3.0Unit3-Acid_and_Bases

Chemistry 2

Unit 3 Unit 3 Acids-Bases and SaltAcids-Bases and Salt

Pajuzi AwangPajuzi AwangIPG KAMPUS DATO’ RAZALI ISMAILIPG KAMPUS DATO’ RAZALI ISMAIL

Acids and BasesAcids and Bases

Bronsted Lowry Acids and BasesBronsted Lowry Acids and Bases Autoionization of WaterAutoionization of Water pHpH Strong Acids and BasesStrong Acids and Bases Weak Acids and BasesWeak Acids and Bases Ionization ConstantsIonization Constants Lewis Acids and BasesLewis Acids and Bases The Common Ion EffectThe Common Ion Effect BuffersBuffers TitrationsTitrations

ReviewReview

Arrhenius AcidArrhenius Acid

– Substance that increases the Substance that increases the concentration of Hconcentration of H++ ions when ions when dissolved in waterdissolved in water

HCl (g) HHCl (g) H++ (aq) + Cl (aq) + Cl-- (aq) (aq)H2O

ReviewReview

Arrhenius BaseArrhenius Base

– Substance that increases the Substance that increases the concentration of OHconcentration of OH-- ions when ions when dissolved in water.dissolved in water.

NaOH (s) NaNaOH (s) Na++ (aq) + OH (aq) + OH-- (aq) (aq)H2O

ReviewReview

The Arrhenius definition of acids and The Arrhenius definition of acids and bases has limitations:bases has limitations:– Limited to aqueous solutionsLimited to aqueous solutions

Two other common definitions for acids Two other common definitions for acids and bases.and bases.– Bronsted-Lowry acids and basesBronsted-Lowry acids and bases– Lewis acids and bases (pg 125)Lewis acids and bases (pg 125)

Bronsted-Lowry Acids & Bases Bronsted-Lowry Acids & Bases (pg 96)(pg 96) Bronsted-Lowry AcidBronsted-Lowry Acid– A substance (ion or molecule) that can A substance (ion or molecule) that can

transfer a proton (Htransfer a proton (H++ ion) to another ion) to another substancesubstance

HCl (g) + HHCl (g) + H22OO (l) H(l) H33OO++ (aq) + Cl (aq) + Cl-- (aq) (aq)

B-L acid

Bronsted-Lowry Acids & Bases Bronsted-Lowry Acids & Bases (pg 96)(pg 96) Bronsted-Lowry BaseBronsted-Lowry Base– A substance (ion or molecule) that can A substance (ion or molecule) that can

accept a proton (Haccept a proton (H++ ion) from another ion) from another substancesubstance

HCl (g) + HHCl (g) + H22OO (l) H(l) H33OO++ (aq) + Cl (aq) + Cl-- (aq) (aq)

B-L acid B-L base

Notice that H2O acts as a base, accepting a proton from HCl.

Bronsted-Lowry Acids & BasesBronsted-Lowry Acids & Bases

Bronsted-Lowry acid:Bronsted-Lowry acid:– Molecule or ionMolecule or ion– Must have H atom in the formula that Must have H atom in the formula that

can be lost as Hcan be lost as H++ ion ion

Bronsted-Lowry Base:Bronsted-Lowry Base:– Molecule or ionMolecule or ion– Must have lone pair of electrons Must have lone pair of electrons

(nonbonding pair) that can bind a H(nonbonding pair) that can bind a H++ ionion


HH++ ion in Water ion in Water

+ e-

H atomH atom HH++ ion ion

+

Lose e- Proton with no surrounding valence electrons


HH+ + ion interacts strongly with the ion interacts strongly with the nonbonding pairs of electrons on water nonbonding pairs of electrons on water moleculesmolecules– Forms hydrated hydrogen ionForms hydrated hydrogen ion

““hydronium ionhydronium ion””

HH+ + + O – H + O – H H – O – H H – O – H ++

HH HH

Hydronium ion


The hydrated proton (hydronium ion) is The hydrated proton (hydronium ion) is responsible for the characteristic responsible for the characteristic properties of aqueous solutions of acids.properties of aqueous solutions of acids.

HH33OO+ + is a more realistic depiction of the is a more realistic depiction of the

hydrogen ion in solutionhydrogen ion in solution– for convenience we often use Hfor convenience we often use H++ (aq) (aq)

to depict the hydrated hydrogen ionto depict the hydrated hydrogen ion

HCl (g) + H20 (l) H3O+ (aq) + Cl- (aq)

HCl (aq) H+ (aq) + Cl- (aq)

Bronsted-Lowry Acids & BasesBronsted-Lowry Acids & Bases Bronsted-Lowry theory is applicable to Bronsted-Lowry theory is applicable to

reactions that occur in both aqueous and reactions that occur in both aqueous and non-aqueous solution.non-aqueous solution.

HH H H ++

H – N + H – Cl H – N + H – Cl H – N – H + Cl H – N – H + Cl--

HH HH

HCl (aq) + NH3 (aq) NH4+ (aq) + Cl- (aq)

HCl (g) + NH3 (g) NH4Cl (s)


Consider the following reactions:Consider the following reactions:

HH22O (l) + HCl (g) O (l) + HCl (g) H H33OO++ (aq) + Cl(aq) + Cl-- (aq) (aq)

NHNH33 (g) + H (g) + H22O (l) O (l) NH NH44++ (aq) + OH (aq) + OH-- (aq) (aq)

acid

base

In the first reaction, HIn the first reaction, H22O acts as a O acts as a base.base.

In the second reaction, HIn the second reaction, H22O acts as an O acts as an

acid.acid.


Water is Water is amphotericamphoteric::– Capable of acting as either an acid or a Capable of acting as either an acid or a

base.base.

In any acid-base equilibrium, both In any acid-base equilibrium, both forward and reverse reactions involve forward and reverse reactions involve proton transfers.proton transfers.

HNOHNO22 (aq)(aq) + H + H22OO (l)(l) NO NO2 2 –– (aq)(aq) + H + H33OO

++ (aq)(aq)




++ (aq)(aq)

Forward Reaction:Forward Reaction:– B-L acid =B-L acid =– B-L base =B-L base =

Reverse Reaction:Reverse Reaction:– B-L acid =B-L acid =– B-L base =B-L base =

HNOHNO22

HH22OO

HH33OO++

NONO22--



++ (aq)(aq)

baseacid acidbase

When HNOWhen HNO22 acts as an acid, it loses a acts as an acid, it loses a

proton and forms NOproton and forms NO22 – –

– NONO22-- acts as a base acts as a base

When HWhen H22O acts as a base, it gains a proton O acts as a base, it gains a proton

and forms Hand forms H33OO++

– HH33OO++ acts as an acid acts as an acid

Bronsted-Lowry Acids & Bases Bronsted-Lowry Acids & Bases (pg 97)(pg 97)HNOHNO22 (aq)(aq) + H + H22OO (l)(l) NO NO2 2

–– (aq)(aq) + H + H33OO++ (aq)(aq)

baseacid acidbase

HNOHNO22 and NO and NO22-- are called a are called a conjugate conjugate

acid-base pair.acid-base pair.

HH22O and HO and H33OO++ are also a are also a conjugate acid-conjugate acid-

base pair.base pair.

Bronsted-Lowry Acids & BasesBronsted-Lowry Acids & Bases Conjugate acid-base pair:Conjugate acid-base pair:– An acid and a base that differ only in An acid and a base that differ only in

the presence or absence of a single the presence or absence of a single protonproton HNOHNO22 and NO and NO22

--

HH33OO++ and H and H22OO HCOHCO33

-- and COand CO33 2-2-

NHNH44++ and NH and NH33


Every acid has a Every acid has a conjugate base:conjugate base:– The base formed by removing a proton The base formed by removing a proton

from an acidfrom an acid Conjugate base of HConjugate base of H22OO–OH OH --

Conjugate base of HConjugate base of H22SOSO44

–HSOHSO44 --

Conjugate base of HClOConjugate base of HClO44

–ClOClO4 4 ––

NOTE: The conjugate base is always NOTE: The conjugate base is always shown on the product side.shown on the product side.


Every base has a Every base has a conjugate acid:conjugate acid:– The acid formed when a base gains a The acid formed when a base gains a

protonproton Conjugate acid of HConjugate acid of H22OO

–HH33O O ++

Conjugate acid of SOConjugate acid of SO4 4 2-2-

–HSOHSO44 --

Conjugate acid of CN Conjugate acid of CN --

–HCNHCN NOTE: The conjugate acid is always NOTE: The conjugate acid is always

shown on the product side.shown on the product side.

Chemistry 2


Pajuzi AwangPajuzi AwangIPG KDRI-Lecture 2IPG KDRI-Lecture 2


HNOHNO22 (aq)(aq) + + HH22OO (l)(l) NONO2 2 –– (aq)(aq) + + HH33OO

++ (aq)(aq)

baseacid conjugatebase

Conjugateacid


Example:Example: What is the conjugate base of: What is the conjugate base of:

HClOHClO44

HH33POPO44

HSOHSO44 --

ClO4 -

H2PO4 -

SO4 2-


Example:Example: What is the conjugate acid of: What is the conjugate acid of:

HSOHSO44 - -

CNCN - -

OH OH --

H2SO4

HCN

H2O


Example:Example: Identify the acid and base on the Identify the acid and base on the reactant side and the conjugate acid and reactant side and the conjugate acid and conjugate base on the product side of the conjugate base on the product side of the following reaction.following reaction.

HSOHSO44–– (aq) + CO (aq) + CO33

2-2- (aq) SO (aq) SO442-2- (aq) + HCO (aq) + HCO33

2-2- (aq) (aq)

baseacid Conjugateacid

Conjugatebase



++ (aq)(aq)


In any acid-base equilibrium, both the In any acid-base equilibrium, both the forward and reverse reactions involve forward and reverse reactions involve proton transfers.proton transfers.


++ (aq)(aq)

How can we predict the position of the How can we predict the position of the chemical equilibrium?chemical equilibrium?

baseacid Conjugateacid

Conjugatebase

Bronsted Lowry Acids & BasesBronsted Lowry Acids & Bases

The relative strengths of the acid and the The relative strengths of the acid and the conjugate acid can be used to predict the conjugate acid can be used to predict the position of the equilibrium.position of the equilibrium.

Equilibrium favors the formation of the Equilibrium favors the formation of the weaker acidweaker acid

– The stronger acid more effectively The stronger acid more effectively loses a proton than the conjugate acidloses a proton than the conjugate acid

ACID-BASE PROPERTIES ACID-BASE PROPERTIES (pg 98)(pg 98)

Autoionization and the Ion-Product of Autoionization and the Ion-Product of WaterWater The pH and pOH ScaleThe pH and pOH Scale

Autoionization of WaterAutoionization of Water

One of the most important properties of One of the most important properties of water is its ability to act as either an acid water is its ability to act as either an acid or a baseor a base

In the presence of an acid, water acts as In the presence of an acid, water acts as a base by accepting a Ha base by accepting a H++

HCl (g) + HHCl (g) + H22OO H H33OO++ (aq) + Cl (aq) + Cl - - (aq) (aq)

In the presence of a base, water acts as In the presence of a base, water acts as an acid, donating a Han acid, donating a H++

NHNH33 (g) + H (g) + H22OO NHNH44++ (aq) + OH (aq) + OH - - (aq) (aq)


Water molecules can also donate a proton to Water molecules can also donate a proton to another water molecule in a process called another water molecule in a process called autoionization.autoionization.

H O + H O H O H H O + H O H O H ++ + OH + OH --

HH HH HH

Autoinization


Autoionization:Autoionization:– the process in which water the process in which water

spontaneously forms low spontaneously forms low concentrations of Hconcentrations of H++ and OH and OH -- ions by ions by proton transfer from one water proton transfer from one water molecule to anothermolecule to another

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


Autoionization of water is a very rapid Autoionization of water is a very rapid processprocess– both forward and reverse reactions both forward and reverse reactions

occur rapidlyoccur rapidly

At any given time only a very small At any given time only a very small number of water molecules are ionizednumber of water molecules are ionized

If every letter in our text represented a water molecule, you would have to look through about 50 texts to find one H3O+!


An equilibrium constant expression can An equilibrium constant expression can be written for the autoionization of water:be written for the autoionization of water:

2 H2 H22O (l) HO (l) H33OO++ (aq) + OH (aq) + OH - - (aq) (aq)

KKww = [H = [H33OO++] [OH] [OH - -] = [H] = [H+ + ] [OH] [OH- - ]]

where Kwhere Kww = ionization constant for water = ionization constant for water

= ion-product constant= ion-product constant

Note: H2O is excluded because it is a pure liquid.


The ionization constant expression for The ionization constant expression for water (and its value) is useful not only for water (and its value) is useful not only for pure water but also for pure water but also for all dilute aqueous all dilute aqueous solutions.solutions.– Use it to calculate the [HUse it to calculate the [H+ + ] and [OH] and [OH-- ] ]

for dilute solutions.for dilute solutions.

For all dilute aqueous solutions:For all dilute aqueous solutions:

KKww = [H = [H+ + ] [OH] [OH-- ] = 1.0 x 10 ] = 1.0 x 10-14-14 at 25at 25ooCC


In a neutral solution:In a neutral solution:– [H[H+ + ] = [OH] = [OH-- ] ]

In an acidic solution:In an acidic solution:– [H[H+ + ] > [OH] > [OH-- ] ]

In a basic solutionIn a basic solution– [H[H+ + ] < [OH] < [OH-- ] ]


Example:Example: Calculate the [HCalculate the [H+ + ] and [OH] and [OH-- ] ] in a neutral in a neutral solution at 25solution at 25ooC.C.

[H[H+ + ] [OH] [OH-- ] = (x) (x) = 1.0 x 10 ] = (x) (x) = 1.0 x 10-14-14

xx22 = 1.0 x 10 = 1.0 x 10-14-14

x = 1.0 x 10x = 1.0 x 10-7 -7 MM

In a neutral solution: [H[H+ + ] = [OH] = [OH-- ] = 1.0 x 10 ] = 1.0 x 10-7 -7 MM

Chemistry 2




In a neutral solution:In a neutral solution:– [H[H+ + ] = [OH] = [OH-- ] = 1.0 x 10 ] = 1.0 x 10-7-7 M at 25 M at 25ooCC

In an acidic solution:In an acidic solution:– [H[H+ + ] > [OH] > [OH-- ] ]– [H[H+ + ] > 1.0 x 10] > 1.0 x 10-7-7 M at 25 M at 25ooCC

In a basic solutionIn a basic solution– [H[H+ + ] < [OH] < [OH-- ] ]– [H[H+ + ] < 1.0 x 10] < 1.0 x 10-7-7 M at 25 M at 25ooCC


Example: Example: What is the [HWhat is the [H+ + ] at 25] at 25ooC for a C for a solution in which [OHsolution in which [OH-- ] = 0.010 M. ] = 0.010 M.

[[H+ ] [OH- ] = 1.0 x 10-14 at 25oC

Given: Given: [OH[OH--] = 0.010 M] = 0.010 M

Find:Find: [H [H+ + ] ]

[H[H+ + ] (0.010) = 1.0 x 10] (0.010) = 1.0 x 10-14-14

[H[H++ ] = ] = 1.0 x 101.0 x 10-14-14 = 1.0 x 10 = 1.0 x 10-12-12 M M0.0100.010


Example: Example: What is the [OHWhat is the [OH-- ] at 25 ] at 25ooC in a C in a solution in which [Hsolution in which [H+ + ] = 2.5 x 10] = 2.5 x 10-6-6 M. M.

[H[H+ + ][OH][OH-- ] = 1.0 x 10 ] = 1.0 x 10-14-14

Given: Given: [H[H+ + ] = 2.5 x 10] = 2.5 x 10-6-6 M M

Find:Find: [OH [OH-- ] ]

2.5 x 102.5 x 10-6-6 [OH [OH-- ] = 1.0 x 10 ] = 1.0 x 10-14-14

[OH[OH-- ] = ] = 1.0 x 101.0 x 10-14-14 = 4.0 x 10 = 4.0 x 10-9-9 M M 2.5 x 102.5 x 10-6-6

Chemistry 2



pHpH

Since the [HSince the [H++] is usually very small in ] is usually very small in aqueous solutions, we normally express aqueous solutions, we normally express the [Hthe [H++] in terms of] in terms of pH pH..

pH = - logpH = - log1010 [H [H++]]

pHpH

Example: Example: Calculate the pH of a neutral Calculate the pH of a neutral solution at 25solution at 25ooC.C.

Given: Given: neutral solutionneutral solution

Find:Find: pH pH

Neutral solution:Neutral solution:[H[H++] = 1.0 x 10] = 1.0 x 10-7-7 M M at 25at 25ooCC

pH = - log [HpH = - log [H++]]

pH = - log (1.0 x 10pH = - log (1.0 x 10-7-7) = - (-7.0) = 7.0) = - (-7.0) = 7.0

pHpH

Example: Example: Calculate the pH of a solution Calculate the pH of a solution with [Hwith [H++] = 2.5 x 10] = 2.5 x 10-5-5..

Given: Given: [H[H++] = 2.5 x 10] = 2.5 x 10-5-5

Find:Find: pH pH

pH = - log [HpH = - log [H++]]

pH = - log (2.5 x 10pH = - log (2.5 x 10-5-5) = - (-4.60) = 4.60) = - (-4.60) = 4.60

pHpH

Example: Example: Calculate the pH of a solution Calculate the pH of a solution with [OHwith [OH--] = 6.5 x 10] = 6.5 x 10-5-5..

Given: Given: [OH[OH--] = 6.5 x 10] = 6.5 x 10-5-5

Find:Find: pH pH

[H[H++] [OH] [OH--] = 1.0 x 10] = 1.0 x 10-14 -14 at 25at 25ooCC

[H[H++] = ] = 1.0 x 101.0 x 10-14-14 = 1.54 x 10 = 1.54 x 10-10 -10 MM6.5 x 106.5 x 10-5-5

pH = - log (1.54 x 10pH = - log (1.54 x 10-10-10) = - (-9.8) = 9.8) = - (-9.8) = 9.8

Find [HFind [H++] first.] first.

pH and pOHpH and pOH

The potential of the hydrogen ion was The potential of the hydrogen ion was defined in 1909 as defined in 1909 as the negative of the the negative of the logarithm of logarithm of [H[H++].].

pH = -log[H3O+] pOH = -log[OH-]

-logKW = -log[H3O+]-log[OH-]= -log(1.0x10-14)

KW = [H3O+][OH-]= 1.0x10-14

pKW = pH + pOH= -(-14)

pKW = pH + pOH = 14

pHpH

The negative log is also used to express The negative log is also used to express the magnitude of other small quantities:the magnitude of other small quantities:

pOH = - log [OHpOH = - log [OH- - ]]

pH and pOH are related by the following pH and pOH are related by the following equation that is derived by taking the equation that is derived by taking the negative log of the expression for Knegative log of the expression for Kww

pH + pOH = 14.00pH + pOH = 14.00

pHpH

Example:Example: Calculate the pOH of a solution Calculate the pOH of a solution with [OH with [OH - - ] = 2.5 x 10] = 2.5 x 10-3-3 M. M.

pOH = - log [OHpOH = - log [OH-- ] ]

pOH = - log (2.5 x 10pOH = - log (2.5 x 10-3-3))

pOH = 2.60pOH = 2.60

pHpH

Example:Example: Calculate the pH of a solution Calculate the pH of a solution with [OH with [OH - - ] = 2.5 x 10] = 2.5 x 10-3-3 M. M.

Given:Given: [OH [OH - - ] = 2.5 x 10] = 2.5 x 10-3-3 M M

Find:Find: pHpH

There are two There are two approaches that can approaches that can be used to solve this be used to solve this problem.problem.

Find pOH, thenFind pOH, thenpH + pOH = 14.00pH + pOH = 14.00

11

Find [HFind [H++], then ], then find pH.find pH.22

pHpH

pOH = - log [OHpOH = - log [OH-- ] ]

pOH = - log (2.5 x 10pOH = - log (2.5 x 10-3-3))

pOH = 2.60pOH = 2.60

pH = 14.00 – pOH = 14.00 – 2.60 = 11.40pH = 14.00 – pOH = 14.00 – 2.60 = 11.40

Given:Given: [OH [OH - - ] = 2.5 x 10] = 2.5 x 10-3-3 M MFind:Find: pHpH

Find pOH, thenFind pOH, then

pH + pOH = 14.00pH + pOH = 14.00

Approach # 1:Approach # 1:

pHpH

[H[H++] [OH] [OH-- ] = 1.00 x 10 ] = 1.00 x 10-14-14

[H[H++] = ] = 1.00 x 101.00 x 10-14-14 = 4.00 x 10 = 4.00 x 10-12-12 M M2.5 x 102.5 x 10-3-3

pH = - log (4.00 x 10pH = - log (4.00 x 10-12-12) = 11.40) = 11.40

Given:Given: [OH [OH - - ] = 2.5 x 10] = 2.5 x 10-3-3 M MFind:Find: pHpH

Find [H+], thenFind [H+], then

pH = -log [HpH = -log [H++]]

Approach # 2:Approach # 2:

pHpH

Given the pH of a solution, you can also Given the pH of a solution, you can also find the [Hfind the [H++] and the [OH] and the [OH--].].

Since pH = - log [HSince pH = - log [H++], ],

[H[H++] = 10 ] = 10 –pH–pH

Since pOH = - log [OHSince pOH = - log [OH--],],

[OH[OH--] = 10 ] = 10 -pOH-pOH

pHpH

Example:Example: What are the [H What are the [H++] and [OH] and [OH--] for a ] for a solution with a pH of 2.50 at 25solution with a pH of 2.50 at 25ooC?C?

pH = -log [HpH = -log [H++]]

[H[H++] = 10] = 10-pH-pH = 10 = 10 -2.5 -2.5 = 3.16 x 10 = 3.16 x 10-3-3 M M

[H[H++] [OH] [OH--] = 1.00 x 10] = 1.00 x 10-14-14

(3.16 x 10(3.16 x 10-3-3 M)[OH M)[OH--] = 1.00 x 10] = 1.00 x 10-14-14

[OH[OH--] = 3.16 x 10] = 3.16 x 10-12-12 M M

Chemistry 2

Unit 3-Next week (19.4 Unit 3-Next week (19.4 STRENGHT OF ACID AND BASE STRENGHT OF ACID AND BASE IN WATER IN WATER


Bronsted-Lowry Acids & Bases Bronsted-Lowry Acids & Bases (pg 103)(pg 103)

Strong Acid:Strong Acid:– acid that ionizes completelyacid that ionizes completely

no undissociated molecules in no undissociated molecules in solutionsolution

– Conjugate base formed by a strong Conjugate base formed by a strong acid is a very weak baseacid is a very weak base has negligible tendency to gain a has negligible tendency to gain a

protonproton– HCl, HHCl, H22SOSO44


Weak Acid:Weak Acid:– acid that ionizes partiallyacid that ionizes partially

aqueous solutions contain a mixture of aqueous solutions contain a mixture of acid molecules and the component acid molecules and the component ionsions

– Conjugate base formed by a weak acid is Conjugate base formed by a weak acid is a weak base.a weak base. slight tendency to gain protonsslight tendency to gain protons

– Acetic acid, citric acid, phosphoric acidAcetic acid, citric acid, phosphoric acid


Substances with negligible aciditySubstances with negligible acidity– substances that contain hydrogen but substances that contain hydrogen but

do not exhibit acidic behaviour in do not exhibit acidic behaviour in waterwater

– conjugate bases are very strong basesconjugate bases are very strong bases react completely with water to form react completely with water to form

OHOH – – ionion

– CHCH44, H, H22, OH, OH--


Summary:Summary:– An inverse relationship exists between An inverse relationship exists between

the strength of an acid and its the strength of an acid and its conjugate base or between a base and conjugate base or between a base and its conjugate acid.its conjugate acid.

– The stronger the acid, the weaker the The stronger the acid, the weaker the conjugate baseconjugate base

– The stronger the base, the weaker the The stronger the base, the weaker the conjugate acidconjugate acid

AcidAcidHClHCl

HH22SOSO44

HNOHNO33

HH33OO++ (aq) (aq)

HSOHSO44 - -

NHNH44++

HH22OO

OHOH--

CHCH44


BaseBaseClCl --

HSOHSO44 --

NONO33 --

HH22OO

SOSO44 2-2-

NHNH33

OH OH --

OO 2-2-

CHCH33 - -

Weak bases

Strongbases

negligible

Weak acids

negligible

Strongacids

The seven most common strong acids:The seven most common strong acids:– HClHCl hydrochloric acidhydrochloric acid– HBrHBr hydrobromic acidhydrobromic acid– HIHI hydroiodic acidhydroiodic acid– HNOHNO33 nitric acidnitric acid– HClOHClO33 chloric acidchloric acid– HClOHClO44 perchloric acidperchloric acid– HH22SOSO44 sulfuric acidsulfuric acid


You must know these acids by name and formula.

Example:Example: Use Fig. 16.4 to predict whether Use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left the equilibrium lies predominantly to the left or to the right:or to the right:


HSOHSO44-- (aq)(aq) + CO + CO33

22-- (aq)(aq) SO SO442-2- (aq)(aq) + HCO + HCO33

-- (aq)(aq)

acid base Conj.base

Conj.acid

HSO4 - is a stronger acid than HCO3

-.Equilibrium favors formation of the weaker acid.

Equilibrium lies to the right.

Strong AcidsStrong Acids

Strong acid:Strong acid:– Strong electrolyteStrong electrolyte– Ionizes completely in aqueous solutionIonizes completely in aqueous solution

HNOHNO33 (aq)(aq) H H ++ (aq)(aq) + NO + NO33 –– (aq)(aq)

The only significant source of HThe only significant source of H++ ion in ion in an aqueous solution of a strong acid is an aqueous solution of a strong acid is usually the strong acid.usually the strong acid.

HNOHNO33 (aq)(aq) H H ++ (aq)(aq) + NO + NO33 –– (aq)(aq)

In a 0.05 M HNOIn a 0.05 M HNO33 (aq) solution, (aq) solution,

[H[H++] = ] = 0.05 mol HNO0.05 mol HNO33 x x 1 mol H1 mol H++ = 0.05 M= 0.05 M

LL 1 mol HNO1 mol HNO33


Consequently, the [HConsequently, the [H++] in a solution of a ] in a solution of a strong monoprotic acid can be strong monoprotic acid can be determined easily using the determined easily using the concentration of the strong acid itself.concentration of the strong acid itself.


Example:Example: What is the pH of a 0.25 M HCl What is the pH of a 0.25 M HCl (aq) solution?(aq) solution?

Given:Given: [HCl] = 0.25 M [HCl] = 0.25 M

Find:Find: pHpH

Find [HFind [H++], ], then pHthen pH

[H[H++] = ] = 0.25 mol HCl0.25 mol HCl x x 1 mol H1 mol H++ = 0.25 M = 0.25 MLL 1 mol HCl1 mol HCl

pH = - log (0.25) = 0.60pH = - log (0.25) = 0.60

Strong BasesStrong Bases

Strong BaseStrong Base– strong electrolytestrong electrolyte– ionizes completely in aqueous solutionionizes completely in aqueous solution

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

Common strong basesCommon strong bases– alkali metal hydroxidesalkali metal hydroxides– hydroxides of Ca, Sr, and Bahydroxides of Ca, Sr, and Ba

Strong Bases Strong Bases

Strongly basic solutions are also formed Strongly basic solutions are also formed when certain basic materials react with when certain basic materials react with water to form the OHwater to form the OH-- ion. ion.

OO2-2- (aq)(aq) + H + H22O O (l)(l) 2 OH 2 OH-- (aq)(aq)

HH-- (aq)(aq) + H + H22O O (l) (l) H H22 (g)(g) + OH + OH-- (aq)(aq)


The pH of an aqueous solution of a The pH of an aqueous solution of a strong base can be determined using the strong base can be determined using the concentration of the strong baseconcentration of the strong base

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

A 0.25 M solution of NaOH has an [OHA 0.25 M solution of NaOH has an [OH--] ] of 0.25 M:of 0.25 M:

0.25 mol NaOH0.25 mol NaOH x x 1 mol OH1 mol OH- - = 0.25 M= 0.25 MLL 1 mol NaOH1 mol NaOH

0.25 M0.25 M


The pH of the base solution can then be The pH of the base solution can then be found in two ways:found in two ways:– Calculate pOH Calculate pOH

use pH + pOH = 14.00 to determine use pH + pOH = 14.00 to determine pHpH

– Calculate [HCalculate [H++]] use [Huse [H++] [OH] [OH--] = 1.00 x 10] = 1.00 x 10-14-14

Then calculate pHThen calculate pH

Step 1: Determine [OHStep 1: Determine [OH--]]

[OH[OH--] = ] = 0.25 mol Ca(OH)0.25 mol Ca(OH)22 x x 2 mol OH2 mol OH--

LL 1 mol Ca(OH)1 mol Ca(OH)22

= 0.50 M= 0.50 M


Example:Example: Calculate the pH of a 0.25 M Calculate the pH of a 0.25 M Ca(OH)Ca(OH)22 (aq) solution. (aq) solution.

Given: Given: [Ca(OH)[Ca(OH)22] = 0.25 M] = 0.25 M

Find:Find: pH pH

Step 1: Determine [OHStep 1: Determine [OH--]]

[OH[OH--] = ] = 0.25 mol Ca(OH)0.25 mol Ca(OH)22 x x 2 mol OH2 mol OH--

LL 1 mol Ca(OH)1 mol Ca(OH)22

= 0.50 M= 0.50 M


Step 2: Calculate pOHStep 2: Calculate pOH

pOH = - log [OHpOH = - log [OH--]]pOH = - log (0.50) = 0.30pOH = - log (0.50) = 0.30

Step 3: Calculate pHStep 3: Calculate pH

pH + pOH = 14.00pH + pOH = 14.00

pH = 14.00 - 0.30 = 13.70pH = 14.00 - 0.30 = 13.70


Example: Example: What is the pH of a solution What is the pH of a solution prepared by mixing 10.0 mL of 0.015 M prepared by mixing 10.0 mL of 0.015 M Ba(OH)Ba(OH)22 and 30.0 mL of 7.5 x 10 and 30.0 mL of 7.5 x 10-3-3 M NaOH? M NaOH?

Given:Given: 10.0 mL of 0.015 M Ba(OH) 10.0 mL of 0.015 M Ba(OH)22

30.0 mL of 7.5 x 1030.0 mL of 7.5 x 10-3 -3 M NaOHM NaOH

Find:Find: pHpH


Step 1: Find the total [OHStep 1: Find the total [OH--]]

[OH[OH--] = ] = moles OHmoles OH-- from NaOH + mol OH from NaOH + mol OH-- from Ba(OH) from Ba(OH)22

total volume in Ltotal volume in L

= = mmol OHmmol OH-- from NaOH + mmol OH from NaOH + mmol OH-- from Ba(OH) from Ba(OH)22

total volume in mLtotal volume in mL

mmol OHmmol OH--NaOH NaOH == 30.0 mL x 7.5 x 10 30.0 mL x 7.5 x 10-3-3M = M = 0.225 mmol0.225 mmol

mmol OHmmol OH--Ba(OH)2Ba(OH)2 = 10.0 mL x 0.015 M x = 10.0 mL x 0.015 M x 2 mol OH2 mol OH

1 mol Ba(OH)1 mol Ba(OH)22

= = 0.300 mmol0.300 mmol


Step 1 (cont)Step 1 (cont)

[OH[OH--] = ] = 0.225 mmol + 0.300 mmol0.225 mmol + 0.300 mmol10.0 mL + 30.0 mL10.0 mL + 30.0 mL

= 1.3 x 10= 1.3 x 10-2-2 M M

Step 2: Find pOHStep 2: Find pOH

pOH = - log [OHpOH = - log [OH--] = - log (1.3 x 10] = - log (1.3 x 10-2-2) = 1.89) = 1.89

Strong BasesStrong Bases

Step 3: Find pHStep 3: Find pH

pH + pOH = 14.00pH + pOH = 14.00

pH = 14.00 - pOH = 14.00 - 1.88 = 12.12pH = 14.00 - pOH = 14.00 - 1.88 = 12.12

4.4 IONIATION CONSTANTS OF 4.4 IONIATION CONSTANTS OF WEAK ACID AND WEAK WEAK ACID AND WEAK BASES (pg 107)BASES (pg 107)

KKaa and K and Kbb as a Measure of Acid and Base as a Measure of Acid and Base

StrenghtStrenght

Weak AcidsWeak Acids

Most acidic substances are Most acidic substances are weak acidsweak acids::– partially ionize in solutionpartially ionize in solution– the solution contains an the solution contains an equilibrium equilibrium

mixturemixture of acid molecules and its of acid molecules and its component ionscomponent ions

CHCH33COCO22HH HH++ (aq) + CH (aq) + CH33COCO22

-- (aq) (aq)

Acetic acidAcetic acid


The extent to which a weak acid ionizes The extent to which a weak acid ionizes can be expressed using an equilibrium can be expressed using an equilibrium constant known as the constant known as the acid-dissociation acid-dissociation constant (Kconstant (Kaa).).

For a general reaction:For a general reaction:HX HX (aq) (aq) H H++ (aq)(aq) + X + X- - (aq)(aq)

KKa a = [H= [H++][X][X--]]

[HX][HX]

Note: The rules for Note: The rules for writing an expression for writing an expression for KKaa are the same as those are the same as those

for Kfor Kcc, K, Kp p and K and Kspsp..


The magnitude of KThe magnitude of Kaa indicates the indicates the

tendency of the hydrogen ion in an acid tendency of the hydrogen ion in an acid to ionize.to ionize.

– The larger the value of KThe larger the value of Kaa, the stronger , the stronger

the acid is.the acid is.

The pH of a weak acid solution can be The pH of a weak acid solution can be calculated using the initial concentration calculated using the initial concentration of the weak acid and its Kof the weak acid and its Kaa..


To calculate the pH of a weak acid To calculate the pH of a weak acid solution:solution:– Write the ionization equilibrium for the Write the ionization equilibrium for the

acid.acid.– Write the equilibrium constant Write the equilibrium constant

expression and its numerical value.expression and its numerical value.– Set up a table showing initial Set up a table showing initial

concentration, change, equilibrium concentration, change, equilibrium concentration.concentration.

– Substitute equilibrium concentrations Substitute equilibrium concentrations into the equilibrium constant into the equilibrium constant expression.expression.


To calculate the pH of a weak acid To calculate the pH of a weak acid solution (cont):solution (cont):– Solve for the change in concentration.Solve for the change in concentration.

AssumeAssume that the change in that the change in concentration is small (i.e. < 5%) concentration is small (i.e. < 5%) compared to the initial compared to the initial concentration of the weak acid.concentration of the weak acid.

– Check the validity of previous Check the validity of previous assumptions.assumptions.

– Calculate the final concentrations and Calculate the final concentrations and pH.pH.


Example: Example: Calculate the pH of a 0.20 M Calculate the pH of a 0.20 M solution of HCN. Ksolution of HCN. Kaa = 4.9 x 10 = 4.9 x 10-10-10

Step 1: Write the equation for the Step 1: Write the equation for the ionization.ionization.

HCN (aq) HHCN (aq) H++ (aq) + CN (aq) + CN-- (aq) (aq)

Step 2: Write the expression for Ka.Step 2: Write the expression for Ka.

KKaa = [H = [H++] [CN] [CN--] = 4.9 x 10] = 4.9 x 10-10-10

[HCN][HCN]


Step 3: Set up a table.Step 3: Set up a table.


Initial 0.20 M 0.00 M 0.00 M

Change -x +x +x

Equil. 0.20 – x x x


Step 4: Substitute equilibrium Step 4: Substitute equilibrium concentrations into the Kconcentrations into the Kaa expression. expression.

KKaa = = (x) (x)(x) (x) = 4.9 x 10 = 4.9 x 10-10-10

(0.20 - x)(0.20 - x)

Step 5: Assume that x << 0.20 M and solve Step 5: Assume that x << 0.20 M and solve for x.for x.

KKaa = = (x) (x)(x) (x) = 4.9 x 10 = 4.9 x 10-10-10

0.200.20xx22 = (0.20) (4.9 x 10 = (0.20) (4.9 x 10-10-10) = 9.8 x 10) = 9.8 x 10-11-11

x = 9.9 x 10x = 9.9 x 10-6-6


Step 6: Check the validity of our Step 6: Check the validity of our assumption.assumption.

Is x << 0.20 M Is x << 0.20 M Is x less than 5% of 0.20 MIs x less than 5% of 0.20 M

9.9 x 109.9 x 10-6-6 x 100 = 0.005 % x 100 = 0.005 % 0.200.20

So the assumption is valid.So the assumption is valid.


Step 7: Substitute value for x into the Step 7: Substitute value for x into the table to find the [Htable to find the [H++].].


Initial0.20 M0.00 M0.00 MChange-9.9 x 10-6+9.9 x 10-6+9.9 x 10-6Equil.~ 0.209.9 x 10-69.9 x 10-6


Step 8: Calculate the pH using the [HStep 8: Calculate the pH using the [H++]]

pH = - log [HpH = - log [H++] = - log (9.9 x 10] = - log (9.9 x 10-6-6) = 5.00) = 5.00

Weak BasesWeak Bases

Many substances behave as weak bases Many substances behave as weak bases in water.in water.

Weak base + HWeak base + H22O O conjugate acid + OHconjugate acid + OH--

The extent to which a weak base reacts The extent to which a weak base reacts with water to form its conjugate acid and with water to form its conjugate acid and OHOH-- ion can be expressed using an ion can be expressed using an equilibrium constant known as the equilibrium constant known as the base-base-dissociation constant (Kdissociation constant (Kbb).).


KKbb always refers to the equilibrium in always refers to the equilibrium in which a base reacts with water to form which a base reacts with water to form its conjugate acid and OHits conjugate acid and OH-- ion. ion.

For the reaction:For the reaction:

NHNH33 (aq)(aq) + H + H22O O (l)(l) NH NH44++ (aq)(aq) + OH + OH-- (aq)(aq)

KKbb = [NH = [NH44++] [OH] [OH--]]

[NH[NH33]]Note: The rules for Note: The rules for writing an expression for writing an expression for KKbb are the same as those are the same as those

for Kfor Kcc, K, Kp p and Kand Kspsp..

Weak BaseWeak Base

To calculate the pH of a weak base To calculate the pH of a weak base solution:solution:– Write the ionization equilibrium for the Write the ionization equilibrium for the

base.base.– Write the equilibrium constant Write the equilibrium constant

expression and its numerical value.expression and its numerical value.– Set up a table showing initial Set up a table showing initial

concentration, change, equilibrium concentration, change, equilibrium concentration.concentration.

– Substitute equilibrium concentrations Substitute equilibrium concentrations into the equilibrium constant into the equilibrium constant expression.expression.

Weak Bases Weak Bases

To calculate the pH of a weak base To calculate the pH of a weak base solution (cont):solution (cont):– Solve for the change in concentration.Solve for the change in concentration.

AssumeAssume that the change in that the change in concentration is small (i.e. < 5%) concentration is small (i.e. < 5%) compared to the initial compared to the initial concentration of the weak base.concentration of the weak base.

– Check the validity of previous Check the validity of previous assumptions.assumptions.

– Calculate the [OHCalculate the [OH--] concentration and ] concentration and pOHpOH

– Use pOH to calculate pH.Use pOH to calculate pH.

Weak Bases Weak Bases

Example:Example: Calculate the pH of a 0.20 M solution of Calculate the pH of a 0.20 M solution of methylamine, CHmethylamine, CH33NHNH22. K. Kbb = 3.6 x 10 = 3.6 x 10-4-4..

Step 1: Write the equation for the ionization.Step 1: Write the equation for the ionization.

CHCH33NHNH22 (aq) + H (aq) + H22O CHO CH33NHNH33++ (aq) + OH (aq) + OH-- (aq) (aq)

Step 2: Write the expression for KStep 2: Write the expression for Kbb..

KKbb = [CH = [CH33NHNH33++] [OH] [OH--] = 3.6 x 10] = 3.6 x 10-4-4

[CH[CH33NHNH22]]


Step 3: Set up a table.Step 3: Set up a table.

CHCH33NHNH2 2 CHCH33NHNH33+ + + OH+ OH--

Initial0.20 M0.00 M0.00 MChange-x+x+xEquil.0.20 – xxx


Step 4: Substitute equilibrium Step 4: Substitute equilibrium concentrations into the Kconcentrations into the Kbb expression. expression.

KKbb = = (x) (x)(x) (x) = 3.6 x 10 = 3.6 x 10-4-4

(0.20 - x)(0.20 - x)

Step 5: Assume that x << 0.20 M and solve Step 5: Assume that x << 0.20 M and solve for x.for x.

KKbb = = (x) (x)(x) (x) = 3.6 x 10 = 3.6 x 10-4-4

0.200.20xx22 = (0.20) (3.6 x 10 = (0.20) (3.6 x 10-4-4) = 7.2 x 10) = 7.2 x 10-5-5

x = 8.5 x 10x = 8.5 x 10-3-3


Step 6: Check the validity of our Step 6: Check the validity of our assumption.assumption.

Is x << 0.20 M Is x << 0.20 M Is x less than 5% of 0.20 MIs x less than 5% of 0.20 M

8.5 x 108.5 x 10-3-3 x 100 = 4.2 % x 100 = 4.2 % 0.200.20

So the assumption is valid.So the assumption is valid.


Step 7: Substitute value for x into the Step 7: Substitute value for x into the table to find the [OHtable to find the [OH--].].

CHCH33NHNH2 2 CHCH33NHNH33+ + + OH+ OH--

Initial0.20 M0.00 M0.00 MChange-8.5 x 10-3+8.5 x 10-3+8.5 x 10-3Equil.~ 0.208.5 x 10-38.5 x 10-3


Step 8: Calculate the pOHStep 8: Calculate the pOH

pOH = - log [OHpOH = - log [OH--] = - log (8.5 x 10] = - log (8.5 x 10-3-3) = 2.07) = 2.07

Step 9: Calculate the pHStep 9: Calculate the pH

pH + pOH = 14.00pH + pOH = 14.00

pH = 14.00 - pOH = 14.00 - 2.07 = 11.93pH = 14.00 - pOH = 14.00 - 2.07 = 11.93

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3.0Unit3-Acid_and_Bases