s) Hg2
2+(aq) + 2 Cl-(aq)
Ksp = 1.1 x 10-18
Now raise [Cl-] to 1.0 M. What is the value of [Hg2
2+] at this point?
Solution[Hg2
2+] = Ksp / [Cl-]2
= Ksp / (1.0)2 = 1.1 x 10-18 M
The concentration of Hg22+ has been
reduced by 1016 !
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Separating Metal Ions Cu2+, Ag+, Pb2+
Ksp ValuesAgCl 1.8 x 10-10
PbCl2 1.7 x 10-5
PbCrO4 1.8 x 10-14
Cu2+ Ag+ Pb2+
Cl- InsolublePbCl2 AgCl
SolubleCuCl2
HeatInsoluble
AgClSolublePbCl2
CrO4-2
InsolublePbCrO4
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Separating Salts by Differences in Ksp
A solution contains 0.020 M Ag+ and Pb2+. Add CrO4
2- to precipitate Ag2CrO4 (red) and PbCrO4 (yellow). Which precipitates first?
Ksp for Ag2CrO4 = 9.0 x 10-12
Ksp for PbCrO4 = 1.8 x 10-14
SolutionThe substance whose Ksp is first exceeded
precipitates first. The ion requiring the smaller amount of
CrO42- ppts. first.
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Separating Salts by Differences in Ksp
[CrO42-] to ppt. PbCrO4 = Ksp / [Pb2+]
= 1.8 x 10-14 / 0.020 = 9.0 x 10-13 M[CrO4
2-] to ppt. Ag2CrO4 = Ksp / [Ag+]2 = 9.0 x 10-12 / (0.020)2 = 2.3 x 10-8 M PbCrO4 precipitates first.
SolutionCalculate [CrO4
2-] required by each ion.
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How much Pb2+ remains in solution when Ag+ begins to precipitate?
SolutionWe know that [CrO4
2-] = 2.3 x 10-8 M to begin to ppt. Ag2CrO4.
What is the Pb2+ conc. at this point?[Pb2+] = Ksp / [CrO4
2-] = 1.8 x 10-14 / 2.3 x 10-8 M
= 7.8 x 10-7 MLead ion has dropped from 0.020 M to < 10-6 M
Separating Salts by Differences in Ksp
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Common Ion EffectAdding an Ion “Common” to an
Equilibrium
PbCl2(s) Pb2+(aq) + 2Cl-(aq)
NaCl Na+(aq) + Cl- (aq)
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0.010 M Ba(NO3)2.
Ksp for BaSO4 = 1.1 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solutiona) Solubility in pure water = [Ba2+] = [SO4
2-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)1/2 = 1.1 x 10-5 M
The Common Ion Effect
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BaSO4(s) Ba2+(aq) + SO42-(aq)
Ksp = 1.1 x 10-10
Solutionb) Now dissolve BaSO4 in water already
containing 0.010 M Ba2+. Which way will the “common ion” shift the
equilibrium? ___ Will solubility of BaSO4 be less than or
greater than in pure water?___
The Common Ion Effect
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BaSO4(s) Ba2+(aq) + SO42-
(aq)Solution [Ba2+] [SO4
2-]initialchangeequilib.
The Common Ion Effect
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Calculate the solubility of BaSO4 in (a) pure water and (b) in 0.010 M Ba(NO3)2.
Ksp for BaSO4 = 1.1 x 10-10
BaSO4(s) Ba2+(aq) + SO42-(aq)
Solution [Ba2+] [SO4
2-]initial 0.010 0change + y + yequilib. 0.010 + y y
The Common Ion Effect
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Ksp = [Ba2+] [SO42-] = (0.010 + y) (y)
Because y << x (1.1 x 10-5 M) 0.010 + y 0.010. Therefore,
Ksp = 1.1 x 10-10 = (0.010)(y)y = 1.1 x 10-8 M =
solubility in presence of added Ba2+ ion.
Le Chatelier’s Principle is followed!
The Common Ion Effect