copyright mcgraw-hill 20091 chapter 17 acid-base equilibria and solubility equilibria insert picture...

Copyright McGraw-Hill 2009 1

Chapter 17

Acid-Base Equilibria and

Solubility Equilibria

Insert picture fromFirst page of chapter


17.1 The Common Ion Effect

• When a compound containing an ion in common with an already dissolved substance is added to a solution at equilibrium, the equilibrium shifts to the left.

• This phenomenon is known as the common ion effect.

• Produced by the addition of a second solute.


Example: 1.0 L of 0.10 M solution of CH3COOH

Add 0.050 mol of CH3COONa

Effect


Effect on equilibrium calculations

522

3

3a 1.8x10

0.10

x

x 0.10

x

COOH][CH

]][HCOO[CH

MM

K

Without the common ion

31.34x10][H 2.871.34x10pH 3 )log(

5a 1.8x10 K

0.10 M

0.10 M − x

− x +x

x x

+x


Effect on equilibrium calculations

5

3

3a 1.8x10

0.10

M) (x)(0.050

x 0.10

x) M (x)(0.050

COOH][CH

]][HCOO[CH

MMK

With the common ion

510 x 3.6][H 4.4410 x 63pH 5 ).log(

4.442.87

5a 1.8x10 K

0.10 M

0.10 M − x

− x + x

x

+ x

0.050 M + x

0.050 M


17.2 Buffer Solutions• Buffer

– Solution containing a weak acid and its conjugate base

– Solution containing a weak base and its conjugate acid

– Resist changes in pH • on addition of small amount of acid• on addition of small amount of base

– Calculation of the pH of a buffer pH is a common ion problem


• Henderson-Hasselbalch equation• Quantitative equation based on the Ka expression• Expressed using -logs

[HA]

]][A[Ha

K

[HA]

][AppH a

logK

acid] [weak

base] [conjugateppH a log K


[HA]

][AppH a

logK

5a 1.8x10 K

4.74p a K

] [0.100

] [0.1004.74pH

M

Mlog

4.74pH

0.100 L of solution


Add 0.001 mol of HCl, a strong acid.

Beforereaction: 0.001 mol 0.010 mol 0.010 mol

After reaction: 0 mol 0.009 mol 0.011 mol

H (aq)(aq) + CH3COO CH3COOH (aq)

] [0.011

] [0.0094.74pH

M

Mlog

4.65pH

0.100 L of solution

mol 0100L 0.100 x L

mol 0.100 .


Beforereaction: 0.001 mol 0.010 mol 0.010 mol

After reaction: 0 mol 0.009 mol 0.011 mol

] [0.009

] [0.0114.74pH

M

Mlog

4.83pH

Add 0.001 mol of NaOH, a strong base.

(aq)CH3COOH +(aq)OH CH3COO (aq)H2O (l) +

mol 0100L 0.100 x L

mol 0.100 .

0.100 L of solution


Without a buffer, the pH changes drastically.water

Add 0.001 mol H+ Add 0.001 mol OH-

L 0.10

mol 0.001pH log L 0.10

mol 0.001pOH log

2.00pH 2.00pOH

12.00pH


• Preparing a Buffer with a Specific pH– Concentration condition

– Effective range

0.1acid] [weak

base] [conjugate10

1 ppH a K


Select an appropriate acid, and describe how you would prepare a buffer with pH 4.5.


possible acids

Using C6H5COOH

COOH]H[C

]COOH[C4.1954

56

-56log.

COOH]H[C

]COOH[C2.04

56

-56


Dissolve 2.04 mol of C6H5COONa and 1.00 mol of C6H5COOH in enough water to form 1.00 L of solution.

COOH]H[C

]COOH[C2.04

56

-56

If the solubility of the substances does not permit these amounts to dissolve, reduce the amounts but maintain the same ratio.

Dissolve 0.408 mol of C6H5COONa and 0.20 mol of C6H5COOH in enough water to form 1.00 L of solution.


17.3 Acid-Base Titrations

• Titration – addition of a solution of accurately known concentration to a solution of unknown concentration until the reaction is complete.– Standard solution - one of known concentration– Equivalence point – the point when

stochiometrically equivalent amounts have been added

– Endpoint – the point in the laboratory when the titration is stopped

– Titrant – the solution that is place in the buret


• Types of titration systems to be considered

– Strong acid – strong base

– Weak acid – strong base

– Strong acid – weak base


• Typical points to measure pH during a titration– Initial pH– Between initial pH and the equivalence point– Equivalence point– After the equivalence point

• Titration curve – a graph of pH as a function of volume of titrant added


• Strong acid –strong base titration• Net ionic equation

• pH at points in the titration is determined by stoichiometry and the excess reagent.

• Titration of 0.100 M NaOH (buret) and 25.0 mL 0.100 M HCl (pH being monitored)

H OH H2O +


1.00 (0.100pH )log M


mmol 2.50ml

HCl mmol 0.100 xml 25.00HCl mmol initial

1.364) (0.0429logpH M

mmol 1.00ml

NaOH mmol 0.100 xml 10.00NaOH mmol initial

H mmol 1.50mmol 1.00 mmol 2.500HCl mmol initial

mmol 0.0429mL 35.0

mmol 1.50 ]H [


M 10 x 1.00][OH ]H [ 7

7.00) 10 x 1.00log(pH 7 M


1.777 (0.0167pOH )log M

mmol 3.50ml

NaOH mmol 0.100 xml 35.00NaOH mmol added

OH mmol 1.00mmol 2.50- mmol 3.50NaOH mmol remaining

mmol 0.0167mL 60.0

mmol 1.00 ]OH [

12.221.777 14.00pH


• Weak acid –strong base titration– More complex equations govern pH due to

equilibrium reactions– pH at points in the titration are determined by

a specific equilibrium reaction. – Titration of 0.100 M NaOH (buret) and 25.0

mL 0.100 M CH3COOH (pH being monitored)

• Overall reaction

• At equivalence point


522

3

3a 1.8x10

0.10

x

x 0.10

x

COOH][CH

]][HCOO[CH

MM

K

M 310 x 1.34][H

2.873pH


COOH][CH

]COO[CHppH

3

3a

logK

4.56mmol 1.5

mmol 1.04.74pH log


mmol 0.050mL 50.0

mmol 2.50 ]COOCH [ 3

10-14

b 5.6x1010 x 1.8

10 x 1.00 5

K


1022

3

3b 5.6x10

0.050

x

x 0.050

x

]COO[CH

COOH]][CH[OH

MM

K

M 610 x 5.3][OH

5.28pOH

7285.2814.00pH .


1.777 (0.0167pOH )log M

mmol 3.50ml

NaOH mmol 0.100 xml 35.00NaOH mmol added

OH mmol 1.00mmol 2.50 mmol 3.50NaOH mmol remaining

mmol 0.0167mL 60.0

mmol 1.00 ]OH [

12.221.777 - 14.00pH

Identical to strong acid-strong base titration


• Strong acid –weak base titration– Similar calculations to Weak acid-strong base

titration– pH at points in the titration are determined by

a specific equilibrium reaction. – Titration of 0.100 M HCl (buret) and 25.0 mL

0.100 M NH3 (pH being monitored)

• Overall reaction

• At equivalence point


522

3

4b 1.8x10

0.10

x

x 0.10

x

][NH

]][OH[NH

MM

K

M 310 x 1.3][OH

2.89pOH

Initial pH

11.112.8914.00pH


mmol 0.050mL 50.0

mmol 2.50 ]NH [ 4

1014

a 5.6x1010 x 1.8

10 x 1.00

5

K

Equivalence point pH

1022

4

3a 5.6x10

0.050

x

x 0.050

x

][NH

]][NH[H

MM

K

M 610 x 5.3][H 5.28pH


• Acid-base indicator– A weak organic acid or base for which the

ionized and un-ionized forms are different colors.

– pH range over which the indicator changes color depends on the magnitude of the Ka or Kb

– Use to signal endpoints during a titration– The pH at the equivalence point must be

within the pH range where the indicator changes color.


An Indicator Derived from Red Cabbage

pH increases


17.4 Solubility Equilibria• Solubility Product Expression and Ksp

– Ksp is the solubility product constant

– Set up like other equilibrium expression– General example;

MXn(s) M⇄ n+(aq) + nX(aq)

– Solids and liquids are not included in equilibrium expressions

nnK ]][X[M -sp


• Molar solubility (s)– number of moles of solute in 1 L of a saturated solution (mol/L)

• Solubility - number of grams of solute in 1 L of a saturated solution (g/L)

• Ksp can be used to determine molar solubility (and solubility)– Handle as an equilibrium problem– Use an equilibrium table

• Molar solubility can be used to determine the value of the Ksp.


Calculate the solubility of SnS in g/L at 25°C.


SnS(s) Sn2+(aq) + S2-(aq)

SnS(s) Sn2+(aq) + S2-(aq)

s

+s

s

+s

2622sp 10 x 1.0]][S[Sn K

2sss ))((10 x 1.0 26

Ms 10 x 1.0 13

L

g10 x 1.5

mol

150.77gx

L

mol 10 x 1.0solubility

1113


The solubility of lead(II) chromate (PbCrO4)

is 4.5 x 105 g/L. Calculate the solubility

product (Ksp) of lead(II) chromate.


Pb2+(aq) + CrO42-(aq)PbCrO4(s)

Ms 75

10 x 1.4g 323.2

mol x

L

g10 x 4.5

]][CrO[Pb 24

22sp

sK

1477sp 10 x 2.0]10 x ][1.410 x [1.4 MMK


• Precipitation can be predicted– Compare the reaction quotient (Q) to the Ksp

where “i” designates the initial concentration

– Q < Ksp no precipitate will form

– Q > Ksp precipitate will form until Q = Ksp

nii

nQ ][X][M

MXn(s) M⇄ n+(aq) + nX(aq)


17.5 Factors Affecting Solubility• The Common Effect – an example of

LeChâtelier’s principle– The presence of a second salt that produces

an ion common to a solubility equilibrium will reduce solubility.

• Example: AgCl in a solution of AgNO3

– The concentration of a product ion is increased forcing the solubility reaction toward reactant, the solid.

• Example: Ag+ from the AgNO3, reverses the solubility reaction.


Calculate the molar solubility of BaSO4 in

0.0010 M Na2SO4.


s s

BaSO4(s) Ba2+(aq) + SO42-(aq)

BaSO4(s) Ba2+(aq) + SO42-(aq)

0.0010 M

0.0010 M + s

+s +s

s

1024

2sp 10 x 1.1 ]][SO[Ba K

1010 x 1.1) (0.00100)() )(0.00100( MssMs

Ms 710 x 1.1


Key Points

• The Common Ion Effect

• Buffer Solutions– pH of a buffer– Preparation of a buffer

• Acid-Base Titrations– Strong acid-strong base– Weak acid-strong base– Strong acid-weak base


– Acid-base indicators

• Solubility Equilibria– Solubility product expression

– Ksp

– Calculation of Ksp

– Calculation of solubility– Predicting precipitation reactions

• Comparison of Q and Ksp

• Factors Affecting Solubility– Common ion effect


• pH• Complex ion formation

–Complex ions–Formation constants

• Separation of Ions Using Solubility Differences– Fractional precipitation– Qualitative analysis of metal ions

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