Ch. 15: Applications Ch. 15: Applications of Aqueous Equilibriaof Aqueous Equilibria

15.4 Titrations and pH curves15.4 Titrations and pH curves

TitrationTitration used to find the concentration of a used to find the concentration of a

solution using a solution of known solution using a solution of known concentrationconcentration

__________________________ solution with unknown concentrationsolution with unknown concentration

__________________________ solution with known concentrationsolution with known concentration

equivalence point is usually marked equivalence point is usually marked by the end point of an indicatorby the end point of an indicator

pH curvepH curve also called a titration curvealso called a titration curve plotting of pH of the analyte as a plotting of pH of the analyte as a

function of the amount of titrant function of the amount of titrant addedadded

millimole (mmol)millimole (mmol) 1/10001/1000thth of a mole of a mole = molarity= molarity

Strong Acid and Strong BaseStrong Acid and Strong Base Titrate 50 mL of 0.2 HNOTitrate 50 mL of 0.2 HNO33 with 0.1 M with 0.1 M

NaOHNaOHA.A. no NaOH- 50.0 mL HNOno NaOH- 50.0 mL HNO33

B.B. 10.0 mL NaOH- 50.0 mL HNO10.0 mL NaOH- 50.0 mL HNO33 HH++ + OH + OH -- H H22OOIICCEE

Strong Acid and Strong BaseStrong Acid and Strong BaseC.C. 20 mL NaOH and 50 mL HNO20 mL NaOH and 50 mL HNO33

D.D. 50 mL NaOH and 50 mL HNO50 mL NaOH and 50 mL HNO33

HH++ + OH + OH -- HH22OO

IICCEE

HH++ + OH + OH -- HH22OO

IICCEE

Strong Acid and Strong BaseStrong Acid and Strong BaseE.E. 100 mL NaOH and 50 mL HNO100 mL NaOH and 50 mL HNO33

F.F. 150 mL NaOH and 50 mL HNO150 mL NaOH and 50 mL HNO33

HH++ + OH + OH -- HH22OO

IICCEE

HH++ + OH + OH -- HH22OO

IICCEE

Strong Acid and Strong BaseStrong Acid and Strong Basevery gradual changes in pH until close to the equivalence pt.

Weak Acid and Strong BaseWeak Acid and Strong Base Titrate 50 mL of 0.10M HCTitrate 50 mL of 0.10M HC22HH33OO22 with 0.1 M with 0.1 M

NaOHNaOHA.A. no NaOH- 50.0 mL no NaOH- 50.0 mL HCHC22HH33OO22

HCHC22HH33OO22 H H++ + + CC22HH33OO22 --

IICCEE

Weak Acid and Strong BaseWeak Acid and Strong BaseB.B. 10mL NaOH and 50 mL HC10mL NaOH and 50 mL HC22HH33OO22 OHOH-- + + HCHC22HH33OO22 H H22O + O + CC22HH33OO22 --

IICCEE HCHC22HH33OO22 H H++ + + CC22HH33OO22

--

IICCEE

Weak Acid and Strong BaseWeak Acid and Strong BaseC.C. 50mL NaOH and 50 mL HC50mL NaOH and 50 mL HC22HH33OO22 OHOH-- + + HCHC22HH33OO22 H H22O + O + CC22HH33OO22 --

IICCEE CC22HH33OO22 -- + H + H22O O HC HC22HH33OO22 + OH- + OH-IICCEE

Weak Acid and Strong BaseWeak Acid and Strong BaseD.D. 75mL NaOH and 50 mL HC75mL NaOH and 50 mL HC22HH33OO22 OHOH-- + + HCHC22HH33OO22 H H22O + O + CC22HH33OO22 --

IICCEEthere are two bases to consider: which will be most important?Can just use the OH- to determine the pH

Comparing ShapesComparing Shapes

Weak Base and Strong AcidWeak Base and Strong Acid

Weak Base and Strong AcidWeak Base and Strong Acid

Ch. 15: Applications Ch. 15: Applications of Aqueous Equilibriaof Aqueous Equilibria

15.5 Indicators15.5 Indicators

Acid-Base Acid-Base IndicatorsIndicators

used to mark end of used to mark end of titration titration (____________________)(____________________)

only works if you choose only works if you choose one whose end point is one whose end point is the same as eq. pointthe same as eq. point

usually are complex usually are complex molecules that act as molecules that act as weak acidsweak acids

change colors when H+ is change colors when H+ is removedremoved

phenolphthalein

IndicatorsIndicatorsHIn HIn H H++ + In + In--

yellow blueyellow blue If more H+ (acid) is added to the solution, the If more H+ (acid) is added to the solution, the

reaction shifts to the left to make it yellowreaction shifts to the left to make it yellow If more OH- (base) is added to the solution, the If more OH- (base) is added to the solution, the

reaction shifts to the right to make it bluereaction shifts to the right to make it blueBromthymol BlueBromthymol Blue

Calculating the pH at end pointCalculating the pH at end point Assume that color change is visible to Assume that color change is visible to

the eye when ratio of [Inthe eye when ratio of [In--]/[HIn] is 1/10]/[HIn] is 1/10 that means that one tenth of the that means that one tenth of the

indicator must have changed forms indicator must have changed forms Using the H-H equation and the one Using the H-H equation and the one

tenth rule, the pH of end point or color tenth rule, the pH of end point or color change is:change is:

Only shows where the color change occurs

Choosing an IndicatorChoosing an Indicator

The weaker the acid being titrated, the smaller the vertical area around eq. pt, that means less flexibility in choosing an indicator.

ExampleExample Estimate the pH of a solution in Estimate the pH of a solution in

which bromcresol green is blue and which bromcresol green is blue and thymol blue is yellow.thymol blue is yellow.

A 0.100 M HCl solution is made. 2 A 0.100 M HCl solution is made. 2 drops of methyl orange are added. drops of methyl orange are added. What color is the solution?What color is the solution?

Ch. 15: Applications Ch. 15: Applications of Aqueous Equilibriaof Aqueous Equilibria

15.6: Solubility Equilibria and 15.6: Solubility Equilibria and Solubility ProductsSolubility Products

SolubilitySolubility As a salt dissolves in water and ions As a salt dissolves in water and ions

are released, they can collide and re-are released, they can collide and re-from the solidfrom the solid

Equilibrium is reached when the rate Equilibrium is reached when the rate of ________________ equals the rate of of ________________ equals the rate of __________________________________________

CaFCaF22(s) ↔ Ca(s) ↔ Ca2+2+(aq) + 2F(aq) + 2F--(aq)(aq) ______________ solution______________ solution

When no more solid can dissolve at When no more solid can dissolve at equilibriumequilibrium

Solubility ProductSolubility Product Solubility product: KSolubility product: Kspsp

CaFCaF22(s) ↔ Ca(s) ↔ Ca2+2+(aq) + 2F(aq) + 2F--(aq)(aq) Ksp=Ksp= Why do we leave out the CaFWhy do we leave out the CaF22?? Adding more solid will not effect the Adding more solid will not effect the

amount of solid that can dissolve at a amount of solid that can dissolve at a certain temperaturecertain temperature

It would increase both reverse and forward It would increase both reverse and forward reaction rates b/c there is a greater amountreaction rates b/c there is a greater amount

KKspsp Values Values

Example 1Example 1 CuBr has a solubility of 2.0x10CuBr has a solubility of 2.0x10-4-4 mol/L at mol/L at

2525°°C. Find the KC. Find the Kspsp value. value. The solubility tells us the amount of The solubility tells us the amount of

solute that can dissolves in 1 L of watersolute that can dissolves in 1 L of water Use ICE chart: solubility tells you x valueUse ICE chart: solubility tells you x value

Ksp=Ksp=

CuBr(s) ↔ CuCuBr(s) ↔ Cu++(aq) + Br(aq) + Br--

(aq) (aq) IICCEE

Example 2Example 2 The KThe Kspsp value for Cu(IO value for Cu(IO33))22 is 1.4x10 is 1.4x10-7-7 at at

2525°°C. Calculate its solubility.C. Calculate its solubility. Solve for solubility = x value using ICE Solve for solubility = x value using ICE

chartchart

KKspsp=[Cu=[Cu2+2+][IO][IO33--]]22

Cu(IOCu(IO33))22(s) ↔ Cu(s) ↔ Cu2+2+(aq) + 2IO(aq) + 2IO33--

(aq)(aq) IICCEE

Comparing Solubilities Comparing Solubilities You can only compare solubilities using You can only compare solubilities using

KKspsp values for compounds containing the values for compounds containing the same number of ionssame number of ions

CaSOCaSO44 > CuCO > CuCO33 > AgI > AgIKKspsp values: 6.1x10 values: 6.1x10-5-5 > 2.5x10 > 2.5x10-10-10 > 1.5x10 > 1.5x10-6-6

Why can we use KWhy can we use Kspsp values to judge values to judge solubility?solubility?

Can only compare using actual solubility Can only compare using actual solubility values (x) when compounds have values (x) when compounds have different numbers of ionsdifferent numbers of ions

Common Ion EffectCommon Ion Effect Solubility of a solid is lowered when a Solubility of a solid is lowered when a

solution already contains one of the solution already contains one of the ions it containsions it contains

Why?Why?

Example 3Example 3 Find the solubility of CaFFind the solubility of CaF22 (s) if the K (s) if the Kspsp is is

4.0 x 104.0 x 10-11-11 and it is in a 0.025 M NaF and it is in a 0.025 M NaF solution.solution.

KKspsp=[Ca=[Ca2+2+][F][F--]]22

CaFCaF22(s) ↔ Ca(s) ↔ Ca2+2+(aq) + 2F(aq) + 2F--

(aq)(aq)IICCEE

pH and solubilitypH and solubility pH can effect solubility because of the pH can effect solubility because of the

common ion effectcommon ion effect Ex: Mg(OH)Ex: Mg(OH)22(s) ↔ Mg(s) ↔ Mg2+2+(aq) + 2OH(aq) + 2OH--(aq)(aq)

How would a high pH effect solubility?How would a high pH effect solubility? High pH = _______ [OH-] High pH = _______ [OH-] solubility solubility

________________________ Ex: AgEx: Ag33POPO44(s) ↔ 3Ag(s) ↔ 3Ag++(aq) + PO(aq) + PO44

3-3-(aq)(aq) What would happen if H+ is added?What would happen if H+ is added? HH++ uses up _____ to make phosphoric acid uses up _____ to make phosphoric acid Eq. shifts to _______ - Solubility __________Eq. shifts to _______ - Solubility __________

Ch. 15: Applications Ch. 15: Applications of Aqueous Equilibriaof Aqueous Equilibria

15.7: Precipitation and 15.7: Precipitation and Qualitative AnalysisQualitative Analysis

PrecipitationPrecipitation Opposite of dissolutionOpposite of dissolution Can predict whether precipitation or Can predict whether precipitation or

dissolution will occurdissolution will occur Use Q: ion productUse Q: ion product

Equals KEquals Kspsp but doesn’t have to be at but doesn’t have to be at ________________________________

Q > K: more reactant will form, Q > K: more reactant will form, ____________ until equilibrium reached____________ until equilibrium reached

Q < K: more product will form, Q < K: more product will form, ________________________

Example 1Example 1 A solution is prepared by mixing 750.0 A solution is prepared by mixing 750.0

mL of 4.00x10mL of 4.00x10-3-3M Ce(NOM Ce(NO33))33 and 300.0 mL and 300.0 mL 2.00x102.00x10-2-2M KIOM KIO33. Will Ce(IO. Will Ce(IO33))33 precipitate precipitate out?out? Calculate Q value and compare to K (on Calculate Q value and compare to K (on

chart)chart) Ce(IOCe(IO33))33(s) ↔ Ce(s) ↔ Ce3+3+(aq) + 3IO(aq) + 3IO33

--(aq)(aq) Q=[CeQ=[Ce3+3+][IO][IO33

--]]33

Example 2Example 2 A solution is made by mixing 150.0 mL A solution is made by mixing 150.0 mL

of 1.00x10of 1.00x10-2-2 M Mg(NO M Mg(NO33))22 and 250.0 mL and 250.0 mL of 1.00x10of 1.00x10-1-1 M NaF. Find concentration M NaF. Find concentration of Mgof Mg2+2+ and F and F-- at equilibrium with solid at equilibrium with solid MgFMgF22 (K (Kspsp=6.4x10=6.4x10-9-9))

MgFMgF22(s) (s) Mg Mg2+2+(aq) + 2F(aq) + 2F--(aq)(aq) Need to figure out whether the Need to figure out whether the

concentrations of the ions are high concentrations of the ions are high enough to cause precipitation firstenough to cause precipitation first

Find Q and compare to KFind Q and compare to K

Example 2Example 2

Q > K so shift to left, precipitation occursQ > K so shift to left, precipitation occurs Will all of it precipitate out??Will all of it precipitate out?? No- No-

we need to figure out how much is created we need to figure out how much is created using stoichiometryusing stoichiometry

then how much ion is left over using ICE chartthen how much ion is left over using ICE chart Like doing acid/base problemLike doing acid/base problem

2-2 ]][F[MgQ

Example 2Example 2

MgMg2+2+(aq) + 2F(aq) + 2F--(aq) ↔ MgF(aq) ↔ MgF22(s)(s)IICCEE

MgFMgF22(s)↔ Mg(s)↔ Mg2+2+(aq) + 2F(aq) + 2F--(aq)(aq)IICCEE

How much will be used if goes to completion?

Some of it dissolved- how much are left in solution?

Example 2Example 2

][

][

104.6]][[

2

922

F

xMg

FMgK sp

Qualitative AnalysisQualitative Analysis Process used to separate a solution Process used to separate a solution

containing different ions using solubilitiescontaining different ions using solubilities A solution of 1.0x10A solution of 1.0x10-4-4 M Cu M Cu++ and 2.0x10 and 2.0x10-3-3 M M

PbPb2+2+. If I. If I-- is gradually added, which will is gradually added, which will precipitate out first, CuI or PbIprecipitate out first, CuI or PbI22?? 1.4x101.4x10-8-8=[Pb=[Pb2+2+][I][I--]]22 = (2.0x10 = (2.0x10-4-4)[I)[I--]]22

[I[I--]= : [I]= : [I--]> than that to cause PbI]> than that to cause PbI22 to ppt to ppt 5.3x105.3x10-12-12=[Cu=[Cu++][I][I--] = (1.0x10] = (1.0x10-4-4)[I)[I--]] [I[I--]= : [I]= : [I--]> than that to cause CuI to ]> than that to cause CuI to

pptppt Takes a much lower conc to cause CuI to ppt so it Takes a much lower conc to cause CuI to ppt so it

will happen firstwill happen first

Qualitative AnalysisQualitative Analysis

Ch. 15: Applications Ch. 15: Applications of Aqueous Equilibriaof Aqueous Equilibria

15.8: Complex Ion Equilibria15.8: Complex Ion Equilibria

Complex Ion EquilibriaComplex Ion Equilibria __________________________________

Charged species containing metal ion Charged species containing metal ion surrounded by ligandssurrounded by ligands

____________________ Lewis bases donating electron pair to Lewis bases donating electron pair to

empty orbitals on metal ionempty orbitals on metal ion Ex: HEx: H22O, NHO, NH33, Cl-, CN-, OH-, Cl-, CN-, OH-

______________________________________________________ Number of ligands attachedNumber of ligands attached

Complex Ion EquilibriaComplex Ion Equilibria Usually, the conc of the ligand is very Usually, the conc of the ligand is very

high compared to conc of metal ion high compared to conc of metal ion in the solutionin the solution

Ligands attach in stepwise fashionLigands attach in stepwise fashion AgAg++ + NH + NH33 Ag(NH Ag(NH33))++

Ag(NHAg(NH33))++ + NH + NH33 Ag(NH Ag(NH33))2+2+

Example 3Example 3 Find the [AgFind the [Ag++], [Ag(S], [Ag(S22OO33))--],], and [Ag(Sand [Ag(S22OO33))22

3-3-]] in solution made with 150.0 mL of 1.00x10in solution made with 150.0 mL of 1.00x10--

33 M AgNO M AgNO33 with 200.0 mL of 5.00 M with 200.0 mL of 5.00 M NaNa22SS22OO33.. AgAg++ + S + S22OO33

2-2- Ag(S Ag(S22OO33))-- KK11=7.4x10=7.4x1088

Ag(SAg(S22OO33))- - + S+ S22OO332-2- Ag(S Ag(S22OO33))22

3-3-KK22=3.9x10=3.9x1044

Because of the difference in conc between Because of the difference in conc between ligand and metal ion, the reactions can be ligand and metal ion, the reactions can be assumed to go to completionassumed to go to completion

Example 3Example 3 AgAg++ + 2S + 2S22OO33

2-2- Ag(S Ag(S22OO33))223-3-

II

CCEE

mLmmol])O[Ag(S

][

-3232

232 mL

mmolOS

])([

109.3]86.2][)([

]1029.4[

109.3]][)([

])([

132

41

32

4

4232

132

3232

2

OSAg

OSAg

OSOSAgOSAgK

][

109.3]86.2][[]108.3[

104.7]][[])([

1

41

9

8232

1

132

1

Ag

Ag

OSAgOSAgK