1 chapter seventeen additional aspects of aqueous equilibria

11

Chapter SeventeenChapter SeventeenAdditional Aspects of Additional Aspects of Aqueous EquilibriaAqueous Equilibria

22

The Common Ion Effect and The Common Ion Effect and Buffer SolutionsBuffer Solutions

Common ion effect - solutions in which the same Common ion effect - solutions in which the same ion is produced by two different compoundsion is produced by two different compounds

Buffer solutions - resist changes in pH when Buffer solutions - resist changes in pH when acids or bases are added to themacids or bases are added to them– due to common ion effectdue to common ion effect

Two common kinds of buffer solutionsTwo common kinds of buffer solutions1 solutions of a weak acid plus a soluble ionic salt of the solutions of a weak acid plus a soluble ionic salt of the

weak acidweak acid2 solutions of a weak base plus a soluble ionic salt of solutions of a weak base plus a soluble ionic salt of

the weak basethe weak base

33


Weak Acids plus Salts of Weak AcidsWeak Acids plus Salts of Weak Acids– acetic acid CHacetic acid CH33COOHCOOH

– sodium acetate NaCHsodium acetate NaCH33COOCOO

reacts with base

CH COOH + H O H O CH COO

Na CH COO Na CH COO

reacts with acid

3 2 3+

3-

3 3-

100%

44


Example: Calculate the concentration of HExample: Calculate the concentration of H++and and the pH of a solution that is 0.15 the pH of a solution that is 0.15 MM in acetic acid in acetic acid and 0.15 and 0.15 MM in sodium acetate. in sodium acetate.

55


Example: Calculate the concentration of HExample: Calculate the concentration of H++and and the pH of a solution that is 0.15 the pH of a solution that is 0.15 MM in acetic acid in acetic acid and 0.15 and 0.15 MM in sodium acetate. in sodium acetate.

MMM

xMxMMx

15.0 15.0 15.0

COOCHNaCOONaCH

)15.0(

COOCHHCOOHCH

3100%

3

-33

66


Substitute these quantities into the ionization Substitute these quantities into the ionization expression.expression.

x

xx

15.0

15.0108.1

COOHCH

COOCHHK 5

3

-3

+

a

77


Apply the simplifying assumptionApply the simplifying assumption

74.4pH

H108.1

108.10.15

0.15

gives sassumption theseMaking

15.015.0 and 15.015.0

5

5

Mx

x

xx

88


Compare the acidity of a pure acetic acid Compare the acidity of a pure acetic acid solution and the buffer we just described.solution and the buffer we just described.

99


Compare the acidity of a pure acetic acid solution and the buffer we just described.Compare the acidity of a pure acetic acid solution and the buffer we just described.

[H[H++] is ] is 8989 times greater in pure acetic acid than in buffer solution. times greater in pure acetic acid than in buffer solution.

Solution [H+] pH0.15 M CH3COOH 1.6 x 10-3 M 2.800.15 M CH3COOH

& 0.15 M NaCH3COO

1.8 x 10-5 M 4.74

1010


General expression for the ionization of a weak General expression for the ionization of a weak monoprotic acid.monoprotic acid.

HA H A+

1111


Its ionization constant expression isIts ionization constant expression is

KH A

HAa

1212


Solve the expression for [HSolve the expression for [H++]]

H KHA

A

acid

salta

1313


Making the assumption that the concentrations Making the assumption that the concentrations of the weak acid and the salt are reasonable.of the weak acid and the salt are reasonable.

The expression reduces toThe expression reduces to

H Kacidsalta

1414


The above relationship is valid for buffers The above relationship is valid for buffers containing a weak monoprotic acid and a soluble, containing a weak monoprotic acid and a soluble, ionic salt.ionic salt.

The relationship changes if the salt’s cation is not The relationship changes if the salt’s cation is not univalent tounivalent to

H Kacid

n salt

where n = charge on cation

a

1515


Simple rearrangement of this equation and Simple rearrangement of this equation and application of algebra yields the application of algebra yields the

Henderson-Hasselbach equationHenderson-Hasselbach equation

log log log

log log log

log

H Kacidsalt

multiply by -1

H Ksaltacid

pH pKsaltacid

a

a

a

1616

Weak Bases plus Salts of Weak Bases plus Salts of Weak BasesWeak Bases

Buffers that contain a weak base plus the salt of Buffers that contain a weak base plus the salt of a weak base - for example - ammonia plus a weak base - for example - ammonia plus ammonium nitrate.ammonium nitrate.

1717


Buffers that contain a weak base plus the salt of Buffers that contain a weak base plus the salt of a weak base - for example - ammonia plus a weak base - for example - ammonia plus ammonium nitrate.ammonium nitrate.

NH H O NH OH

NH NO NH NO

NH

NH

3 2 4+

4 3 4+

3

4+

3

100%

518 10KOH

b .

1818


Example: Calculate the concentration of Example: Calculate the concentration of OH- and the pH of the solution that is 0.15 OH- and the pH of the solution that is 0.15 MM in aqueous ammonia, NH in aqueous ammonia, NH33, and 0.30 , and 0.30 MM in in

ammonium nitrate, NHammonium nitrate, NH44NONO33..

MMM

xMxMMx

30.0 30.0 30.0

NONHNONH

)15.0(

OHNH OH NH

3+4

%10034

+423

1919


Substitute these values into the ionization Substitute these values into the ionization expression for ammonia and solve algebraically.expression for ammonia and solve algebraically.

8.95pH and 5.05pOH

OH100.9

108.115.0

30.0K

assumption gsimplifyin apply the

108.115.0

30.0K

108.1NH

OHNHK

6

5b

5b

5

3

4b

Mx

x

x

xx

2020


Let’s compare the aqueous ammonia Let’s compare the aqueous ammonia concentration to that of the buffer described concentration to that of the buffer described above.above.

2121


Let’s compare the aqueous ammonia concentration to that of the buffer described above.Let’s compare the aqueous ammonia concentration to that of the buffer described above.

The [OHThe [OH--] in aqueous ammonia is ] in aqueous ammonia is 180180 times greater than in the buffer.times greater than in the buffer.Solution [OH-] pH

0.15 M NH3 1.6 x 10-3 M 11.200.15 M NH3

& 0.15 M NH4NO3

9.0 x 10-6 M 8.95

2222


Derive a general relationship for buffer solutions Derive a general relationship for buffer solutions that contain a weak base plus a salt of a weak that contain a weak base plus a salt of a weak base.base.– Ionization equationIonization equation

:B H O BH OH

where B represents a weak base2

2323


Ionization expressionIonization expression– general formgeneral form

KBH OH

Bb

2424


OH Kbasesaltb

OH Kbase

n salt

where n = charge on anion

b

2525


Simple rearrangement of this equation and Simple rearrangement of this equation and application of algebra yields the application of algebra yields the

Henderson-Hasselbach equationHenderson-Hasselbach equation

log log log

log log log

log

OH Kbasesalt

multiply by -1

OH Ksaltbase

pOH pKsaltbase

b

a

b

2626

Buffering ActionBuffering Action

Buffer solutions resist changes in pH.Buffer solutions resist changes in pH.

Example: If 0.020 mole of HCl is added to 1.00 Example: If 0.020 mole of HCl is added to 1.00 liter of solution that is 0.100 liter of solution that is 0.100 MM in aqueous in aqueous ammonia and 0.200 ammonia and 0.200 MM in ammonium chloride, in ammonium chloride, how much does the pH change? Assume no how much does the pH change? Assume no volume change due to addition of the gaseous volume change due to addition of the gaseous HCl.HCl.

1 Calculate the pH of the original buffer solutionCalculate the pH of the original buffer solution

2727


OHNH

NH Cl

OH

OH

pOH 5.05 pH 8.95

- 3

4

-

-

K

MM

M

b

18 100100 20

9 0 10

5

6

...

.

2828


2 Now we calculate the concentration of all Now we calculate the concentration of all species after the addition of HCl.species after the addition of HCl.– HCl will react with some of the ammoniaHCl will react with some of the ammonia

HCl NH NH Cl

Initial 0.020 mol 0.100 mol 0.200 mol

Change - 0.020 mol - 0.020 mol + 0.020 mol

After rxn. 0 mol 0.080 mol 0.220 mol

mol1.0 L

mol1.0 L

3 4

NH

NH Cl

3

4

M M

M M

0 0800 080

0 2200 220

..

..

2929


3 Now that we have the concentrations of our Now that we have the concentrations of our salt and base, we can calculate the pH.salt and base, we can calculate the pH.

OH KNH

NH Cl

OH

OH

pOH 5.19 pH 8.81

b3

4

18 100 0800 220

6 5 10

5

6

...

.

MM

M

3030


4 Calculate the change in pH.Calculate the change in pH.

pH = 8.81- 8.95 = -0.14

3131

Preparation of Buffer SolutionsPreparation of Buffer Solutions

Example: Calculate the concentration of HExample: Calculate the concentration of H++ and the pH of and the pH of the solution prepared by mixing 200 mL of 0.150 the solution prepared by mixing 200 mL of 0.150 MM acetic acetic acid and 100 mL of 0.100 acid and 100 mL of 0.100 MM sodium hydroxide solutions. sodium hydroxide solutions.

Determine the amounts of acetic acid and sodium Determine the amounts of acetic acid and sodium hydroxide (before reaction)hydroxide (before reaction)

3232


Example: Calculate the concentration of HExample: Calculate the concentration of H++ and the pH of and the pH of the solution prepared by mixing 200 mL of 0.150 the solution prepared by mixing 200 mL of 0.150 MM acetic acetic acid and 100 mL of 0.100 acid and 100 mL of 0.100 MM sodium hydroxide solutions. sodium hydroxide solutions.

Determine the amounts of acetic acid and sodium Determine the amounts of acetic acid and sodium hydroxide (before reaction)hydroxide (before reaction)

? mmol CH COOH = 200 mL0.15 mmol

L mmol CH COOH

? mmol NaOH = 100 mL0.100 mmol

L mmol NaOH

3 3

30 0

10 0

.

.

3333


Sodium hydroxide and acetic acid react in a 1:1 Sodium hydroxide and acetic acid react in a 1:1 mole ratio.mole ratio.

NaOH + CH COOH Na CH COO + H O

Initial 10.0 mmol 30.0 mmol

Change -10.0 mmol -10.0 mmol +10.0 mmol

After rxn. 0 20.0 mmol 10.0 mmol

3 3 2

3434


After the two solutions are mixed, the total After the two solutions are mixed, the total volume is 300 mL (100 + 200), and the volume is 300 mL (100 + 200), and the concentrations are:concentrations are:

M M

M M

CH COOH 3

NaCH COO 3

3

3

mLmL

CH COOH

mLmL

NaCH COO

20 0300

0 0667

10 0300

0 0333

..

..

3535


Substitution into the ionization constant Substitution into the ionization constant expression (or Henderson-Hasselbach equation) expression (or Henderson-Hasselbach equation) givesgives

KH CH COO

CH COOH

H

pH

a3

3

18 10

18 10 0 0667

0 03333 6 10

4 44

5

55

.

. .

..

.

M

3636

Preparation of Buffer SolutionsPreparation of Buffer SolutionsFor biochemical situations, it is sometimes important to For biochemical situations, it is sometimes important to prepare a buffer solution of a given pH.prepare a buffer solution of a given pH.

Example: Calculate the number of moles of solid ammonium Example: Calculate the number of moles of solid ammonium chloride, NHchloride, NH44Cl, that must be used to prepare 1.00 L of a Cl, that must be used to prepare 1.00 L of a

buffer solution that is 0.10 buffer solution that is 0.10 MM in aqueous ammonia, and that in aqueous ammonia, and that has a pH of 9.15.has a pH of 9.15.– Because pH=9.15Because pH=9.15

3737

Preparation of Buffer SolutionsPreparation of Buffer SolutionsFor biochemical situations, it is sometimes important to For biochemical situations, it is sometimes important to prepare a buffer solution of a given pH.prepare a buffer solution of a given pH.

Example:Calculate the number of moles of solid ammonium Example:Calculate the number of moles of solid ammonium chloride, NHchloride, NH44Cl, that must be used to prepare 1.00 L of a Cl, that must be used to prepare 1.00 L of a

buffer solution that is 0.10 buffer solution that is 0.10 MM in aqueous ammonia, and that in aqueous ammonia, and that has a pH of 9.15.has a pH of 9.15.– Because pH=9.15Because pH=9.15

pOH = 14.00 - 9.15 = 4.85

OH- 10 14 104 85 5. . M

3838


Appropriate equations and equilibria Appropriate equations and equilibria representations are:representations are:

xMxMxM

MMM

Cl NH ClNH

104.1 104.1 104.110.0

OH NH OH + NH

44

555

-423

3939

Preparation of Buffer SolutionsPreparation of Buffer SolutionsSubstitute into the ionization constant Substitute into the ionization constant expression (or Henderson-Hasselbach expression (or Henderson-Hasselbach equation) for aqueous ammoniaequation) for aqueous ammonia

g/L 9.6mol

g 53

L

mol 13.0

L

ClNH g ?

ClNH=ClNH 13.0

108.110.0

104.1K

assumption gsimplifyin apply the

104.110.0

104.1104.1K

108.1NH

OHNHK

4

original44

55

b

5

55

b

5

3

4b

Mx

x

x

4040

Acid-Base IndicatorsAcid-Base Indicators

Equivalence point - point at which chemically Equivalence point - point at which chemically equivalent amounts of acid and base have reactedequivalent amounts of acid and base have reacted

End point - point at which chemical indicator End point - point at which chemical indicator changes colorchanges color

HIn H In

Color 1 Color 2

4141


Equilibrium constant expression for an indicator Equilibrium constant expression for an indicator would bewould be

KH In

HIna

4242


Rearrange this expression to get a feeling for Rearrange this expression to get a feeling for range over which indicator changes color.range over which indicator changes color.

In

HInK

H

-a

4343

Acid-Base IndicatorsAcid-Base IndicatorsSome Acid-Base IndicatorsSome Acid-Base Indicators

IndicatorColor in

acidic range pH rangeColor in

basic rangeMethylviolet

yellow 0-2 purple

Methylorange

pink 3.1-4.4 yellow

Litmus red 4.7-8.2 bluePhenol-

phthaleincolorless 8.3-10.0 red

4444

Strong Acid/Strong Base Strong Acid/Strong Base Titration CurvesTitration Curves

Plot pH vs. Volume of acid or base added in titration.Plot pH vs. Volume of acid or base added in titration.

Consider the titration of 100.0 mL of 0.100 Consider the titration of 100.0 mL of 0.100 MM perchloric acid with 0.100 perchloric acid with 0.100 MM potassium hydroxide. potassium hydroxide.– Plot pH vs. mL of KOH addedPlot pH vs. mL of KOH added– 1:1 mole ratio1:1 mole ratio

OHKClOKOHHClO 244

4545


Before titration starts the pH of the HClOBefore titration starts the pH of the HClO44

solution is solution is 1.00.1.00.– Remember perchloric acid is a strong acidRemember perchloric acid is a strong acid

00.1log(0.100)pH

100.0H

10000.100 100.0

ClOHHClO 4%100

4

M

M.M M

4646


After 20.0 mL of 0.100 After 20.0 mL of 0.100 MM KOH has been added KOH has been added the pH is the pH is 1.171.17..

1.17pH 067.0H

067.0mL 120

HClO mmol 8.0

rxn.

mmol 2.0 mmol 0.0 mmol 8.0 After

mmol 2.0 mmol 2.0- mmol 2.0- :Change

mmol 2.0 mmol 10.0 :Start

OHKClO KOH HClO

4HClO

244

4

M

MM

4747



1.48pH 033.0H

033.0mL 150

HClO mmol 5.0

rxn.




OHKClO KOH HClO

4HClO

244

4

M

MM

4848



2.28pH 0053.0H

0053.0mL 190

HClO mmol 1.0

rxn.




OHKClO KOH HClO

4HClO

244

4

M

MM

4949



7.00pH

neutral baseor acid No

rxn.




OHKClO KOH HClO 244

5050


We have calculated only a few points on the titration We have calculated only a few points on the titration curve. Similar calculations for remainder of titration curve. Similar calculations for remainder of titration show clearly the shape of the titration curve.show clearly the shape of the titration curve.

5151

Weak Acid/Strong Base Weak Acid/Strong Base Titration CurvesTitration Curves

Consider the titration of 100.0 mL of 0.100 Consider the titration of 100.0 mL of 0.100 MM acetic acid, CHacetic acid, CH33 COOH, with 0.100 COOH, with 0.100 MM KOH. KOH.– react in a 1:1 mole ratioreact in a 1:1 mole ratio

1 mol 1mol 1mol

CH COOH + KOH K CH COO + H O

1mmol 1mmol 1mmol3

+3

-2

5252


Before the equivalence point is reached , both Before the equivalence point is reached , both CHCH33COOH and K CHCOOH and K CH33COO are present in solution COO are present in solution

forming a buffer.forming a buffer.

weak acid / salt of weak acid

form a buffer solution

5353


1 Determine the pH of the acetic acid solution before Determine the pH of the acetic acid solution before titration is begun.titration is begun.

CH COOH CH COO H

3 3

-

010. x M xM xM

5454



CH COOH CH COO H

KH CH COO

CH COOH

K

apply assumption

3 3-

a

+3

-

3

a

010

18 10

01018 10

5

5

.

.

..

x M xM xM

x xx

5555



CH COOH CH COO H

KH CH COO

CH COOH

K

apply assumption

= 1.3 10

1.3 10 pH

3 3-

a

+3

-

3

a

2 -3

-3

010

18 10

01018 10

18 10

2 89

5

5

6

.

.

..

.

.

x M xM xM

x xx

x x

H

5656


2 After 20.0 mL of KOH solution has been added, the After 20.0 mL of KOH solution has been added, the pH is:pH is:

KOH + CH COOH K CH COO H O

Initial: 2.00 mmol 10.0 mmol

Chg. due to rxn:-2.00 mmol - 2.00 mmol + 2.00 mmol

After rxn: 0.00 mmol 8.00 mmol 2.00 mmol

3+

3-

2

5757


2 After 20.0 mL of KOH solution has been added, the After 20.0 mL of KOH solution has been added, the pH is:pH is: KOH + CH COOH K CH COO H O


Chg. due to rxn:-2.00 mmol - 2.00 mmol + 2.00 mmol


8.0 mmol120 mL

2.0 mmol120 mL

3+

3-

2

CH COOH

CH COO

3

3-

M M

M M

0 067

0 017

.

.

5858



KH CH COO

CH COOH

HCH COOH

CH COO

H

pH

a3

3

3

3

18 10

18 10

18 100 0670 017

7 1 10

4 15

5

5

5 5

.

.

...

.

.

M

5959



Similarly for all other cases before the equivalence Similarly for all other cases before the equivalence point is reached.point is reached.

KH CH COO

CH COOH

HCH COOH

CH COO

H

pH

a3

3

3

3

18 10

18 10

18 100 0670 017

7 1 10

4 15

5

5

5 5

.

.

...

.

.

M

6060


At the equivalence point, the solution is 0.500 At the equivalence point, the solution is 0.500 M M in KCH in KCH33COO, the salt of a strong base and a COO, the salt of a strong base and a

weak acid which hydrolyzes to give a basic solution.weak acid which hydrolyzes to give a basic solution.– Both processes make the solution basicBoth processes make the solution basic

Cannot have a pH=7.00 at equivalence point.Cannot have a pH=7.00 at equivalence point.

Let us calculate the pH at the equivalence point.Let us calculate the pH at the equivalence point.

6161


1 Set up the equilibrium reaction:Set up the equilibrium reaction:



Chg. due to rxn:-10.0 mmol -10.0 mmol +10.0 mmol


3+

3-

2

6262


2 Determine the concentration of the salt in solution.Determine the concentration of the salt in solution.

M =10.0 mmol

200 mL

M CH COO

KCH COO

KCH COO 3

3

3

0 0500

0 0500 0 0500

.

. .

M

M M

6363


3 Perform a hydrolysis calculation for the potassium Perform a hydrolysis calculation for the potassium acetate in solution.acetate in solution.

CH COO H O CH COOH OH

K =CH COOH OH

CH COO

K =

OH

pOH 5.28 pH 8.72

3 2 3

b3

-

3

b

0 0500

5 6 10

0 0500 0 05005 6 10

2 8 10 17 10

11

211

2 12 6

.

.

. ..

. .

x M xM xM

x xx

x

x x

6464


After the equivalence point is reached, the pH is After the equivalence point is reached, the pH is determined by the excess KOH - as in Strong determined by the excess KOH - as in Strong Acid/Strong Base example.Acid/Strong Base example.





3+

3-

2

6565


After the equivalence point is reached, the pH is After the equivalence point is reached, the pH is determined by the excess KOH - as in Strong determined by the excess KOH - as in Strong Acid/Strong Base example.Acid/Strong Base example.





mmol mmol

OH

pOH 2.32 and pH = 11.68

3+

3-

2

KOH KOH

M M

M

10210

4 8 10

4 8 10

3

3

..

.

6666


We have calculated only a few points on the titration We have calculated only a few points on the titration curve. Similar calculations for remainder of titration curve. Similar calculations for remainder of titration show clearly the shape of the titration curve.show clearly the shape of the titration curve.

6767

Strong Acid/Weak BaseStrong Acid/Weak BaseTitration CurvesTitration Curves

Titration curves for Strong Acid/Weak Bases look Titration curves for Strong Acid/Weak Bases look similar to Strong Base/Weak Acid but they are similar to Strong Base/Weak Acid but they are inverted.inverted.

6868

Weak Acid/Weak BaseWeak Acid/Weak BaseTitration CurvesTitration Curves

Titration curves have very short vertical sections.Titration curves have very short vertical sections.

Solution is buffered both before and after the equivalence Solution is buffered both before and after the equivalence point.point.

Visual indicators cannot be used.Visual indicators cannot be used.

6969

Solubility Product ConstantsSolubility Product ConstantsSilver chloride, AgCl,is rather insoluble in water.Silver chloride, AgCl,is rather insoluble in water.

Careful experiments show that if solid AgCl is Careful experiments show that if solid AgCl is placed in pure water and vigorously stirred, a placed in pure water and vigorously stirred, a small amount of the AgCl dissolves in the water.small amount of the AgCl dissolves in the water.

aqaq ClAgClAg

7070

Solubility Product ConstantsSolubility Product Constants

10sp 101.8ClAgK

7171


Solubility product constant for a compound is the product of Solubility product constant for a compound is the product of the concentrations of the constituent ions, each raised to the the concentrations of the constituent ions, each raised to the power that corresponds to the number of ions in one formula power that corresponds to the number of ions in one formula unit of the compound.unit of the compound.

Consider the dissolution of silver sulfide in water. Consider the dissolution of silver sulfide in water.

Ag S 2 Ag + S2+ 2-

H O+ 2-2 100%

7272


Its solubility product expression isIts solubility product expression is

K Ag Ssp

2 2 10 10 49.

7373


The dissolution of solid calcium phosphate in water is The dissolution of solid calcium phosphate in water is represented asrepresented as

Ca PO 3 Ca 2 PO3

2+43

2 s

H O2

43

100%

2

7474


Its solubility product constant expression isIts solubility product constant expression is

K Ca POsp2 3

43 2

10 10 25.

7575


In general, the dissolution of a slightly soluble compound and In general, the dissolution of a slightly soluble compound and its solubility product expression are represented asits solubility product expression are represented as

KKspsp has a fixed value for a given system at a given temperature has a fixed value for a given system at a given temperature

M Y r M s Y

K M Y

r s s

H Os r

100%

sps r r s

2

7676


The same rules apply for compounds that have The same rules apply for compounds that have more than two kinds of ions.more than two kinds of ions.

An example is calcium ammonium phosphate.An example is calcium ammonium phosphate.

CaNH PO Ca NH PO

K Ca NH PO

4 4 s2

4 43

sp2

4 43

7777

Determination of Solubility Determination of Solubility Product ConstantsProduct Constants

Example: One liter of saturated silver chloride solution Example: One liter of saturated silver chloride solution contains 0.00192 g of dissolved AgCl at 25contains 0.00192 g of dissolved AgCl at 25ooC. C. Calculate the molar solubility of, and KCalculate the molar solubility of, and Ksp sp for, AgCl.for, AgCl.

Molar solubility can be calculated from the data:Molar solubility can be calculated from the data:

? .

.

mol AgClL

g AgClL

mol AgCl143 g AgCl

mol AgClL

0 00192 1

134 10 5

7878


The equation for the dissociation of silver chloride and The equation for the dissociation of silver chloride and its solubility product expression areits solubility product expression are

AgCl Ag Cl

1.34 10 1.34 10

K Ag Cl

s

-5 -5

sp

M M

7979


Substitution into the solubility product Substitution into the solubility product expression givesexpression gives

K Ag Clsp

134 10 134 10

18 10

5 5

10

. .

.

8080

Uses of Solubility Uses of Solubility Product ConstantsProduct Constants

We can use the solubility product constant to calculate the solubility of We can use the solubility product constant to calculate the solubility of a compound at 25a compound at 25ooC.C.

Example: Calculate the molar solubility of barium sulfate, BaSOExample: Calculate the molar solubility of barium sulfate, BaSO 44, in , in pure water and the concentration of barium and sulfate ions in pure water and the concentration of barium and sulfate ions in saturated barium sulfate at 25saturated barium sulfate at 25ooC. C.

KKspsp= 1.1 x 10= 1.1 x 10-10-10..

8181


Example: Calculate the molar solubility of barium Example: Calculate the molar solubility of barium sulfate, BaSOsulfate, BaSO44, in pure water and the concentration of , in pure water and the concentration of

barium and sulfate ions in saturated barium sulfate at barium and sulfate ions in saturated barium sulfate at 2525ooC. KC. Kspsp= 1.1 x 10= 1.1 x 10-10-10..

BaSO Ba SO

Ba SO

4 s aq2+

4 aq

242

2

1011 10

xM xM xM

Ksp .

8282


Substitute into solubility product expression and Substitute into solubility product expression and solve for x, giving the ion concentrations.solve for x, giving the ion concentrations.

x x

x M

11 10

10 10

10

5

.

.

Ba SO242

8383


Now we can calculate the mass of BaSONow we can calculate the mass of BaSO44 in 1.00 L of in 1.00 L of

saturated solution.saturated solution.

?

.

g BaSOL

1.0 10 molL

234 gmol

g

45

BaSOL

4

2 3 10 3

8484

The Reaction Quotient in The Reaction Quotient in Precipitation ReactionsPrecipitation Reactions

Use solubility product constants to calculate the concentration of Use solubility product constants to calculate the concentration of ions in a solution and whether or not a precipitate will form.ions in a solution and whether or not a precipitate will form.

Example: We mix 100 mL of 0.010 M potassium sulfate, KExample: We mix 100 mL of 0.010 M potassium sulfate, K22SOSO44, ,

and 100 mL of 0.10 M lead (II) nitrate, Pb(NOand 100 mL of 0.10 M lead (II) nitrate, Pb(NO33))22 solutions. Will a solutions. Will a

precipitate form?precipitate form?

8585


Example: We mix 100 mL of 0.010 M potassium Example: We mix 100 mL of 0.010 M potassium sulfate, Ksulfate, K22SOSO44, and 100 mL of 0.10 M lead (II) nitrate, , and 100 mL of 0.10 M lead (II) nitrate,

Pb(NOPb(NO33))22 solutions. Will a precipitate form? solutions. Will a precipitate form?

K SO 2 K SO

Pb NO Pb NO

Will PbSO precipitate?

2 4H O 100% +

42

3 2H O 100% 2+

3-

4

2

2

2

8686


Calculate the QCalculate the Qspsp for PbSO for PbSO44..– Solution volumes are additive.Solution volumes are additive.– Concentrations of the important ions are:Concentrations of the important ions are:

MM

M

MM

M

Pb2+

SO 42-

2+

42-

mL 0.10200 mL

Pb

mL 0.010200 mL

SO

1000 050

1000 0050

.

.

8787


Finally, we calculate QFinally, we calculate Qspsp for PbSO for PbSO44..

Q Pb SO

K for PbSO

Q K therefore solid forms

sp2

42

sp 4

sp sp

0 050 0 0050

2 5 10

18 10

4

8

. .

.

.

8888

Fractional PrecipitationFractional PrecipitationFractional precipitation is a method of Fractional precipitation is a method of precipitating some ions from solution while precipitating some ions from solution while leaving others in solution.leaving others in solution.– Look at a solution that contains CuLook at a solution that contains Cu++, Ag, Ag++, and Au, and Au++ – We could precipitate them as chloridesWe could precipitate them as chlorides

8989

Fractional PrecipitationFractional PrecipitationFractional precipitation is a method of Fractional precipitation is a method of precipitating some ions from solution while precipitating some ions from solution while leaving others in solution.leaving others in solution.– Look at a solution that contains CuLook at a solution that contains Cu++, Ag, Ag++, and Au, and Au++ – We could precipitate them as chloridesWe could precipitate them as chlorides

13

sp

10sp

7sp

100.2ClAuKCl AuAuCl

108.1ClAgKCl AgAgCl

109.1ClCuKClCuCuCl

9090

Fractional PrecipitationFractional PrecipitationExample: If solid sodium chloride is slowly added to a solution Example: If solid sodium chloride is slowly added to a solution that is 0.010 that is 0.010 MM each in Cu each in Cu++, Ag, Ag++, and Au, and Au++ ions, which ions, which compound precipitates first? Calculate the concentration of Clcompound precipitates first? Calculate the concentration of Cl-- required to initiate precipitation of each of these metal I required to initiate precipitation of each of these metal I chlorides.chlorides.

9191


AuCl. eprecipitat torequired Cl ofion concentrat theCalculate

last. eprecipitat willCuCl reasoning, same By the

first. esprecipitatit so Ksmallest thehas AuCl

-

sp

9292


M11

1313

sp13

-

sp

100.2010.0

100.2

Au

100.2Cl

K100.2ClAu

AuCl. eprecipitat torequired Cl ofion concentrat theCalculate

last. eprecipitat willCuCl reasoning, same By the

first. esprecipitatit so Ksmallest thehas AuCl

9393

Fractional PrecipitationFractional PrecipitationRepeat the calculation for silver chloride.Repeat the calculation for silver chloride.

K Ag Cl

ClAg

sp

18 10

18 10 18 100 010

18 10

10

10 10

8

.

. ..

. M

9494

Fractional PrecipitationFractional PrecipitationFor copper (I) chloride to precipitate.For copper (I) chloride to precipitate.

K Cu Cl

ClCu

sp+

+

19 10

19 10 19 100 010

19 10

7

7 7

5

.

. ..

. M

9595

Fractional PrecipitationFractional Precipitation

We have calculated the [ClWe have calculated the [Cl--] required ] required

to precipitate AuCl, [Clto precipitate AuCl, [Cl--] >2.0 x 10] >2.0 x 10-11-11 M M

to precipitate AgCl, [Clto precipitate AgCl, [Cl--] >1.8 x 10] >1.8 x 10-8 -8 M M

to precipitate CuCl, [Clto precipitate CuCl, [Cl--] >1.9 x 10] >1.9 x 10-5 -5 M M

We can calculate the amount of AuWe can calculate the amount of Au++ precipitated precipitated before Agbefore Ag++ begins to precipitate, as well as the begins to precipitate, as well as the amounts of Auamounts of Au++ and Ag and Ag++ precipitated before Cu precipitated before Cu++ begins to precipitate.begins to precipitate.

9696

Fractional PrecipitationFractional PrecipitationExample: Calculate the percent of AuExample: Calculate the percent of Au++ ions that precipitate ions that precipitate before AgCl begins to precipitate.before AgCl begins to precipitate.

Use the [ClUse the [Cl--] from before to determine the [Au] from before to determine the [Au++] ] remaining in remaining in solution just before AgCl begins to precipitatesolution just before AgCl begins to precipitate..

9797

Fractional PrecipitationFractional PrecipitationExample: Calculate the percent of AuExample: Calculate the percent of Au++ ions that precipitate ions that precipitate before AgCl begins to precipitate.before AgCl begins to precipitate.

Use the [ClUse the [Cl--] from before to determine the [Au] from before to determine the [Au++] ] remaining in remaining in solution just before AgCl begins to precipitatesolution just before AgCl begins to precipitate..

Au Cl

AuCl

Au Au unprecipitated

2 0 10

2 0 10 2 0 10

18 10

11 10

13

13 13

8

5

.

. .

.

.

9898

Fractional PrecipitationFractional PrecipitationThe percent of AuThe percent of Au++ ions ions unprecipitated unprecipitated just just before AgCl precipitates is before AgCl precipitates is

% Au

Au

Auunprecipitated+ unprecipitated

original

unprecipitated

100%

11 100 010

100 0 00011%8.

..

9999

Fractional PrecipitationFractional PrecipitationThe percent of AuThe percent of Au++ ions ions unprecipitated unprecipitated just before AgCl precipitates isjust before AgCl precipitates is

Therefore, Therefore, 99.99989%99.99989% of the Au of the Au++ ions precipitates before AgCl begins to precipitate. ions precipitates before AgCl begins to precipitate.

% Au

Au

Auunprecipitated+ unprecipitated

original

unprecipitated

100%

11 100 010

100 0 00011%8.

..

100100

Fractional PrecipitationFractional PrecipitationSimilar calculations for the concentration of AgSimilar calculations for the concentration of Ag++ ions ions unprecipitated unprecipitated beforebefore CuCl begins to precipitate gives CuCl begins to precipitate gives

atedunprecipit Ag105.9Ag

109.1

108.1

Cl

108.1Ag

108.1ClAg

6

5

1010

10

101101

Fractional PrecipitationFractional PrecipitationThe percent of AuThe percent of Au++ ions ions unprecipitated unprecipitated just just before AgCl precipitates is before AgCl precipitates is

atedunprecipit

6

original

atedunprecipit+atedunprecipit

%095.0100010.0

105.9

%100Ag

AgAg %

102102

Fractional PrecipitationFractional PrecipitationThe percent of AuThe percent of Au++ ions ions unprecipitated unprecipitated just before AgCl precipitates isjust before AgCl precipitates is

Thus, Thus, 99.905%99.905% of the Ag of the Ag++ ions precipitates before CuCl begins to precipitate. ions precipitates before CuCl begins to precipitate.

atedunprecipit

6

original

atedunprecipit+atedunprecipit

%095.0100010.0

105.9

%100Ag

AgAg %

103103

Factors that Affect SolubilityFactors that Affect Solubility

CaFCaF22 (s) (s) Ca Ca+2+2 (aq)+ (aq)+ 2F2F-- (aq) (aq)

Addition of a Common ion (F- from NaF)Addition of a Common ion (F- from NaF)– Solubility decreasesSolubility decreases– Equilibrium shifts to leftEquilibrium shifts to left

Changes in pH (H+ reacts with F-)Changes in pH (H+ reacts with F-)– Solubility increases (with increasing pH)Solubility increases (with increasing pH)– Equilibrium shifts to rightEquilibrium shifts to right

104104


AgAg++(aq) + 2NH(aq) + 2NH33(aq) (aq) Ag(NH Ag(NH33))22++ (aq) (aq)

complex ioncomplex ion

Formation of a complex ionFormation of a complex ion– Lewis Acid base chemistryLewis Acid base chemistry

Calculate KCalculate Kf f Formation ConstantFormation Constant

105105


AgAg++(aq) + 2NH(aq) + 2NH33(aq) (aq) Ag(NH Ag(NH33))22++ (aq) (aq)

complex ioncomplex ion

AgCl(s)AgCl(s) AgAg++(aq) + Cl(aq) + Cl--(aq)(aq)

– In formation of complex ionIn formation of complex ion– Removed AgRemoved Ag++ from the equilibrium from the equilibrium – Equilibrium shifts to rightEquilibrium shifts to right– Favors dissolving AgClFavors dissolving AgCl

106106

Synthesis QuestionSynthesis Question

Bufferin is a commercially prepared Bufferin is a commercially prepared medicine that is literally a buffered medicine that is literally a buffered aspirin. How could you buffer aspirin? aspirin. How could you buffer aspirin? Hint - what is aspirin?Hint - what is aspirin?

107107


Aspirin is acetyl salicylic acid. So to buffer Aspirin is acetyl salicylic acid. So to buffer it all that would have to be added is the it all that would have to be added is the salt of acetyl salicylic acid.salt of acetyl salicylic acid.

108108

Group QuestionGroup Question

Blood is slightly basic, having a pH of 7.35 Blood is slightly basic, having a pH of 7.35 to 7.45. What chemical species causes to 7.45. What chemical species causes our blood to be basic? How does our our blood to be basic? How does our body regulate the pH of blood?body regulate the pH of blood?

109109


Most kidney stones are made of calcium Most kidney stones are made of calcium oxalate, Ca(Ooxalate, Ca(O22CCOCCO22). Patients who have ). Patients who have

their first kidney stones are given an their first kidney stones are given an extremely simple solution to stop further extremely simple solution to stop further stone formation. They are told to drink six stone formation. They are told to drink six to eight glasses of water a day. How does to eight glasses of water a day. How does this stop kidney stone formation? this stop kidney stone formation?

110110


stone. solida as than

rather solution in ions towardsright, theto

mequilibriu theshifts water more Drinking

COOCaOHCa(COO)

oxalate calciumfor

expressionproduct Solubility

2aq2

2aq2s2

111111

Group QuestionGroup Question

The cavities that we get in our teeth are a The cavities that we get in our teeth are a result of the dissolving of the material our result of the dissolving of the material our teeth are made of, calcium hydroxy apatite. teeth are made of, calcium hydroxy apatite. How does using a fluoride based toothpaste How does using a fluoride based toothpaste decrease the occurrence of cavities?decrease the occurrence of cavities?

1 chapter seventeen additional aspects of aqueous equilibria

Documents