Addition of Acid Addition of Acid or Base to or Base to SolutionSolution

CE 541CE 541

When adding Base or Acid to Solution:When adding Base or Acid to Solution:Interaction between different chemical species occurInteraction between different chemical species occurChemical equilibrium establishedChemical equilibrium establishedNature of changes depends on Nature of changes depends on

Nature of speciesNature of species Concentration of speciesConcentration of species

Changes can be predicted by solving equilibrium Changes can be predicted by solving equilibrium problemproblemIt is cumbersomeIt is cumbersomeAcid-base titration can be used to help understand Acid-base titration can be used to help understand the processthe process

What is Titration?What is Titration?It is a procedure by which a measured amount It is a procedure by which a measured amount of a chemical is added to solution in order to of a chemical is added to solution in order to bring about the desired change.bring about the desired change.

Titration of Strong Acids Titration of Strong Acids and Basesand Bases

Strong Base Titrated with Strong AcidStrong Base Titrated with Strong Acid High initial pH (12-13)High initial pH (12-13) Slowly declines to pH ≈ 10Slowly declines to pH ≈ 10 Rapid decline to pH ≈ 4Rapid decline to pH ≈ 4 Slow declineSlow decline Between pH 10 – 4 curve is verticalBetween pH 10 – 4 curve is vertical Equivalence point between pH 4 and 10Equivalence point between pH 4 and 10

Equivalence point is reached when Equivalence point is reached when equivalents of base (or acid) added equivalents of base (or acid) added equal to equivalents of base (or acid) equal to equivalents of base (or acid) initially presentinitially present

Strong Acid Titrated with Strong Acid Titrated with Strong BaseStrong Base

Titration ProblemsTitration ProblemsCan be solved using charge balance. If Can be solved using charge balance. If strong HCl is titrated with strong NaOHstrong HCl is titrated with strong NaOH

NaOH + HCl → HNaOH + HCl → H22O + NaO + Na++ + Cl + Cl--

Charge balanceCharge balance

[Na[Na++] + [H] + [H++] = [OH] = [OH--] + [Cl] + [Cl--]]

to get the exact pH, activities must be used. to get the exact pH, activities must be used. However, concentration can produce close However, concentration can produce close solution.solution.

IfIf

CC00 = molar concentration of HCl = molar concentration of HCl

VV00 = volume of HCl to be titrated = volume of HCl to be titrated

C = molar concentration of NaOHC = molar concentration of NaOH

V = volume of NaOH addedV = volume of NaOH added

Then:Then:

VV

CVOHH

VV

VC

0

00

0

][][

Re-arrangeRe-arrange

if C >> Cif C >> C00, then V can be ignored., then V can be ignored.

Study Example 4.18Study Example 4.18

1 f point, eequivalencat function; eequivalenc thecalled is

][][

1

][][1

00

0

00

000

00

VC

CVf

HH

Kf

VV

CV

HOHVC

CV

VV

CV

w

Titration of Weak Acids and Titration of Weak Acids and BasesBases

Weak Acid Titrated with Strong BaseWeak Acid Titrated with Strong Base Curve character depends on acidCurve character depends on acid

Monobasic → yields one HMonobasic → yields one H++ ion ion Polybasic → yields more than one HPolybasic → yields more than one H++

ionion Initial pH depends on degree of Initial pH depends on degree of

ionizationionization Most of natural waters contain weak Most of natural waters contain weak

acids and basesacids and bases

Titration of Monobasic Titration of Monobasic Weak Acid with Strong Weak Acid with Strong

BaseBaseAcid ionizes as follows:Acid ionizes as follows:

HA ↔ HHA ↔ H+ + + A+ A--

The equilibrium relationships are:The equilibrium relationships are:

)2(]][[

)1(][

]][[

w

A

KOHH

KHA

AH

when weak acid is titrated with when weak acid is titrated with strong basestrong base

HA + B + OHHA + B + OH-- → B → B++ + A + A-- + H + H22OO

BB++ is the cation associated with the is the cation associated with the strong base.strong base.

The charge balance isThe charge balance is

[H[H++] + [B] + [B++] = [A] = [A--] + [OH] + [OH--] (3)] (3)

[H[H++] can be determined by solving ] can be determined by solving equations (1), (2), and (3)equations (1), (2), and (3)

IfIf

CC00 = molar concentration of weak = molar concentration of weak acidacid

VV00 = volume of weak acid to be = volume of weak acid to be titratedtitrated

C = molar concentration of strong C = molar concentration of strong basebase

V = volume of strong base addedV = volume of strong base added

Then:Then:VV

VCAHA

0

00][][

mass balance of strong base cationmass balance of strong base cation

[B+] = (CV) / (V[B+] = (CV) / (V00 + V) + V)

combining mass balance and combining mass balance and equilibrium relationship of weak equilibrium relationship of weak acidacid

1

01

1

00

][1

][

][1

][

A

A

K

H

C

A

H

K

C

HA

The charge balance becomesThe charge balance becomes

RearrangeRearrange

The equation can be used to The equation can be used to generate titration curvegenerate titration curve

Study Example 19Study Example 19

VV

VCOHH

VV

CV

0

001

0

][][

00

01

00

][][ VC

VVH

H

Kf

VC

CV w

Beginning of TitrationBeginning of Titration

In the initial solution containing In the initial solution containing the weak acidthe weak acid

[B[B++] = 0] = 0 pH is low pH is low [OH [OH--] << [H] << [H++]]

So, charge balance becomesSo, charge balance becomes

[H[H++] ≈ [A] ≈ [A--]]

OriginallyOriginally

[H[H++] + [B] + [B++] = [A] = [A--] +[OH] +[OH--]]

Substitute in Substitute in

and solve for [Hand solve for [H++]]

At the beginning of titration (most At the beginning of titration (most of cases)of cases)

[A] << C[A] << C00

AKHA

AH

][

]][[

][][][ 0 ACKHAKH AA

so,so,

then,then,

Study Examples 20-21Study Examples 20-21

0][ CKH A

00 log2

1log

2

1log

2

1]log[ CpKCKHpH AA

Midpoint of TitrationMidpoint of Titration

As titration proceedsAs titration proceeds

HA + BHA + B++ + OH + OH-- → B → B++ + A + A-- + H + H22OO [A[A--] increases] increases [HA] decreases[HA] decreases

when neutralization reaches 50%when neutralization reaches 50% [B[B++] ≈ (1/2) C] ≈ (1/2) C00

[A[A--] increases significantly] increases significantly

so,so,

[H[H++] << [B] << [B++]]

[OH[OH--] << [A] << [A--]]

From charge balance From charge balance

([H([H++] + [B] + [B++] = [A] = [A--] + [OH] + [OH--])])

[B[B++] ≈ [A] ≈ [A--] ≈ (1/2) C] ≈ (1/2) C00

and since and since

[HA] + [A[HA] + [A--] = C] = C00

then then

[HA] ≈ (1/2) C[HA] ≈ (1/2) C00

Substitute inSubstitute inAKHA

AH

][

]][[

A

AA

pKpH

or

KHKC

CH

][

2

12

1][

0

0

Equivalence Point of Equivalence Point of TitrationTitration

At that point, equivalents of base At that point, equivalents of base added are equal to equivalents of added are equal to equivalents of acid in the original solution (f = 1)acid in the original solution (f = 1)So,So,

[B[B++] = C] = C00

The pH value is high enough to The pH value is high enough to assume assume [H[H++] << [OH] << [OH--]]

CC00 ≈ [A ≈ [A--] + [OH] + [OH--]]

CC00 – [A – [A--] ≈ [OH] ≈ [OH--]]

Since Since

CC00 – [A – [A--] = [HA]] = [HA]

Thus, Thus,

Substitute inSubstitute in

GivesGives

ww KOHHH

KOHHA

]][[....

][][][

AKHA

AH

][

]][[

21

][

][

1

AwKK

A

H

At the end of titration At the end of titration

[A[A--] ≈ C] ≈ C00

then,then,

wA pKpKCpH 0log2

1

When titrating weak base with When titrating weak base with strong acid, pH at various points strong acid, pH at various points can be determined using the same can be determined using the same approach.approach.

Beginning pointBeginning point

pH ≈ pKpH ≈ pKww – (1/2) pK – (1/2) pKBB + (1/2) log C + (1/2) log C00

MidpointMidpoint

pH ≈ pKpH ≈ pKww – pK – pKBB

Equivalence pointEquivalence point

pH ≈ (1/2) (pKpH ≈ (1/2) (pKww – pK – pKBB – logC – logC00))

And when titrating weak acid with And when titrating weak acid with strong basestrong base

Beginning of titrationBeginning of titration

pH ≈ (1/2)(pKpH ≈ (1/2)(pKAA – log C – log C00))MidpointMidpoint

pH ≈ pKpH ≈ pKAA

Equivalence pointEquivalence point

pH ≈ (1/2) (log CpH ≈ (1/2) (log C00 + pK + pKAA + pK + pKww))

Study Examples 22 and 23Study Examples 22 and 23

Calculate the pH of the equivalence point in Calculate the pH of the equivalence point in the titration of solutions containing the the titration of solutions containing the following concentrations of sodium following concentrations of sodium bicarbonate with sulfuric acid: (a) 10 mg/l; (b) bicarbonate with sulfuric acid: (a) 10 mg/l; (b) 100 mg/l; © 1000 mg/l100 mg/l; © 1000 mg/l

A 500ml solution contains 100 mg of NaOH. A 500ml solution contains 100 mg of NaOH. Calculate the followingCalculate the following1. initial pH of the solution1. initial pH of the solution2. pH after addition of 2 ml of 1 N H2. pH after addition of 2 ml of 1 N H22SOSO44

3. pH after addition of 4 ml of 1 N H3. pH after addition of 4 ml of 1 N H22SOSO44

4. ml of 1 N H4. ml of 1 N H22SOSO44 required to reach the required to reach the equivalence pointequivalence point

BuffersBuffers

are substances in solution that offer are substances in solution that offer resistance to changes in pH as acids resistance to changes in pH as acids or bases are added or formed within or bases are added or formed within the solutionthe solution

buffer solutions usually contain buffer solutions usually contain mixtures of weak acids and their mixtures of weak acids and their salts or weak bases and their acids salts or weak bases and their acids (conjugates)(conjugates)

ifif ][

]][[

HA

AHK A

RearrangeRearrange

So, pH of a buffer solution So, pH of a buffer solution depends on the ratio of salt depends on the ratio of salt concentration to acid concentration to acid concentrationconcentration

Study Example 24Study Example 24

][

][log

][

][1

][

1

acid

saltpKpH

thus

HA

A

KH

A

A

Calculate the pH of a buffer solution prepared Calculate the pH of a buffer solution prepared by mixing 2.4 g of acetic acid (CHby mixing 2.4 g of acetic acid (CH33COOH) and COOH) and 0.73 g of sodium acetate (CH0.73 g of sodium acetate (CH33COONa) in 1 COONa) in 1 liter of water.liter of water.

Buffering Index or Buffering Buffering Index or Buffering Capacity can be determined by:Capacity can be determined by:

where β is the buffering capacity where β is the buffering capacity or index of weak acid plus its or index of weak acid plus its conjugate salt.conjugate salt.If we have weak salt (base) plus its If we have weak salt (base) plus its conjugate acid, thenconjugate acid, then

KKAA is replaced by (K is replaced by (Kww / K / KBB))

Study Example 25Study Example 25

][

][][

][303.2

HK

HKCH

H

K

A

ATw