chemistry 102(01) spring 2009

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

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?

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

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

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

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) •

• 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

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

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)

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

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

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

Calcualtion of pH of BuffersHenderson Hesselbach Equation [ACID]pH = pKa - log --------- [BASE] [BASE]pH = pKa + log --------- [ACID]

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

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

Adding an Acid or Baseto a Buffer

Buffer Systems

Titrations ofAcids and Bases

• Titration• Analyte• Titrant

analyte + titrant => products

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

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

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

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

pH and Color of Indicators

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)

• Acid-base indicators are weak acids that undergo a color change at a known pH.

phenolphthalein

pH

Indicator examples

Titration Apparatus

Buret delivering base to a flask containing an acid. The pink color in the flask is due to the phenolphthalein indicator.

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

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]

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

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.

Indicator and Titration Curve 0.1000 M HCl vs 0.1000 M NaOH

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


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



Va Ma = Vb Mb , equivalence point

at equivalence point of a strong acid - strong base titration

pH º 7.0000



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

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

Titration of Weak Acid with Strong Base

Effect of Acid Strength on Titration Curve

Titration of Weak Base with Strong Acid

Titration of Diprotic Weak

Acid with Strong Base

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)

Acid Rain

acid rain is defined as rain with a pH < 5.6

pH = 5.6 for rain in equilibrium with atmospheric carbon dioxide

Sulfuric Acid from Sulfur burning

SO2

S + O2 => SO2

SO3

2 SO2 + O2 => 2 SO3

Sulfuric AcidSO3 + H2O => H2SO4

Nitric Acid2 NO2(g) + H2O(l) => HNO3(aq) + HNO2(aq)

How Acid Precipitation Forms

Acid Precipitation in U.S.

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

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]

Solubility Product

Constant Values

Dissolving Slightly Soluble Salts Using Acids

Insoluble salts containing anions of Bronsted-Lowry bases can be dissolved in solutions of low pH

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

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

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

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

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)

Sodium Thiosulfate Dissolves Silver Bromide

Formation Constants

Amphoterism

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

Will Precipitation Occur?

Kidney Stones

Kidney stones are normally composed of:calcium oxalate

calcium phosphatemagnesium ammonium phsphate

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

PRECIPITATION REACTIONS

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


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


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

Because this is the product of “solubilities”, we call it

Ksp = solubility product constant See Table in the Text

Ksp = solubility product

Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-

(aq) Ksp = 1.9 x 10-5

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

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?)

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.

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

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 !

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

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.


[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.

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


Common Ion EffectAdding an Ion “Common” to an

Equilibrium

PbCl2(s) Pb2+(aq) + 2Cl-(aq)

NaCl Na+(aq) + Cl- (aq)

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


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?___


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

(aq)Solution [Ba2+] [SO4

2-]initialchangeequilib.


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


Solution [Ba2+] [SO4

2-]initial 0.010 0change + y + yequilib. 0.010 + y y


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!


chemistry 102(01) spring 2009

Documents

x x x h

x x2 kanh4

solution of weak acid

ka haaph

typesmixture of weak

acidbase titrations17

aq h3o

acid rain17