15 - 1 Brønsted-Lowry Definitions The Ion Product for Water The pH and Other “p” Scales Aqueous...

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Brønsted-Lowry DefinitionsBrønsted-Lowry DefinitionsThe Ion Product for WaterThe Ion Product for Water

The pH and Other “p” ScalesThe pH and Other “p” ScalesAqueous Solutions of Acids and BasesAqueous Solutions of Acids and Bases

HydrolysisHydrolysisThe Common Ion EffectThe Common Ion Effect

Buffer SolutionsBuffer SolutionsIndicators and TitrationsIndicators and Titrations

Polyprotic AcidsPolyprotic Acids

Acids and Bases Topic # 14 Acids and Bases Topic # 14

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Arrhenius definitionsArrhenius definitions

AcidAcid Anything that produces hydrogen ions in a water solution.

HCl (aq) H+ + Cl-

BaseBase Anything that produces hydroxide ions in a water solution.

NaOH (aq) Na+ + OH-

Arrhenius definitions are limited to aqueous solutions.

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Brønsted-Lowry definitionsBrønsted-Lowry definitions

Expands the Arrhenius definitions

AcidAcid Proton donor

BaseBase Proton acceptor

This definition explains how substances like ammonia can act as bases.

NH3(g) + H2O(l) NH4+ + OH-

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Strong acids and basesStrong acids and bases•considered to ionize completely.

HCl(aq) + H2O(l) H3O+ (aq) + Cl-(aq)

NaOH(aq) + H2O(l) Na+(aq) + OH-(aq)

Weak acids and basesWeak acids and bases•do not ionize completely.

HC2H3O2 (aq) + H2O(l) H3O+(aq) +C2H3O2-(aq)

NH3 (aq) + H2O(l) NH4+ (aq) + OH-

(aq)

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When a Brønsted-Lowry acid dissolves in water, the water acts as a base.

HC2H3O2 (aq) + 2H2O(l) H3O+(aq) + C2H3O2-

(aq)

Acid Base Acid Base

H3O+ (aq) is called the hydronium ion.

However, the exact number of water moleculesrequired to hydrate a proton is not known.

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Conjugate acid-base pairs.Conjugate acid-base pairs. Acids and bases that are related by loss or gain of H+ as H3O+ and H2O.

Examples.Examples. Acid Base

H3O+ H2O

HC2H3O2 C2H3O2-

NH4+ NH3

H2SO4 HSO4-

HSO4- SO42-

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Common acids and basesCommon acids and bases

AcidsAcids Formula Molarity*nitric HNO3 16

hydrochloric HCl 12sulfuric H2SO4 18acetic HC2H3O2 18

BasesBasesammonia NH3(aq) 15sodium hydroxide NaOH solid

*undiluted.

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Common Acids and basesCommon Acids and bases

AcidicAcidic BasicBasic

Citrus fruits Baking sodaAspirin DetergentsCoca Cola Ammonia

cleanersVinegar Tums and

RolaidsVitamin C Soap

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Water is an amphotericamphoteric substance that can act either as an acid or a base,

HC2H3O2 (aq) + H2O(l) H3O+(aq) + C2H3O2-

(aq) acid base acid base

H2O(l) + NH3(aq) NH4+(aq) + OH-(aq)

acid base acid base

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Auto-ionization of waterAuto-ionization of water

Auto-ionizationAuto-ionization When water molecules react with each another to form ions.

H2O(l) + H2O(l) H3O+(aq) + OH-

(aq)

(10-7M) (10-7M)

Kw = [ H3O+ ] [ OH- ]

= 1.0 x 10-14 at 25oC

Note:Note: [H2O] is not included in expression because H2O is a pure liquid!

ion productof water

ion productof water

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Ion product for waterIon product for water

Kw is a temperature dependent equilibrium constant. We commonly use 25oC as the standard temperature.

Temperature, oC Kw

0 1.153 x 10-15

20 6.87 x 10-15

25 1.012 x 10-14

30 1.459 x 10-14

50 5.31 x 10-14

Temperature, oC Kw

0 1.153 x 10-15

20 6.87 x 10-15

25 1.012 x 10-14

30 1.459 x 10-14

50 5.31 x 10-14

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Autoionization of waterAutoionization of water

[H+] and [OH-] are always present in aqueous solutions. Only for a neutral solution are they at the same concentration.

• Neutral solution.Neutral solution.[H+] = 10-7 M = [OH-]

• Acidic solution.Acidic solution.[H+] > 10-7 M > [OH-]

• Basic solution.Basic solution.[H+] < 10-7 M < [OH-]

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pH and other “p” scalespH and other “p” scalesWe need to measure and use acids and

bases over a very large concentration range. And many times knowing those

concentrations is very important.

pH and pOH are systems to keep track of these very large ranges.

pH = -log[H3O+]

pOH = -log[OH-]

pH + pOH = 14

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pH calculationspH calculations

Determine the following. pH = -log[H+]

pH of 6.7 x 10-3 M H+

= 2.2

pH of 5.2 x 10-12 M H+

= 11.3

[H+], if the pH is 4.5

= 3.2 x 10-5 M H+

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pOH examplespOH examples

Determine the following. pOH = -log[OH-] = 14 - pH

pOH of 1.7 x 10-4 M NaOH

pOH = 3.8 pH = 10.2

pOH of 5.2 x 10-12 M H+

pH = 11.3 pOH = 2.7

[OH-] , if the pH is 4.5

pOH = 9.5

[OH-] = 3.2 x 10-10 M

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pH scalepH scale

A logarithmic scale used to keep track of the large changes in [H+].

14 7 0

10-14 M 10-7 M 1 M Very Neutral VeryBasic Acidic

When you add an acid, the pH gets smaller.

When you add a base, the pH gets larger.

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pH of somepH of somecommon materialscommon materials

Substance pH

1 M HCl 0.0Gastric juices 1.0 - 3.0Lemon juice 2.2 - 2.4Classic Coke 2.5Coffee 5.0Pure Water 7.0Blood 7.35 - 7.45Milk of Magnesia 10.5Household ammonia 12.01M NaOH 14.0

Substance pH

1 M HCl 0.0Gastric juices 1.0 - 3.0Lemon juice 2.2 - 2.4Classic Coke 2.5Coffee 5.0Pure Water 7.0Blood 7.35 - 7.45Milk of Magnesia 10.5Household ammonia 12.01M NaOH 14.0

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Acid and Base StrengthAcid and Base Strength

Strong acidsStrong acids Ionize completely in water. HCl, HBr, HI, HClO3, HNO3, HClO4, H2SO4.

Weak acids Weak acids Partially ionize in water. Most acids are weak.

Strong basesStrong bases Ionize completely in water. Strong bases are metal

hydroxides - NaOH, KOH

Weak basesWeak bases Partially ionize in water.

no Ka ! 100% Ionization!

no Kb ! 100% Ionization!

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LiOH lithium hydroxide

NaOH sodium hydroxide

KOH potassium hydroxide

RbOH rubidium hydroxide

CsOH cesium hydroxide

*Ca(OH)2 calcium hydroxide

*Sr(OH)2 strontium hydroxide

*Ba(OH)2 barium hydroxide

These bases are considered to be strong.

* Completely dissociated in solutions of 0.01 M or less. These are insoluble bases which ionize 100%. The other five in the list can easily make solutions of 1.0 M and are 100% dissociated at that concentration.

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Acid and Base StrengthAcid and Base Strength

For strong acids and bases, we can directly calculate the pH or pOH if we know the molar concentration.

Examples.Examples.0.15 M HCl would produce 0.15 M H+. pH = -log(0.15) = 0.82 pOH = 14.00 - pH = 14.00 - 0.82 = 13.18

0.052 M NaOH produces 0.052 M OH-. pOH = -log(0.052) = 1.28 pH = 14.00 - pOH = 14.00 - 1.28 =

12.72

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Acid dissociation constant, Acid dissociation constant, KKaa

The ionization of a weak acid can be expressed as an equilibrium.

HA (aq) + H2O(l) H3O+(aq) + A- (aq)

The strength of a weak acid is related to its equilibrium constant, Ka.

Ka = We omit water.

It’s a pure liquid.We omit water.

It’s a pure liquid.

[A-][H3O+][HA]

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Weak acid equilibriaWeak acid equilibria

ExampleExampleDetermine the pH of a 0.10 M benzoic acid solution at 25 oC if Ka = 6.28 x 10-5

HBz(aq) + H2O(l) H3O+(aq) + Bz-(aq)

The first step is to write the equilibrium expression.

Ka = [H3O+][Bz-]

[HBz]

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HBz H3O+ Bz-

Initial conc., M 0.10 0.00 0.00

Change, M -x x x

Eq. Conc., M 0.10 - x x x

[H3O+] = [Bz-] = x

We’ll assume that [Bz-] is negligible compared to [HBz]. The contribution of H3O+ from water is also negligible.

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Solve the equilibrium equation in terms of Solve the equilibrium equation in terms of xx

Ka = 6.28 x 10-5 =

x = (6.28 x 10-5 )(0.10)

= 0.0025 M

pH = 2.60

x2

0.10

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Dissociation of bases, Dissociation of bases, KKbb

The ionization of a weak base can also be expressed as an equilibrium.

B (aq) + H2O(l) BH+(aq) +OH- (aq)

The strength of a weak base is related to its equilibrium constant, Kb.

Kb =[OH-][BH+]

[B]

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KKaa and and KKbb values values

For weak acids and bases

Ka and Kb always have values that are smaller than one.

•Acids with a larger Ka are stronger than ones with a smaller Ka.

•Bases with a larger Kb are stronger than ones with a smaller Kb.

•For a conjugate acid:base pair, pKa + pKb = 14 = pKw

Most acids and bases are weak.

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Ionization constants at 25Ionization constants at 25ooC.C.

Acid pKa pKb

Acetic acid 4.77 9.23Ammonium ion 9.25 4.75 Benzoic acid 4.19 9.81Formic acid 3.74 10.26Lactic acid 3.86 10.14Phenol 9.89 4.11

Acid pKa pKb

Acetic acid 4.77 9.23Ammonium ion 9.25 4.75 Benzoic acid 4.19 9.81Formic acid 3.74 10.26Lactic acid 3.86 10.14Phenol 9.89 4.11

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HydrolysisHydrolysis

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.


•The 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+.

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Al(H2O)63+(aq) + H2O(l) Al(H2O)5(OH)2+(aq) +

H3O+(aq)

These metal ions are able to pull electrons from the H-OH bond. If the pull is strong enough, water is split.

One of the surrounding water molecules will take on the H+ and form H3O+.

There is only a small # of cations with a high enough positive charge that can act as acids in hydrolysis.

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Common ion effectCommon ion effect

(Just an example of Le Châtelier’s principle.)

Common ion effectCommon ion effectThe shift in equilibrium caused by the addition of an ion formed from the solute.

Common ionCommon 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.

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ExampleExampleWhat is the concentration of hydrogen ions in a solution formed by adding 0.097 mol of sodium acetate to one liter of 0.099 M acetic acid (HAc). (Ac = C2H3O2

-)

(assume that the volume of the solution does not change when the salt is added)

[Ac-] [H+][HAc]

Ka = = 1.7 x 10-5

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HAc Ac- H+

Init. Conc., M 0.099 0.097 0.000

Change, M -x +x +x

Eq. Conc., M 0.099 - x 0.097 + x x

To solve for x, substitute the equilibrium values into the expression.

Ka [HAc][Ac-]

(1.7 x 10-5) (0.099-x)(0.097 +x)

x = =

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If we assume that x is negligible compared to 0.099 and 0.097, we can simplify the problem to:

x =

x = 1.7 x 10-5 = [H+]

Note.Note. When a solution contains equal concentrations of an acid and its conjugate base, [H+] = Ka.

(1.7 x 10-5) (0.099)(0.097)

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Salt Hydrolysis

A salt is formed between the reaction of an acid and a base. HCl + NaOH NaCl + H2O

Usually, a neutral salt is formed when a strong acid and a strong base are neutralized in the reaction.

Ions of Neutral Salts Ions of Neutral Salts

CATIONS

Na+ K+ Rd+ Cs+

Mg+2 Ca+2 Sr+2 Ba+2

ANIONS

Cl- Br- I-

ClO4- BrO4- ClO3- NO3-

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In this NEUTRAL environment, all the ions are spectator ions from the strong acid and strong base reaction. There are no insoluble particles. These ions have little tendency to react with water, they just stay dissociated into 100% ions.

So, salts consisting of these ions are called neutral salts and the resulting solution has a pH of 7.

For example: NaCl, KNO3, CaBr2, CsClO4 are neutral salts.

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Acidic Ions

NH4+ Al+3 Pb+2 Sn+2

Transition Metal Ions

HSO4- H2PO4

-

Basic Ions

F- C2H3O2- NO2- HCO3-

CN- CO3 -2 S-2 SO4

-2

HPO4-2 PO4

-3

When weak acids and weak bases react, the strength of the stronger ion conjugate acid or conjugate base in the salt determines the pH of its solutions.

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The resulting salt solution can be acidic, neutral or basic.

A salt formed between a strong acid and a weak base is an acid salt, for example NH4Cl:

( HCl + NH3 NH4Cl ) Strong weak

The NH4Cl is soluble and reacts with H2O:

NH4+ + H2O NH3 + H3O+

The act of the salt particles reacting with water molecules in the ‘salt solution’ is HYDROLYSIS. The act of the salt particles reacting with water molecules in the ‘salt solution’ is HYDROLYSIS.

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A salt formed between a weak acid and a strong base is a basic salt, for example NaCH3COO.

(CH3COOH + NaOH NaCH3COO + H2O)

CH3COO- + H2O CH3COOH + OH-

pH > 7

A salt formed between a weak acid and a strong base is a basic salt, for example NaCH3COO.

(CH3COOH + NaOH NaCH3COO + H2O)

CH3COO- + H2O CH3COOH + OH-

pH > 7

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Deciding whether a salt is acidic, basic or neutral!

******Decide which ion in the salt could produce a stronger acid or stronger base. Example:

NaF ---- NaOH or HF NaOH is the stronger base so

NaF is a basic salt

NaClO4 ---- NaOH or HClO4

BOTH are strong so neutral salt!

Fe(NO3)2 ---- Fe(OH)2 or HNO3 HNO3 is the stronger compound

Fe(NO3)2 is an acidic salt

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What if both ions are from weak acids and weak bases?

Salt: NH4NO2 NH4OH and HNO2 weak base and weak acid

Can compare Kb for NH4OH and Ka for HNO2: Kb NH4OH = 1.6 x10-5 Ka for HNO2 = 4.5 x 10-4

Ka > Kb so the salt is acidic!

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Another example: ( qt.1h. on Salt Hydrolysis WS) (NH4)2CO3 ---- NH4OH and HCO3

- weak base and weak acid ion

Kb for NH4OH = 1.6 x 10 -5

Ka2 for H2CO3 = 4.8 x 10 -11

So (NH4)2CO3 is basic!

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Calculate the pH of a 0.500 M solution of KCN. ( Ka for HCN is 5.8 x 10-10)

Calculations !!!

CN- + H2O HCN + OH-

0.5M +x +x

Kb = [HCN] [OH-] [CN-]

Kb = Kw Ka

1 x10-14

5.8 x 10-10= = 1.7 x10-5

Salt is a basic salt, so need to use Kb….

Salt is a basic salt, so need to use Kb….

- ion makes OH- !

- ion makes OH- !

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1.7 x10-5= x2 0.5 –x

x = 2.9 x 10-3 M

pOH = -log(2.9 x 10-3) = 2.54 14 - 2.54 = 11.46 = pH

/ /

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BuffersBuffers

Solutions that resist pH change when small amounts of acid or base are added.

Two typesTwo types•weak acid and its salt•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.

Attacks HA

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A solution of a weak acid and its’ salt!

Acetic Acid and Sodium Acetate: HC2H3O2 + NaC2H3O2

Mix these two compounds in water gives a solution with this organization:

HC2H3O2 H+ + C2H3O2-

Concentration of the C2H3O2- would be greater than just a solution of low soluble HC2H3O2 !!!!

BuffersBuffers

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Ethanoic acid and ethanoate:

CH3CO2- + H3O+ HCH3CO2 + H2O

The ethanoate ions will be able to absorb any excess acid that is added

HCH3CO2 + OH- CH3CO2- + H2O

The ethanoic acid will be able to absorb any excess base that is added

Both: CH3CO2 - AND CH3CO2H are present in solution

How does a buffer work?

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BuffersBuffersThe 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.

Calculating pH of an acidic buffer use:

pH = pKa + log

Calculating the pH of a basic buffer use:

pH = 14 - ( pKb + log )[HA][A-]

[A-][HA]

Henderson-Hasselbalch equationHenderson-Hasselbalch equation

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BuffersBuffersThe beauty of a buffer….The beauty of a buffer….

A total of 100 ml of a 1.0 M HCl is add in 10 ml increments to 100 ml of each of the following:

Solution A: Pure water - pH = 7 (no acid in it ! )

Solution B: A solution containing

1.0 M HA (acid in it ), 1.0 M A- (conjugate base in it )

with a pKa of 7.0 Calculate the pH after each addition.

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BuffersBuffersInitially, each sample is at pH 7.00

After adding 10 ml of 1.0 M HCl we have:

Pure water[H3O+] = =

0.091

pH = 1.04

From pH 7 to pH 1.04…..This is a pretty big jump!

(10 ml)(1.0 M)(110 ml)

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BuffersBuffersAdd 10 ml of 1.0 M HCl to our buffered system.

.10 moles +.01 moles .10 moles - .01 moles + 0.01 moles .09 0 .11 molesWe started with 0.10 moles of both the acid and

conjugate base forms.

The HCl in our first 10 ml can be expected to react with the conjugate base, converting it to the acid.

After addition, we would have 0.09 moles of the base form and 0.11 moles of the acid form.

CH3CO2- + H3O+ HCH3CO2 + H2O

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BuffersBuffersChecking the pH:Checking the pH:

Since all we need to be concerned about is the ratio of A- to HA in the 10-90% region then:

pH = pKA + log

= 7.00 + log

= 6.91

A change of only 0.09 (Buffer solution)compared to 5.96 ( in pure water) !

[A-][HA]

0.09 mol0.11 mol

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BuffersBuffers

ml HCl pHadded unbuffered buffered 0 7.00 7.00 10 1.04 6.91 20 0.78 6.82 30 0.64 6.73 40 0.54 6.63 50 0.48 6.52 60 0.43 6.40 70 0.39 6.25 80 0.35 6.05 90 0.32 5.72

Matter of fact……..

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BuffersBuffers

ml HCl added

pH buffered

unbuffered

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IndicatorsIndicators

Acid-base indicators are highly colored weak acids or bases.

HIndic Indic-

color 1color 1 color 2color 2

They may have more than one color transition.

Example.Example. Thymol blue Red - Yellow - Blue

One of the forms may be colorless - phenolphthalein (colorless to pink)

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Indicator color change.Indicator color change.

Indicators are commonly used to detect the endpoint of an acid-base titration.

The indicator should not start to undergo a color change until about one pH unit after the equivalence point.

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Selection of an indicator.Selection of an indicator.Under these conditions, the transition range for our indicator is pKa + 1.

Ideally then, you want your indicator pKa to be one above the pKa for an acid sample.

For bases, it should be one lower.

You can seldom find a ‘perfect’ indicator so should use one that is a maximum of one unit away.

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pH transition Indicator range color

Bromophenol Blue 6.2 - 7.6 yellow-blueMethyl Orange 3.2 - 4.4 red-yellowMethyl Red 4.8 - 6.0 red-yellowBromothymol Blue 6.0 - 7.6 yellow-blueCresol Purple 7.4 - 9.0 yellow-purplePhenolphthalein 8.2 - 10.0 colorless - pinkThymolphthaleine 9.4 - 10.6 colorless - blueAlizerin Yellow GG 10.0 - 12.0 yellow - red

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Indicator examplesIndicator examples

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

phenolphthalein

pH

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Indicator examplesIndicator examples

methyl red

bromthymol blue

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Titrations revisitedTitrations revisited

Methods based on measurement of volume.

•If the concentration of an acid is known, the amount of a base can be found.

•If we know the concentration of the base, then we can determine the amount of acid.

•All that is needed is some calibrated glassware and either an indicator or pH meter.

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TitrationsTitrations

BuretBuret - volumetric glassware used for titrations.It allows you to add a known amount of your titrant to the solution you are testing.

If a pH meter is used, the equivalence point can be measured.

An indicator will give you the endpoint.

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TitrationsTitrations

Note the color change which indicates that the ‘endpoint’ has been reached.

Start End

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Titration curvesTitration curves

Acid-base titration curveAcid-base titration curve

A 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.

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Titration curvesTitration curvesp

H EquivalencePoint

% titration or ml titrant

Buffer region

Overtitration

IndicatorTransition

Four regions of titration curve

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Strong acid titrated with a strong base.Strong acid titrated with a strong base.

This is pretty straight forward since the net reaction is:

H3O+ + OH- 2H2O

Prior to equivalence pointPrior to equivalence pointThe pH indicates the amount of sample present after accounting for dilution.

[H3O+] =moles acid - moles base

total volume in liters

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Equivalence pointEquivalence point

[H3O+] = [OH-]

pKW = 14 = pH + pOH

pH = 7.00

So the equivalence point for strong acid/base problems is always at pH=7.00

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OvertitrationOvertitration

Past the equivalence point, we don’t have any acid remaining. All that we are doing is diluting our titrant by adding more solution volume and more OH- ions.

[OH-] =

pH = 14 - pOH

moles excess total volume in liters

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Example.Example.Construct a titration curve for the titration of 100 ml of 0.10 M HCl with 0.10 M NaOH

A. 0% titration0% titrationpH = -log[0.10] = 1

B. After 10 ml NaOHB. After 10 ml NaOH

[H3O+] =

= 0.082 M, pH = 1.09

(100 ml)(0.10 M) - (10 ml)(0.10 M)100 ml + 10 ml

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ml titrant total ml [H3O+] pH 0 100 0.10 1.00

10 110 0.082 1.0920 120 0.067 1.1730 130 0.054 1.2840 140 0.043 1.3750 150 0.033 1.4860 160 0.025 1.6070 170 0.018 1.7480 180 0.011 1.9690 190 0.0053 2.28

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0 30 60 90ml NaOH

pH

Let’s add 10ml more!

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ml NaOH

pH

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Reached the Reached the Equivalence point Equivalence point or 100 % titration or 100 % titration

(100ml acid)(0.10M acid) = (100ml base)( 0.10M base)(100ml acid)(0.10M acid) = (100ml base)( 0.10M base)

pH MUST be 7 1 x10-14 = (1 x10-7)2

Note that for the first 90 ml of our titration, we only saw a change of 1.28 pH units.

Now we have a jump of 4.72 pH units.

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Titration curvesTitration curvesAnything over the equilvalence point isAnything over the equilvalence point isOvertitration: (no more acid there!)Overtitration: (no more acid there!)

Describing the dilution of our titrant:

10 ml overtitration10 ml overtitration

[OH-] = 0.10 M

= 0.0048 M

pOH = 2.32

pH = 14.00 - 2.32 = 11.68

(.10M)10 ml210 ml

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ml titrant total volume [OH-] pH110 210 0.0048 11.68120 220 0.0091 11.96130 230 0.013 12.11140 240 0.017 12.23150 250 0.020 12.30160 260 0.023 12.36170 270 0.026 12.41180 280 0.029 12.46190 290 0.031 12.49200 300 0.033 12.52

Continue the calculations:

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ml NaOH

pH

Over titration

Just adding more OH-

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Titration of a strong base with a strong Titration of a strong base with a strong acid.acid.

This is not significantly different from our earlier example.

If you plot pOH rather that pH, the results would look identical.

Typically we still plot pH versus ml titrant so the curves are inverted.

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Titration curvesTitration curvesp

H

ml titrant

basic sample (base in the beaker)

acidic sample (acid in the beaker)

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Titration of weak acids or basesTitration of weak acids or bases

Titration of a weak acid or base with a strong titrant is a bit more complex than the strong acid/strong base example.

We must be concerned with conjugate acid/base pairs and their equilibria.

ExampleExampleHA + H2O H3O+ + A-

acidacid basebase

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First, we’ll only be concerned about the titration of a weak acid with a strong base or a weak base with a strong acid.

• We still have the same four general regions for our titration curve.

• The calculation will require that you use the appropriate Ka or Kb relationship.

• We’ll start by reviewing the type of calculations involved and then work through an example.

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0% titration0% titrationIf your sample is an acid then use

Ka =

At this point [H3O+] = [A-].

You can solve for [H3O+] by using either the quadratic or approximation approach.

Finally, calculate the pH.

[H3O+][A-][HA]

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0% titration0% titrationIf your sample is an base then use

Kb =

At this point [OH-] = [HA].

You can solve for [OH-] by using either the quadratic or approximation approach.

Then determine pH as 14 - pOH.

[OH-][HA][A-]

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5 - 95% titration5 - 95% titrationIn this region, the pH is a function of the K value and the ‘ratio’ of the acid and base forms of our answer.

A common format for the equilibrium expression used in this region is the Henderson-Hasselbalch equationHenderson-Hasselbalch equation.

We can present it in two forms depending on the type of material we started with.

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5 - 95% titration5 - 95% titration

Starting with an acidStarting with an acid

pH = pKa + log

Starting with a baseStarting with a base

pH = 14 - ( pKb + log )

( We’re just determining the pOH and then converting it to pH. )

[HA][A-]

[A-][HA]

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Another approach that can be taken is to simply use the % titration values.

For an acidic sample, you would use:

pH = pKa + log % titration

100 - % titration

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These equations have their limits and may break down if:

Ka or Kb > 10-3

You are working with very dilute solutions.

In those cases, you need to consider the equilibrium for water

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100% titration - the equivalence point.100% titration - the equivalence point.At the point, we have converted all of our sample to the conjugate form.

If your sample was an acid, now solve for the pH using the Kb relationship -- do the opposite if you started with a base.

Remember that pRemember that pKKaa + p + pKKbb = 14 = 14

You must also account for dilution of your sample as a result of adding titrant.

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Overtitration (> 100%)Overtitration (> 100%)

These calculations are identical to those covered in our strong acid/strong base example.

You simply need to account for the amount of excess titrant added and how much it has been diluted.

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ExampleExampleA 100 ml solution of 0.10 M benzoic acid is titrated with 0.10 M NaOH. Construct a titration curve.

For benzoic acidFor benzoic acidKa = 6.28 x 10-5

pKa = 4.20

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Before adding any base:Before adding any base:

Ka =

[H3O+] = [A-]

[HA] + [A-] = 0.10 M

(We’ve assumed that [A-] is negligible compared to [HA].)

[H3O+][A-][HA]

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Initial pH, before adding any base:Initial pH, before adding any base:

Ka = 6.28 x 10-5 =

x = (6.28 x 10-5 )(0.10)

= 0.0025 M

pH = 2.60

x2

0.10

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adding 10ml increments of base: adding 10ml increments of base: Here we can use the Henderson-Hasselbalch equation in ml titration format

pH = pKa + log

pH = 4.20 + log (10 / 90)

= 3.25

We can calculate other points by repeating this process.

(increments of 10ml)(100ml – total added)

(Showing the first increment of 10ml)

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ml added pH 0 2.6010 3.2420 3.6030 3.8340 4.0250 4.2060 4.3870 4.5780 4.8090 5.15

Note: At 50 ml titration, pH = pKa

Also, the was only a change of 1.91 pH units as we went from 10 to 90 ml titration. (5.15 – 3.24)

Note: At 50 ml titration, pH = pKa

Also, the was only a change of 1.91 pH units as we went from 10 to 90 ml titration. (5.15 – 3.24)

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% titration

pH

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100% titration (EQUILVALENCE POINT)100% titration (EQUILVALENCE POINT)

At this point, virtually all of our acid has been converted to the conjugate base - benzoate.

Need to use the Kb relationship to solve for this point! (have no acid left!)

Kb =

Kb = Kw / Ka = 1.59 x 10-10

[OH-] [HA][A-]

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EQUILVALENCE POINTEQUILVALENCE POINT

At the equivalence point:

[HA] = [OH-](We’ve diluted

the [A-] = 0.05 M sample and the

total volume at this point is 200 ml)

Finally, assume that [benzoic acid] is negligible compared to [benzoate].

.10m .200 liters

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100% titration (100% titration ( EQUILVALENCE POINT)EQUILVALENCE POINT)

Kb = 1.59 x 10-10 =

x = (1.59 x 10-10 )(0.050)

= 2.81 x 10-6 M

pOH = 5.55 pH = 14 - pOH = 8.45

x2

0.050

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% titration

pH

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OvertitrationOvertitrationAll we need to do here is to account for the dilution of our titrant.

10 % overtitration (10 ml excess)10 % overtitration (10 ml excess)

[OH-] = 0.10 M

= 0.0048 MpOH = 2.32pH = 14.00 - 2.32 = 11.68

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ml titrant total volume [OH-] pH110 210 0.0048 11.68120 220 0.0091 11.96130 230 0.013 12.11140 240 0.017 12.23150 250 0.020 12.30

This is identical to what we obtained for our strong acid/strong base example

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Titration of weak acidsTitration of weak acids

% titration

pH

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Polyprotic acidsPolyprotic acids

A number of acids exist that have more that one ionizable hydrogen.

ExamplesExamplesH3PO4 Phosphoric AcidH2SO4 Sulfuric AcidH2C2O4 Oxalic acidH2CO3 Carbonic acid

Each H+ will ionize with its own constant. Also, it becomes more difficult to remove subsequent H+.

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Stepwise equilibriumStepwise equilibrium

ExampleExample - H3PO4

Each hydrogen is capable of dissociation but not to the same extent at any given pH.

H3PO4 H2PO4- HPO4

2- PO43-

Since the removal of any one H+ affects how easily subsequent H+ are removed, each step has its own Ka value.

Ka3Ka2Ka1

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Stepwise equilibriumStepwise equilibrium

[H3O+] [PO43-]

[HPO42-]

KA3 =

[H3O+] [HPO42-]

[H2PO4-]

KA2 =

[H3O+] [H2PO4-]

[H3PO4]KA1 =

Note:[H3O+] is the samefor each expression.

The relative amountsof each species canbe found if the pHis known.

The actual amountscan be found if pHand total H3PO4 isknown.

Note:[H3O+] is the samefor each expression.

The relative amountsof each species canbe found if the pHis known.

The actual amountscan be found if pHand total H3PO4 isknown.

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Titration Curve for Diprotic Acid

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