Applications of Aqueous Applications of Aqueous EquilibriaEquilibria

Chapter 15

Common Ion Effect CalculationsCommon Ion Effect Calculations

Calculate the pH and the percent dissociation of Calculate the pH and the percent dissociation of a .200M HCa .200M HC22HH33OO22 (K (Kaa=1.8x10-5)=1.8x10-5)

BufferBuffer

Calculate the pH and the percent dissociation of Calculate the pH and the percent dissociation of a .200M HCa .200M HC22HH33OO2 2 in the presence of in the presence of .500 M .500 M

NaCNaC22HH33OO22

Common Ion Effect CalculationsCommon Ion Effect CalculationsContinuedContinued

ICEICE

[HC[HC22HH33OO22] ] [C [C22HH33OO22--] ]

[H[H++] ]

Initial (mol/L)Initial (mol/L) .200 .200 0 0 0 0

Change (mol/L)Change (mol/L) - x - x + x + x + x + x

Equil. (mol/L)Equil. (mol/L) .200 - x .200 - x x x x x

Common Ion Effect CalculationsCommon Ion Effect CalculationsContinuedContinued

[HC[HC22HH33OO22] ] [C [C22HH33OO22--] ]

[H[H++] ]

Initial (mol/L)Initial (mol/L) . 200 . 200 .500.500 0 0

Change (mol/L)Change (mol/L) - x - x + x + x + + xx

Equil. (mol/L)Equil. (mol/L) .200 - x .200 - x .500+x.500+x x x

A Buffered SolutionA Buffered Solution

. . . resists change in its pH when either H. . . resists change in its pH when either H++ or OH or OH are added.are added.

1.0 L of 0.50 M H1.0 L of 0.50 M H33CCOOH CCOOH

+ 0.50 M H+ 0.50 M H33CCOONaCCOONa

pH = 4.74pH = 4.74

Adding 0.010 mol Adding 0.010 mol solid NaOH solid NaOH raises the raises the pHpH of of the solution to the solution to 4.764.76, a very minor change., a very minor change.

Preparation of Buffered SolutionsPreparation of Buffered Solutions

Buffered solution can be made from:Buffered solution can be made from:

1.1. a weak acid and its salt (e.g. HCa weak acid and its salt (e.g. HC22HH33OO22 & &

NaCNaC22HH33OO22).).

2. a weak base and its salt (e.g. NH2. a weak base and its salt (e.g. NH33 & NH & NH44Cl).Cl).

Other examples of buffered pairs are:Other examples of buffered pairs are:

HH22COCO3 3 & NaHCO& NaHCO33 HH33POPO44 & NaH & NaH22POPO44

Buffer CalculationsBuffer CalculationsA buffered solution contains 0.50 M acetic A buffered solution contains 0.50 M acetic

acid and 0.50 M sodium acetate. acid and 0.50 M sodium acetate. Calculate the pH of this solution. Ka= Calculate the pH of this solution. Ka= (1.8 x 10(1.8 x 10-5-5) )

Henderson-Hasselbalch EquationHenderson-Hasselbalch Equation

A buffered solution contains 0.50 M acetic A buffered solution contains 0.50 M acetic acid and 0.50 M sodium acetate. Calculate acid and 0.50 M sodium acetate. Calculate the pH of this solution. Ka= the pH of this solution. Ka= (1.8 x 10(1.8 x 10-5-5) )

pH = pKpH = pKaa + log([A + log([A--]/[HA])]/[HA])

pH= -log (1.8 x 10pH= -log (1.8 x 10-5-5) + log ([.50]/[.50])) + log ([.50]/[.50])

pH = 4.74 + 0 = 4.74pH = 4.74 + 0 = 4.74

Henderson-Hasselbalch Henderson-Hasselbalch EquationEquation

- Useful for calculating pH when the Useful for calculating pH when the [A[A]/[HA] ratios are known.]/[HA] ratios are known.

pH p log( A HA

p log( base acid

a

a

K

K

/ )

/ )

Calculate the pH of a solution that Calculate the pH of a solution that contains .250M formic acid HCOOH contains .250M formic acid HCOOH

(Ka=1.8x10-4) And .100M sodium formate (Ka=1.8x10-4) And .100M sodium formate HCOONaHCOONa

Buffer practice Buffer practice

Calculate the pH of a solution that Calculate the pH of a solution that contains .500M formic acid HCOOH contains .500M formic acid HCOOH

(Ka=1.8x10-4) And .200M sodium formate (Ka=1.8x10-4) And .200M sodium formate HCOONaHCOONa

Textbook774 23 a,b,D

24 a,b,c,d

Titration (pH) CurveTitration (pH) Curve

A plot of pH of the solution being analyzed as A plot of pH of the solution being analyzed as a function of the amount of titrant added.a function of the amount of titrant added.

Equivalence (stoichiometric) pointEquivalence (stoichiometric) point: Enough : Enough titrant has been added to titrant has been added to react exactly react exactly with the with the solution being analyzed.solution being analyzed.

15_327

01.0

Vol NaOH added (mL)

50.0

7.0

13.0

pH

100.0

Equivalencepoint

Titration curve for a strong base added to a strong acid -- the equivalence point has a pH of 7.

15_329

Vol NaOH added (mL)

25 50

3.0

9.0

12.0 Equivalencepoint

pH

Titration curve for the addition of a strong base to a weak acid-- pH is above 7.00.

15_332

0Vol 0.10 M HCl (mL)

10

2

20 30 40 50 60 70

4

6

8

10

12

EquivalencepointpH

Titration curve for the addition of a strong acid to a weak base -- the pH at equivalence is below 7.00.

Strong Acid - Strong Base TitrationStrong Acid - Strong Base Titration

• Before equivalence point, [HBefore equivalence point, [H++] is determined ] is determined by dividing number of moles of Hby dividing number of moles of H++ remaining remaining by by total volume of solutiontotal volume of solution in L. in L.

• At equivalence point, pH is 7.00.At equivalence point, pH is 7.00.

• After equivalence point [OHAfter equivalence point [OH--] is calculated ] is calculated by dividing number of moles of excess OHby dividing number of moles of excess OH- - by by total volume of solution total volume of solution in L.in L.

15_327

01.0

Vol NaOH added (mL)

50.0

7.0

13.0

pH

100.0

Equivalencepoint

Titration curve for a strong base added to a strong acid -- the equivalence point has a pH of 7.

15_330

Vol NaOH

Strong acid

pH

Weak acid

The equivalence point is defined by the stoichiometry, not the pH.

Determining the End Point in a Determining the End Point in a TitrationTitration

Two methods are used:Two methods are used:

pH meterpH meter

acid-base indicatoracid-base indicator

Acid-Base IndicatorAcid-Base Indicator. . . marks the . . . marks the end point end point of a titration by changing of a titration by changing color. The color change will be sharp, occurring color. The color change will be sharp, occurring with the addition of a single drop of titrant. with the addition of a single drop of titrant.

The The equivalence point equivalence point is not necessarily the same as is not necessarily the same as the the end pointend point..

Indicators give a visible color change will occur at a Indicators give a visible color change will occur at a pH where: pH where:

pH = pKpH = pKaa 1 1

15_3340 1 2 3 4 5 6 7 8 9 10 11 12 13

The pH ranges shown are approximate. Specific transition ranges depend on the indicator solvent chosen.

pH

Crystal Violet

Cresol Red

Thymol Blue

Erythrosin B

2,4-Dinitrophenol

Bromphenol Blue

Methyl Orange

Bromcresol Green

Methyl Red

Eriochrome* Black T

Bromcresol Purple

Alizarin

Bromthymol Blue

Phenol Red

m - Nitrophenol

o-Cresolphthalein

Phenolphthalein

Thymolphthalein

Alizarin Yellow R

* Trademark CIBA GEIGY CORP.

The useful pH ranges of several common indicators -- theuseful range is usually pKa 1. Why do some indicators have two pH ranges?

15_335AB

pH

00

Vol 0.10 M NaOH added (mL)

2

4

6

8

10

12

14

20 40 60 80 100 120

Equivalencepoint pH

00

Vol 0.10 M NaOH added (mL)

2

4

6

8

10

12

14

20 40 60 80 100 120

Equivalencepoint

Phenolphthalein

Methyl red

Phenolphthalein

Methyl red

On the left is the pH curve for the titration of a strong acid and a strong base. On the right is the curve for a weak acid and a strong base.

Solubility ProductSolubility Product

For solids dissolving to form aqueous solutions.For solids dissolving to form aqueous solutions.

BiBi22SS33((ss) ) 2Bi 2Bi3+3+((aqaq) + 3S) + 3S22((aqaq))

KKspsp = solubility product constant = solubility product constant

andand KKspsp = [Bi = [Bi3+3+]]22[S[S22]]33

Why is BiWhy is Bi22SS3(s) 3(s) not included in the solubilty not included in the solubilty

product expression?product expression?

Ba(OH)2(s)

Ag2CrO4)(s)

Ca3(PO4)2 (s)

Solubility ProductSolubility Product• Relative solubilities can be predicted by Relative solubilities can be predicted by

comparing Kcomparing Ksp sp values values onlyonly for salts that for salts that produce the same total number of ions.produce the same total number of ions.

AgI(s) Ksp = 1.5 x 10-16

CuI(s) Ksp = 5.0 x 10-12

CaSO4(s) Ksp= 6.1 x 10-5

CaSO4(s) > CuI(s) > AgI(s)

Solubility ProductSolubility ProductCuS(s) Ksp = 8.5 x 10-45

Ag2S(s) Ksp = 1.6 x 10-49

Bi2S3(s) Ksp= 1.1 x 10-73

Bi2S3(s) > Ag2S(s) > CuS(s)

Why does this order from most to least soluble appear to be out of order?

Solubility ProductSolubility Product

““Solubility” = Solubility” = ss = concentration of Bi = concentration of Bi22SS33 that that

dissolves. dissolves. The [BiThe [Bi3+3+] is 2s and the [S] is 2s and the [S22] is 3s.] is 3s.

Note:Note: KKspsp is constant (at a given temperature) is constant (at a given temperature)

ss is variable (especially with a common is variable (especially with a common ion present)ion present) Solubility product is an equilibrium

constant and has only one value for a given solid at a given temperature. Solubility is an equilibrium position.

Solubility Product CalculationsSolubility Product CalculationsCupric iodate has a measured solubility of 3.3 x 10Cupric iodate has a measured solubility of 3.3 x 10 --

33 mol/L. What is its solubility product? mol/L. What is its solubility product?

Cu(IOCu(IO33))2(s) 2(s) <---> Cu<---> Cu2+2+(aq)(aq) + 2 IO + 2 IO33

--(aq)(aq)

3.3 x 103.3 x 10-3 -3 M ---> 3.3 x 10M ---> 3.3 x 10-3 -3 M + 6.6 x 10M + 6.6 x 10-3 -3 MM

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

KKspsp = [3.3 x 10 = [3.3 x 10-3-3][6.6 x 10][6.6 x 10-3-3]]22

KKspsp = 1.4 x 10 = 1.4 x 10-7-7

Solubility from Ksp

NiCO3 Ksp 1.4x10-7

Ba3(PO4)2 Ksp=6x10-39

PhBr2

Ksp=4.6x10-6

AgI(s) Ksp = 1.5 x 10-16

CuBr(s) Ksp = 5.0 x 10-12

MgSO4(s) Ksp= 6.1 x 10-5

AgCl Ksp 1.5x10AgCl Ksp 1.5x10-10-10

AgAg22CrO4 Ksp=9.0x10CrO4 Ksp=9.0x10-12-12

Ag3PO4 Ksp 1.8 x10Ag3PO4 Ksp 1.8 x10-18-18

Common Ion EffectCommon Ion Effect

CaFCaF2(s)2(s) <---> Ca <---> Ca2+2+(aq)(aq) + 2F + 2F--

(aq)(aq)

What will be the effect on this equilibrium if What will be the effect on this equilibrium if solid sodium fluoride is added? Explain.solid sodium fluoride is added? Explain.

Equilibrium will shift to the left, due to Le Equilibrium will shift to the left, due to Le Chatelier’s Principle. Solubility product Chatelier’s Principle. Solubility product must stay constant, so the amount of Camust stay constant, so the amount of Ca2+ 2+

& F& F-- must decrease by forming solid CaF must decrease by forming solid CaF22..

Solubility Product CalculationsSolubility Product Calculations

Cu(IOCu(IO33))2(s) 2(s) <---> Cu<---> Cu2+2+(aq)(aq) + 2 IO + 2 IO33

--(aq)(aq)

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

If solid cupric iodate is dissolved in HOH; double & If solid cupric iodate is dissolved in HOH; double & square the iodate concentration. square the iodate concentration.

If mixing two solutions, one containing CuIf mixing two solutions, one containing Cu2+2+ and the and the other IOother IO33

--, then use the concentration of iodate , then use the concentration of iodate

and only square it. and only square it.

Ion Product, QIon Product, Qspsp

If 750.0 mL of 4.00 x 10If 750.0 mL of 4.00 x 10-3-3 M Ce(NO M Ce(NO33))33 is added to is added to 300.0 mL of 2.00 x 10300.0 mL of 2.00 x 10-2 -2 M KIOM KIO33, will Ce(IO, will Ce(IO33))33 precipitate?precipitate?

[Ce[Ce3+3+] = (4.00 x 10] = (4.00 x 10-3 -3 M)XM)X (750.0 mL)(750.0 mL)

(750.0 mL + 300.0 mL)(750.0 mL + 300.0 mL)

[Ce[Ce3+3+] = 2.86 x 10] = 2.86 x 10-3-3 M M

[IO[IO33--] = (2.00 x 10] = (2.00 x 10-2 -2 M) x (300.0 mL)M) x (300.0 mL)

(750.0 mL + 300.0 mL)(750.0 mL + 300.0 mL)

[IO[IO33--] = 5.71 x 10] = 5.71 x 10-3-3 M M

Ion Product, QIon Product, Qspsp

ContinuedContinued

QQspsp = [Ce = [Ce3+3+]]00[IO[IO33--]]oo

33

QQspsp = [2.86 x 10 = [2.86 x 10-3-3][5.72 x 10][5.72 x 10-3-3]]33

QQspsp = 5.32 x 10 = 5.32 x 10-10-10

QQspsp > K > Ksp sp Ce(IOCe(IO33))33 will precipitate. will precipitate.

Ksp = 1.9 x 10 -10

A 200.ml solution of 1.3x10A 200.ml solution of 1.3x10-3-3 M M AgNOAgNO3 3 is mixed with 100ml of a is mixed with 100ml of a

4.5 x104.5 x10-5-5M NaM Na22S will precipitation S will precipitation

occur?occur?Ksp=1.6x10Ksp=1.6x10-49-49

pH & SolubilitypH & Solubility

If a solid precipitate has an anion XIf a solid precipitate has an anion X - - that is an that is an effective base (HX is a weak acid), then effective base (HX is a weak acid), then the salt MX will show increased solubility the salt MX will show increased solubility in an acidic solution.in an acidic solution.

Salts containing OHSalts containing OH--, S, S2-2-, CO, CO332-2-, C, C22OO44

2-2-, & , &

CrOCrO442-2- are all soluble in acidic solution. are all soluble in acidic solution.

Limestone caves are made up of insoluble Limestone caves are made up of insoluble CaCOCaCO33, but dissolve in acidic rain water , but dissolve in acidic rain water

(H(H22COCO33).).

Solubility Product CalculationsSolubility Product Calculations

If a 0.010 M solution of sodium iodate is mixed If a 0.010 M solution of sodium iodate is mixed with a 0.0010 M cupric nitrate, will a with a 0.0010 M cupric nitrate, will a precipitate form? precipitate form?

2 NaIO2 NaIO3(aq)3(aq) + Cu(NO + Cu(NO33))2(aq) 2(aq) ---> Cu(IO---> Cu(IO33))2(s) 2(s) + 2 NaNO+ 2 NaNO3(aq)3(aq)

Cu(IOCu(IO33))2(s) 2(s) <---> Cu<---> Cu2+2+(aq)(aq) + 2 IO + 2 IO33

--(aq)(aq)

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

QQspsp = [1.0 x 10 = [1.0 x 10-3-3][1.0 x 10][1.0 x 10-2-2]]22

QQspsp = 1.0 x 10 = 1.0 x 10-7-7 QQspsp < K < Kspsp no precipitate forms. no precipitate forms.