Acids and bases Different concepts Calculations and scales.

Acids and basesAcids and bases

Different conceptsDifferent conceptsCalculations and scalesCalculations and scales

Learning objectivesLearning objectives

You will be able to:You will be able to:– Identify acids and bases according to Identify acids and bases according to

Arrhenius and Bronsted definitionsArrhenius and Bronsted definitions– Identify conjugate acid base pairsIdentify conjugate acid base pairs– Calculate pH for solutionsCalculate pH for solutions– Write dissociation constant expression KWrite dissociation constant expression Kaa

– Determine pH of weak acid/base solutionsDetermine pH of weak acid/base solutions– Predict pH of salt solutionsPredict pH of salt solutions– Identify factors that influence acid strengthIdentify factors that influence acid strength

ACIDS AND BASESACIDS AND BASES……for it cannot be for it cannot be But I am pigeon-liver’d and lack gall But I am pigeon-liver’d and lack gall To make oppression bitter…To make oppression bitter…HamletHamlet

Definitions of acids and basesDefinitions of acids and basespH scalepH scaleWeak acids and equilibriaWeak acids and equilibriaLewis acids and basesLewis acids and bases

ACIDS AND BASESACIDS AND BASES

The meaning of acid and base has The meaning of acid and base has evolved evolved

Arrhenius acid is one that Arrhenius acid is one that generatesgenerates protons when dissolved in waterprotons when dissolved in water

HA(aq) = HHA(aq) = H++(aq) + A(aq) + A--(aq)(aq)

Arrhenius base is one that Arrhenius base is one that generatesgenerates hydroxide ions when dissolved in waterhydroxide ions when dissolved in water

MOH(aq) = MMOH(aq) = M++(aq) + OH(aq) + OH--(aq)(aq)

BrBrøønsted and Lowrynsted and Lowry

A broader definition of acids and basesA broader definition of acids and bases

In the reaction NHIn the reaction NH33 + HCl = NH + HCl = NH44Cl has all Cl has all

the elements of acid-base neutralization the elements of acid-base neutralization but no Hbut no H22O as would be required in the O as would be required in the

Arrhenius definitionArrhenius definition

BrBrøønsted acid nsted acid donatesdonates a proton a proton

BrBrøønsted base nsted base acceptsaccepts a proton a proton

HH33OO++, H, H22O and OHO and OH-- all have the same all have the same

number of valence electrons – they are number of valence electrons – they are isoelectronicisoelectronic

NeutralizationNeutralization

The mixing of an acid with a base:The mixing of an acid with a base:

ACID + BASE = SALT + WATERACID + BASE = SALT + WATER

The reaction of carbonic acid (COThe reaction of carbonic acid (CO22 in H in H22O) O)

to give limestone:to give limestone:

HH22COCO33 + Ca(OH) + Ca(OH)22 = CaCO = CaCO33 + 2H + 2H22OO

The essence of neutralizationThe essence of neutralization

Elimination of the components of acid and Elimination of the components of acid and base by combination to give Hbase by combination to give H22OO

HH++ + OH + OH-- H H22OO

ACID BASE

BrBrøønsted acidnsted acid

HCl + HHCl + H22O = HO = H33OO++ + Cl + Cl--

DONOR ACCEPTOR

BrBrøønsted basensted base

HH22O + NHO + NH33 = NH = NH44++ + OH + OH--

waterwaterNHNH33 + HCl = NH + HCl = NH44

++ClCl--

No waterNo water

DONOR

ACCEPTOR

The products are themselves acids The products are themselves acids and basesand bases

HA + B HA + B ↔ BH↔ BH++ + A + A--

DONOR DONORACCEPTOR ACCEPTOR

Equilibrium: solution contains Equilibrium: solution contains mixture of all componentsmixture of all components

ACCEPTOR DONORDONOR ACCEPTOR

Conjugate acids and bases: follow Conjugate acids and bases: follow the protonthe proton

The difference is a protonThe difference is a proton

HHAA + + BB ↔ ↔ AA-- + H + HBB++

Conjugate acid-base pair Conjugate acid-base pair

Conjugate base Conjugate acid

base

acid

Substances can be both acids and Substances can be both acids and bases – depends on environmentbases – depends on environment

Note that in one instance HNote that in one instance H22O behaves O behaves

like a base – like a base – acceptingaccepting protons, and in protons, and in another, behaves like an acid – another, behaves like an acid – donatingdonating protonsprotons

HCl + HCl + HH22OO = H = H33OO++ + Cl + Cl--

In presence of an acid HIn presence of an acid H22O is a baseO is a base

NHNH33 + + HH22OO = NH = NH44++ + OH + OH--

In presence of a base HIn presence of a base H22O is an acidO is an acid

It’s a competition for protonsIt’s a competition for protons

The substance that is a stronger proton The substance that is a stronger proton donor becomes the aciddonor becomes the acid

HCl HCl + H+ H22O = HO = H33OO++ + Cl + Cl--

The substance that is the stronger proton The substance that is the stronger proton acceptor becomes the baseacceptor becomes the base

NHNH33 + H + H22O = NHO = NH44++ + OH + OH--

Strength and concentrationStrength and concentration

Not all acids completely donate the Not all acids completely donate the protons to water molecules in solutionprotons to water molecules in solution

HA + HHA + H22O O A A-- + H + H33OO++

The degree of ionization is described by The degree of ionization is described by strengthstrength

The total number of moles per unit volume The total number of moles per unit volume is described by concentrationis described by concentration

Changing concentration does not Changing concentration does not change strengthchange strength

StrengthStrength refers to degree of ionization: refers to degree of ionization:– Strong is completely ionized (100 %)Strong is completely ionized (100 %)– Weak is partly ionized (1 % - 1:10Weak is partly ionized (1 % - 1:1066))

ConcentrationConcentration refers to number of moles refers to number of moles per unit volumeper unit volume

An acid (or base) can be strong and An acid (or base) can be strong and concentrated, weak and concentrated, concentrated, weak and concentrated, strong and dilute, weak and dilutestrong and dilute, weak and dilute

Hydronium ion is the active ingredient Hydronium ion is the active ingredient of an acid in aqueous solutionof an acid in aqueous solution

Protons do not exist in solutionProtons do not exist in solution

CHCH33COCO22H + HH + H22O = O = HH33OO++ + CH + CH33COCO22--

Vinegar in water produces hydronium ionsVinegar in water produces hydronium ions

Hydroxide ion is the active ingredient of Hydroxide ion is the active ingredient of a base – in aqueous solutiona base – in aqueous solution

NHNH33 + H + H22O = NHO = NH44++ + + OHOH--

Ammonia, a base, dissolves in water and Ammonia, a base, dissolves in water and produces hydroxide ionsproduces hydroxide ions

AmphotericityAmphotericity

A substance that behaves as an acid and a base A substance that behaves as an acid and a base is amphoteric (amphiprotic). Water is a good is amphoteric (amphiprotic). Water is a good exampleexample

Acid Base

Ionization of waterIonization of water

Even in pure water, a fraction of the Even in pure water, a fraction of the molecules are ionized and the molecules are ionized and the concentrations of OHconcentrations of OH-- and H and H33OO++ are equal are equal

HH22O + HO + H22O = HO = H33OO++ + OH + OH--

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

In all aqueous solutions, product of In all aqueous solutions, product of concentrations is a constantconcentrations is a constant

[H[H33OO++][OH][OH--] = K] = Kww

At 25At 25°C, °C, [H[H33OO++] = 1 x 10] = 1 x 10-7-7MM

Increasing [HIncreasing [H33OO++] decreases [OH] decreases [OH--] ]

(acidic conditions)(acidic conditions)

Increasing [OHIncreasing [OH--] decreases [H] decreases [H33OO++]]

(basic conditions)(basic conditions)

Calculating [OHCalculating [OH--] from [H] from [H33OO++]]

[H[H33OO++][OH][OH--] = K] = Kww

[OH[OH--] = K] = Kww//[H[H33OO++]]

Example: if [HExample: if [H33OO++] = 1 x 10] = 1 x 10-3-3 M M

Then Then [OH[OH--] = 10] = 10-14-14/10/10-3-3 M M

= 10= 10-11 -11 MM

The pH scale – reduces large The pH scale – reduces large range of numbers to smallrange of numbers to small

In fact, [HIn fact, [H33OO++][OH][OH--] = 10] = 10-14-14 M M22

pH = - logpH = - log1010[H[H33OO++]]– Range of [HRange of [H33OO++] from 10 – 10] from 10 – 10-14-14 M M

– Range of pH from -1 - 14Range of pH from -1 - 14

LowLow pH = acid; pH = acid; highhigh pH = basic pH = basicFor example, 10For example, 10-1-1 M HCl has pH = M HCl has pH = 11

Pure water has [HPure water has [H33OO++] = 10] = 10-7-7 M, pH = M, pH = 77

Ammonia has [HAmmonia has [H33OO++] =10] =10-11-11 M, pH = M, pH = 1111

Note: change of 1 unit in pH is factor of ten in MNote: change of 1 unit in pH is factor of ten in M

Deconstructing the pHDeconstructing the pH

pH = - logpH = - log1010[H[H33OO++] = -log] = -log1010((aa x 10 x 10--bb))

= -log= -log1010aa + + bb

pH = pH = 33..0000

[H[H33OO++] = ] = 1.01.0 x 10 x 10--33 M M

pH = pH = 22..595595

[H[H33OO++] = ] = 2.552.55 x 10 x 10--33 MM

Exponent

Prefix – 2 S.F.

Exponent

Prefix – 3 S.F.

Indicating pHIndicating pH

All indicators involve a conjugate acid-base pair All indicators involve a conjugate acid-base pair which have different colourswhich have different colours

HIn(aq) + HHIn(aq) + H22O = HO = H33OO++(aq) + In(aq) + In--(aq)(aq)

Colour A Colour BColour A Colour B

Strong acids and bases and pHStrong acids and bases and pH

Monoprotic acids are completely ionizedMonoprotic acids are completely ionized

HCl + HHCl + H22O O →→ H H33OO++(aq)(aq) + + ClCl--(aq)(aq)

Polyprotic acids are not completely ionized Polyprotic acids are not completely ionized even if strong (stay tuned)even if strong (stay tuned)

Strong bases are completely dissociatedStrong bases are completely dissociated

NaOH + HNaOH + H22O O → Na→ Na++(aq) + OH(aq) + OH--(aq)(aq)

CaO + CaO + HH22OO → Ca → Ca22++(aq) + (aq) + 22OHOH--(aq)(aq)

Weak acids and equilibriaWeak acids and equilibria

HA(aq) + HHA(aq) + H22O O ↔ A↔ A--(aq) + (aq) + HH33OO++(aq) (aq)

pKpKaa = -log = -log1010KKaa

][

]][[ 3

HA

AOHKa

If the pH of 0.250 M HF is 2.036, what is KIf the pH of 0.250 M HF is 2.036, what is Kaa??

[[HH33OO++] = 9.20 x 10] = 9.20 x 10-3-3 M (antilog -2.036) M (antilog -2.036)

[F[F--] = ] = [[HH33OO++] = 9.20 x 10] = 9.20 x 10-3-3 M M

[HF] = 0.250 – [HF] = 0.250 – [[HH33OO++] = 0.250 - 9.20 x 10] = 0.250 - 9.20 x 10-3-3 M = 0.241 M M = 0.241 M

][

]][[ 3

HF

FOHKa

433

1052.3241.0

1020.91020.9

xxx

Ka

Pathway to pH in a weak acidPathway to pH in a weak acid

Step 1: species present before Step 1: species present before dissocationdissocation

HCN and HHCN and H22OO

Step 2: possible proton transfer processesStep 2: possible proton transfer processes

1.1. HCN(aq) + HHCN(aq) + H22O = HO = H33OO++(aq) + CN(aq) + CN--(aq)(aq)

KKaa = 4.9 x 10 = 4.9 x 10-10-10

2.2. HH22O + HO + H22O = HO = H33OO++(aq) + OH(aq) + OH--(aq)(aq)

KKww = 1.0 x 10 = 1.0 x 10-14-14

Step 3: identify principal processStep 3: identify principal process

Largest KLargest Kaa is principal process, the rest is principal process, the rest

are subsidiaryare subsidiaryAssume all HAssume all H33OO++(aq) is derived from (aq) is derived from

principal process (in this case dissociation principal process (in this case dissociation of HCN)of HCN)

OHOH-- is derived from subsidiary process (in is derived from subsidiary process (in this case dissociation of water)this case dissociation of water)

Step 4: ICE AgeStep 4: ICE Age

Principal Principal processprocess

HCN(aq) + HHCN(aq) + H22O = HO = H33OO++(aq) + CN(aq) + CN--(aq)(aq)

IInitial concnitial conc 0.100.10 00 00

CChangehange -x-x xx xx

EEquilibrium quilibrium concconc

0.10 - x0.10 - x xx xx

Step 5: Substitute for x and solveStep 5: Substitute for x and solve

Obtain quadratic expression in xObtain quadratic expression in xBut…x But…x « 0.01, so 0.10 – x ≈ 0.10« 0.01, so 0.10 – x ≈ 0.10

Assumption x << 0.1 simplifies solutionAssumption x << 0.1 simplifies solutionxx22 = 4.9 x 10 = 4.9 x 10-11-11, x = 7.0 x 10, x = 7.0 x 10-6-6

Need to check assumption: 7.0x10Need to check assumption: 7.0x10-6-6<<0.1<<0.1

x

xx

HCN

CNOHxKa 10.0][

]][[109.4 310

10.010.0

109.42

10 x

x

xxx

X << 0.1

Step 6: from x calculate Step 6: from x calculate concentrationsconcentrations

[[HH33OO++] = [CN] = [CN--] = ] = 7.0 x 107.0 x 10-6-6 M M

[HCN] = 0.10 – x = 0.10 - 7.0 x 10[HCN] = 0.10 – x = 0.10 - 7.0 x 10-6-6 = = 0.10 M0.10 M

Step 7: concentration of species Step 7: concentration of species from subsidiary processesfrom subsidiary processes

OHOH-- is derived from dissociation of H is derived from dissociation of H22OO

[OH[OH--] = K] = Kww//[[HH33OO++]]= 1.0x10= 1.0x10-14-14//7.0 x 107.0 x 10-6-6

= 1.4 x 10= 1.4 x 10-9-9 M M

So, if [OHSo, if [OH--] = 1.4 x 10] = 1.4 x 10-9-9 M, M, [[HH33OO++] from ] from dissociation of Hdissociation of H22O = 1.4 x 10O = 1.4 x 10-9-9 M M

Check: Check: [[HH33OO++] from HCN ] from HCN > > [[HH33OO++] from ] from HH22O (O (7.0 x 107.0 x 10-6 -6 > > 1.4 x 101.4 x 10-9-9))

Step 8: Calculate pHStep 8: Calculate pH

pH = -logpH = -log1010[[HH33OO++] = -log] = -log1010(7.0 x 10(7.0 x 10-6-6))

= 5.15= 5.15