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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 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
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))