Denise Lowinsohn [email protected] ...3 was prepared from 0.85 mL of the concentrated acid in 250 mL of distilled water. What is the pH of this prepared solution? The concentrated

Analytical chemistry

Denise Lowinsohn [email protected]

http://www.ufjf.br/nupis

2018

Date Activities

21/03/18 -Aqueous-solution chemistry

28/03/18 -Neutralization tritations

04/04/18 -Precipitation

11/04/18 -Precipitation tritimetry

18/04/18 -Oxidation/reduction

25/04/18 -Oxidation/reduction titration

02/05/18 -Complex-formation titrations: part 1

09/05/18-Complex-formation titrations: part 2

-Preparing samples for analysis

16/05/18 1º Test (100 points)

23/05/18 -Preparing samples for analysis

30/05/18-Preparing samples for analysis: activity delivery (100 points)

-Application of Statistics to Analytical Chemistry

06/06/18 -Application of Statistics to Analytical Chemistry

13/06/18 Holiday

20/06/18-Application of Statistics to Analytical Chemistry

-Sampling experiment (100 points)

27/06/18 2º Test (100 points)

04/07/18 Oral presentation (100 points)

SCHEDULE

Aqueous-solution chemistry

References

Brown, LeMay e Bursten, Química - A ciência central, 9ª edição, Editora

Pearson – Prentice Hall, 2005.

Daniel C. Harris, Análise Química Quantitativa, Editora LTC, 5a edição, 2001.

Skoog, West, Holler, Fundamentals of analytical chemistry, 7th edition, 1995.

The behaviour of acids and bases is very important in all areas of Chemistry and

others areas of Science.

Industrial processes,

Laboratory and

Biological

Effect of pH - The pH of the medium is an extremely important parameter for many

reactions in Analytical Chemistry.

Acid-base equilibrium

Acid and base: a brief review

Acid: taste sour and cause colour changes in pigments.

Base: bitter taste and slippery feeling.

Arrhenius: In aqueous medium, acids are defined as substances that

increase [H+] and bases increase [OH-]

Acids = substances that produce H3O+ (H+) ions, when dissolved in water

Bases = substances that produce OH- ions, when dissolved in water

Arrhenius: acid + base salt + water.

Problem: the definition applies only to aqueous solutions.

Brønsted-Lowry Theory

Brønsted-Lowry: acid – donates protons and base - accepts protons

Transference of “H+” ion between two substances

Conjugate acid: is the species formed when a base accepts a proton.

Conjugate base: is the species formed when an acid loses a proton.

A1 + B2 ⇌ A2 + B1 (conjugate acids and bases)

species that

donates

protons

(acid 1)

species that

accepts

protons

(base 2)

derived from

base 2

(acid 2)

derived from

acid 1

(base 1)

The most used concept in Analytical Chemistry.

The ion H+ in water

• The H+ ion is a proton without electrons.

• In water, the H+(aq) formes clusters.

• The H+ ion interacts with the nonbonding electron pair of the H2O

molecules to form the hydrogen ions hydrates: hydronium ion

• The most simple cluster is formed by interaction of one proton with one

H2O molecule.

• We can use both: H+(aq) or H3O+(aq).

Acids: can be uncharged molecules (HCl), anions (HSO4-), cations (NH4

+)

Bases: can be uncharged molecules (NH3), anions (Cl-)

Amphoteric (or Amphiprotic) substances: behave as acids or as bases (H2O)

Examples:

HNO2 + H2O ⇌ NO2- + H3O

+

species that

donates

H+

(acid 1)

species that

accepts

protons

(base 2)

derived from

base 2

(acid 2)

derived from

acid 1

(base 1)

H2O + NH3 ⇌ OH- + NH4+

species that

donates H+

(acid 1)

species that

accepts

protons

(base 2)

derived from

base 2

(acid 2)

derived from

acid 1

(base 1)

Brønsted-Lowry Theory

• The stronger the acid, the weaker the conjugate base.

• H+ is the strongest acid that can exist in equilibrium in aqueous solution.

• OH- is the strongest base that can exist in equilibrium in aqueous solution.

Strength of acids and bases

Strong and weak acids/bases

Acids

Strong totally dissociated (ex: HCl, HNO3)

Weak partially dissociated (ex: H3PO4, CH3COOH)

HCl(aq) + H2O(l) ⇾ H3O+(aq) + Cl-(aq)

Bases

Strong totally dissociated (ex: NaOH)

Weak partially dissociated (ex: NH3)

NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq)

Amphiprotic substances

Substances that have both acidic and basic properties. They behave as acids or

bases depending on the medium.

Ex.: H2PO4-, HCO3

-, H2O

Amphiprotic solvents: solvents, depending of the medium, behave as acid or base.

Protic solvent: solvents with H+ reactive.

All protic solvent suffers autoprotolysis.

Aprotic solvent: solvents without H+ reactive.

Autoprotolysis or autoionization: involves the spontaneous reaction of molecules of a

substance to give a pair of ions.

Ion product constant for water

Aqueous solutions contain small amount of hydronium and hydroxide ions as

a consequence of the dissociation reaction:

2H2O(l) ⇌ H3O+(aq) + OH-(aq)

At 25oC

14

3

2

23

2

2

3

101]].[[

].[]].[[

][

]].[[

xKOHOH

OHKOHOH

KOH

OHOH

w

eq

eq

Ion product constant for water

The concentration of

water in dilute aqueous

solutions is enormous

when compared with

the concentration of

hydrogen and

hydroxide ions.

CONSTANT

Lewis Theory

Acid = accepts a pair of electrons

Base = donates a pair of electrons

Lewis acids and bases don’t need contain protons.

Example:

Fe3+(aq) + SCN-

(aq) ⇌ Fe(SCN)2+(aq)

Lewis acid:

accepts a pair of electrons

Lewis base:

donates a pair of electrons

The defintion of Lewis is the most general definition of acids and bases.

pH scale pH = -log[H3O+]

Neutral solution: [H3O+] = [OH-]

[H3O+] = [OH-] = 1.0 x 10-7 mol L-1

Acid solution: [H3O+] > [OH-]

[H3O+] > 1.0 x 10-7 mol L-1 and

[OH-] < 1.0 x 10-7 mol L-1

Alkaline solution: [H3O+] < [OH-]

[H3O+] < 1.0 x 10-7 mol L-1 and

[OH-] > 1.0 x 10-7 mol L-1

In the most

solutions, the

[H+(aq)] is

very small.

•Most of the pH and pOH values are between 0 and 14.

•There are no theoretical limits on pH or pOH values. (for example, pH from 2.0 mol L-1 HCl

solution is -0.301.)

SÖRENSEN introduced in 1909, the

concept of pH, a conveniente way of

expressing acidity – the negative

logarithm of hydrogen ion concentration.

C)(25 14pOH pH pK

]].[[

0

w

3

OHOHKw

[H+] (mol L-1) pH pOH Example

1x10-0 0 14 Battery acid

1x10-1 1 13 Gastric acid

1x10-2 2 12 Lemon juice

1x10-3 3 11 Orange juice, soda

1x10-4 4 10 Tomato juice, acid rain

1x10-5 5 9 Black coffee, bananas

1x10-6 6 8 Urine, milk

1x10-7 7 7 Pure water

1x10-8 8 6 Sea water, eggs

1x10-9 9 5 Baking soda

1x10-10 10 4 Milk of magnesia

1x10-11 11 3 Ammonia solution

1x10-12 12 2 Soapy water

1x10-13 13 1 Bleach, oven cleaner

1x10-14 14 0 Liquid drain cleaner

Practicing.....

What are the molar concentration of H+ and the pH in:

a) 0.010 mol L-1 KOH?

b) 1.8x10-9 mol L-1 NaOH?

A sample of lemon juice with [H+] = 3.8x10-4 mol L-1. What is the pH?

A solution for cleaning windows is commonly available with [H+] = 5.3x10-9 mol L-1.

What is the pH of this solution?

A sample of apple juice freshly squeezed has pH = 3.76. What is [H+]?

A solution formed by the dissolution of an antacid tablet has pH = 9.18. What is

[H+]?

Strong acids

•The strongest common acids are HCl, HBr, HI, HNO3, HClO3, HClO4, e H2SO4.

•Strong acids are strong electrolytes.

•All strong acids are totally dissociated in aqueous solutions. No undissociated

solute molecules. The equilibrium of the reaction is totally shifted towards the

products:

HNO3(aq) + H2O(l) H3O+(aq) + NO3

-(aq)

Calculation: pH of 0.010 mol L-1 strong

acid solution

[ ] The concentration expressed in brackets represents the concentration (mol L-1)

at equilibrium.

C Analytical concentration, represents the real amount of the substance added in

certain solvent to form a solution of known concentration “C”.

HNO3(aq) H+(aq) + NO3-(aq)

Initial 0.010 mol L-1 - -

Equilibrium - 0.010 mol L-1 0.010 mol L-1

CHNO3 = 0.010 mol L-1 total amount of HNO3 present in solution

Concentration at equilibrium: [H3O+] [NO3

-] = 0.010 mol L-1 not considering

autoionization of H2O


acid solution

[ ] The concentration expressed in brackets represents the concentration (mol L-1)

at equilibrium.

C Analytical concentration, represents the real amount of the substance added in

certain solvent to form a solution of known concentration “C”.

HNO3(aq) H+(aq) + NO3-(aq)



CHNO3 = 0.010 mol L-1 total amount of HNO3 present in solution

Concentration at equilibrium: [H3O+] [NO3

-] = 0.010 mol L-1 not considering


pH = - log[H3O+]

[H3O+] = [NO3

-] = CHNO3

pH = -log(C) = -log 0.010

pH = 2.0

What is the pH of 0.040 mol L-1 HClO4 solution?

HNO3, pH = 2.34. What is the molar acid concentration?

A solution of HNO3 was prepared from 0.85 mL of the concentrated acid in 250

mL of distilled water. What is the pH of this prepared solution?

The concentrated acid has 69.5% m/m and density 1.40 g cm-3. (M.W. = 63 g

mol-1)

What is the [H+] and pH of each solutions?

a) 0.0020 mols of HCl in 500 mL of solution

b) 0.15 g de HNO3 (M.W. = 63 g mol-1) in 300 mL of solution

c) 10.0 mL de HCl 15 mol L-1 in 750 mL of solution

Practicing.....

Strong bases

•Most ionic hydroxides are strong bases (for example, NaOH, KOH, e Ca(OH)2).

•Strong bases are strong electrolytes and totally dissociated in solution.

•To be a base an hydroxide need to be soluble.

•The bases don’t need to contain OH- ion:

O2-(aq) + H2O(l) 2OH-(aq)

H-(aq) + H2O(l) H2(g) + OH-(aq)

N3-(aq) + 3H2O(l) NH3(aq) + 3OH-(aq)

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



CNaOH = 0.010 mol L-1 total amount of NaOH present in solution

Concentration at equilibrium: [Na+] [OH-] = 0.010 mol L-1 not considering


pOH = - log[OH-]

[Na+] = [OH-] = CNaOH

pOH = -log(C) = -log 0.010

pOH = 2.0

pKw = pH + pOH

14.0 = pH + 2.0

pH = 12.0


base solution

What is the pH of:

a) 0.028 mol L-1 NaOH solution?

b) 0.0011 mol L-1 Ca(OH)2 solution?

What is the molar concentration of:

a) KOH, pH = 11.89?

b) Ca(OH)2, pH = 11.68?

Practicing.....

Considerations

If the concentration of the strong acid (C) or the strong base (C) is:

1) C 10-6 mol L-1 - simplified calculation:

pH = -log C (strong acid) or pOH = -log C (strong base)

2) C 10-8 mol L-1 - autoionization of water.

3) 10-6 mol L-1 C 10-8 mol L-1 – Effect of autoionization of solvent and acid

or base are comparable – systematic calculation

Weak acids

•Partially dissociated in solution.

•React with the solvent (H2O) by donating a proton.

•Contain significant quantities of both the parent acid and its conjugate base.

•Accordingly, the weak acids are in equilibrium.

HA(aq)+ H2O(l) ⇌ H3O+(aq) + A-(aq) or HA(aq) ⇌ H+(aq) + A-(aq)

][

]][[ 3

HA

AOHKa

EQUILIBRIUM CONSTANT EXPRESSION Very small value

][

]][[

HA

AHKa

or

Note that [H2O] is omitted in the

expression of Ka. (H2O is a pure

liquid)The higher Ka, the stronger is the acid (in this

case, more ions are present in equilibrium

with respect to non-ionized molecules).

Ka >> 1, the acid is totally dissociated and the acid is strong.

Weak acids

Mass-balance and charge-balance

Mass-balance – relate the equilibrium concentrations of various species in a

solution to one another and to the analytical concentrations of the various

solutes.

Charge-balance – electrolyte solutions are electrically neutral even though they

may contain many millions of charged ions. Molar concentration of positive

charge = Molar concentration of negative charge

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

Initial C - - -

Equilibrium C-x - x x

Equilibrium: CB: [H3O+] = [A-] + [OH-]

MB: C = [HA] + [A-]

Practicing.....

Write mass-balance expressions for a 0.0100 mol L-1 solution of HCl that is

in equilibrium with an excess of solid BaSO4.

Write mass-balance expression for the system formed when a 0.010 mol L-1

NH3 solution is saturated with AgBr.

Write a charge-balance equation for 0.100 mol L-1 solution of sodium

chloride.

Write a charge-balance equation for 0.100 mol L-1 solution of magnesium

chloride.

Relations

C/Ka > 102 simplified calculation (by simplifying the calculation: the error

will be less than 5%)

C/Ka ≤ 102 systematic calculation (quadratic equation)

Use Ka to calculate pH

9.2)104.1log(]log[

104.1][][

10.0][

]][[1085.1

3

13

25

xHpH

molLxOAcHx

x

x

HOAc

OAcHxKa

HOAc(aq) ⇌ H+(aq) + OAc-

(aq)

Initial 0.10 - -

Equilibrium 0.10-x x x

C / Ka > 102 simplified calculation

Problems

Find the pH of 0.25 mol L-1 propanoic acid (dissociation constant = 1.3x10-5).

Find the pH of 1000 mmol L-1 HCN solution. Ka = 4.9x10-10

Ka for niacin is 1.6x10-5. What is the pH of 1 mmol L-1 niacin solution?

0.01 mol L-1 weak monoprotic acid solution has pH = 2.38 at 250C. What is the

value of dissociation constant (Ka) of this acid?

HA(aq)+ H2O(l) ⇌ H3O+(aq) + A-(aq) or HA(aq) ⇌ H+(aq) + A-(aq)

][

]][[ 3

HA

AOHKa

degree of dissociation

cTOTAL = [HA] + [A-]

[H+] = [A-] = c

1)1(

. 22 c

c

cc

cc

xccKa

Weak acids

][

]][[

HA

AHKa

or

EQUILIBRIUM CONSTANT EXPRESSION Very small value

Weak acids – degree of dissociation

Degree of dissociation is the fraction of a mole of the reactant that underwent

dissociation:

][][

][][

AHA

A

C

A

TOTAL

The higher the degree of dissociation, the stronger is the acid.

amount of substance of reactant dissociated

amount of substance of reactant present initially

% Ionization = strength of acid or base

HOAc(aq) ⇌ H+(aq) + OAc-

(aq)

Initial 0.10 - -

Equilibrium 0.10-x x x

%4.1100][

][%

104.1][][

10.0][

]][[1085.1

0

.

13

25

xHOAc

Hionization

molLxOAcHx

x

x

HOAc

OAcHxK

equ

a

WEAK

ACID

C / Ka > 102 simplified

calculation

Acid acetic solution pH is 3.26, what is the acid concentration? Calculate the %

of ionization . Ka = 1.8x10-5

Calculate the percentage of HF molecules (Ka = 6.8x10-4) ionized in:

a) 0.10 mol L-1 HF solution

b) 0.010 mol L-1 HF solution

Problems

Ionic Strength

Systematic studies have shown that the effect of added electrolyte on equilibria is

independent of the chemical nature of the electrolyte but depends upon a property

of the solution called the IONIC STRENGTH.

2

2

1iizc

µ = ionic strength

ci = ionic concentration

zi = charge of the ion

It is a measure of the ions total concentration in solution.

The more charged an ion, the higher its participation in the calculation.

The ionic strength is, however, greater than the molar concentration if the solution

contains ions with multiple charges.

Activity

In order to describe quantitatively the effective concentration of participants in an

equilibrium at any given ionic strengh, chemists use a term called activity, a.

(aA+.aB-) / (aAB) = KdissociationDISSOCIATION CONSTANT

ACTIVITY = CONCENTRATION X ACTIVITY COEFFICIENT

aA+ = [A+].A+ aB- = [B-].B- aAB = [AB].AB

ondissociatiBA K

AB

BA

AB

..

][

]].[[

(activity coefficient) varies with concentration!!!!!!

AB ⇌ A+ + B-

• In very dilute solutions, where the ionic strength is minimal and the activity

coefficient is unity. Under such circumstances, the activity and the molar

concentration of the species are identical.

• In solutions that are not too concentrated, the activity coefficient for a given

species is independent of the nature of the electrolyte and dependent only

upon the ionic strength.

• The activity coefficient of an uncharged molecule is approximately unity,

regardless of ionic strength.

Activity - properties

The Debye-Hückel Equation

In 1923, Peter Debye and Erich Hückel developed a theory that would allow us to

calculate the mean ionic activity coefficient of the solution, .

The Debye-Hückel theory is based on three assumptions of how ions act in

solution:

1-Electrolytes completely dissociate into ions in solution.

2-Solutions of electrolytes are very dilute, on the order of 0.01 mol L-1.

3-Each ion is surrounded by ions of the opposite charge, on average.

X

XX

z

3,31

509.0log

2

X = activity coeficient of the species X

µ = ionic strength of the solution

zX = charge on the species X

αX = effective diameter os the hydrated ion X in nm

Calculate the ionic strength of:

a) 0.1 mol L-1 solution of KNO3

b) 0.1 mol L-1 solution of Na2SO4

What is the ionic strength of a solution that is 0.05 mol L-1 in KNO3 and 0.1 mol L-1

in Na2SO4?

Calculate the activity coefficient for Hg2+ in a solution that has an ionic strength of

0.085. Use 0.5 nm for the effective diameter of the ion.

Use activities to calculate the H3O+ ion concentration in a 0.120 mol L-1 solution of

HNO2 that is also 0.05 mol L-1 in NaCl. Find the relative error introduced by

neglecting activities in this calculation.

αH3O+ = 0.9 nm, αNO2

- = 0.3 nm e Ka = 5.1x10-4

Problems

Weak bases

B(aq) + H2O(l) ⇌ BH+(aq) + OH-(aq)

][

]][[

B

OHBHKb

][][

][][

BHB

BH

C

BH

TOTAL

•Partially dissociated in solution.

•React with the solvent (H2O) by accepting a proton.

•Contain significant quantities of both the parent base and its conjugate acid.

Weak bases - types

•Bases usually have solitary pairs or negative charges to attack protons.

•Neutral weak bases contain nitrogen.

•The amines are ammonia-related and have one or more N-H bonds substituted by

N-C bonds (for example, CH3NH2 is methylamine).

• Anions of weak acids are also weak bases. Example: OCl- is the conjugate base

of HOCl (weak acid):

ClO-(aq) + H2O(l) ⇌ HClO(aq) + OH-(aq) Kb = 3.3x10-7

Cocaine is an example of weak base with Kb = 2.6 x 10-6. Calculate the pH of

0.0372 mol L-1 solution of this base.

C

H

O

H

O C

OCH3

O

N

CH3.. CH3

N

O

OCH3

H

O C

C

H

O

H+

+ OH-(aq)

0372.00372.0106.2

][

]][[ 226 x

x

xx

B

OHBHKb

0.0372 / 2.6 x 10-6 > 102 YES

141011.3][OH molLx 51.3pOH 49.10pH

Use Kb to calculate pH

Problems

Calculate the concentration of OH- of the 0.15 mol L-1 NH3 solution. Kb = 1.8x10-5

Which of the following compounds should produce the highest pH as a solution

of 0.05 mol L-1: pyridine (Kb = 1.7x10-9), methylamine (Kb = 4.4x10-4) or nitrous

acid (Ka = 4.0x10-4)?

Ephedrine is a medication and stimulant. It is often used to prevent low blood

pressure during spinal anesthesia. This compound is a weak organic base:

C10H5ON(aq) + H2O(l) ⇌ C10H5ONH+(aq) + OH-(aq)

0.035 mol L-1 ephedrine solution has pH = 11.33.

a) Calculate the concentration at equilbrium of C10H5ON, C10H5ONH+ and OH-

b) Find Kb for the ephedrine.

pKa = -log Ka or pKb = -log Kb

Strength of acid or base

Compound pKa

Acetic 4.76

Boric 9.24

Oxalic 1.27 and 4.27

Amonnia 9.24

The higher pKa, the weaker is the acid and the stronger is the base.

Fractional composition

Fraction of HA in the form A-:

Fraction of HA in the form HA:

][][

][

][][][

][][

HAA

B

KH

K

HAA

A

c

AB

a

a

TA

][][

][

][

][

][][

][][

HAA

BH

KH

H

HAA

HA

c

HABH

aT

HA

][

][][

]][[][

][

]][[

H

HAKAor

K

AHHA

HA

AHK a

a

a

HA(aq) ⇌ H+(aq) + A-(aq)

B(aq) + H2O(l) ⇌ BH+(aq) + OH-(aq)

Problems

pKa for benzoic acid (HA) is 4.20. Find the concentration of A- at pH 5.31 if the

concentration of HA is 0.0213 mol L-1.

Ka for the ammonium ion, NH4+ is 5.70x10-10. Find the fraction in the form BH+ at

pH = 10.38.

The acid HA has pKa = 3.00. Find the fraction in the form HA and the fraction in

the form A- at pH = 2.00, pH = 3.00, and pH = 4.00. Compute the quotient

[HA]/[A-] at each pH.

Relation between Ka and Kb

An important relationship exists between Ka and Kb of a conjugate acid-base

pair in aqueous solution. We can derive this result with the acid HA and its

conjugate base A-.

HA(aq) ⇌ H+(aq) + A-(aq)

][

]][[

HA

AHKa

A-(aq) + H2O(l) ⇌ HA(aq) + OH-(aq)

][

]][[

A

OHHAKb

H2O(l) ⇌ H+(aq) + OH-(aq)

+

baw KxKK

baw pKpKpK

Problems

Which is the weaker acid: acetic (Ka = 1.58 x 10-5) or lactic (Ka = 1.58 x 10-4)?

The quinoline base has the following structure:

The respective conjugated acid is described in the chemistry books as having a

pKa of 4.90. What is the basic dissociation constant of quinoline?

Neutralization

25.00 mL of 0.020 mol L-1 NaOH solution added to 10.00 mL of 0.025 mol L-1

HCl solution, find the pH of final solution.

NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)

Initial 5x10-4mol 2,5x10-4mol - -

Neutralization





Reacted 2.5x10-4mol 2.5x10-4mol 2.5x10-4mol 2.5x10-4mol

Neutralization





Reacted 2.5x10-4mol 2.5x10-4mol 2.5x10-4mol 2.5x10-4mol

Final 2.5x10-4mol - 2.5x10-4mol 2.5x10-4mol

85.1115.2 pHpOH

13

3

4

101.71035

105.2][

Lmolx

x

xOH

Problems

25.00 mL of 0.010 mol L-1 KOH solution added to 10.00 mL of 0.050 mol L-1 HCl

solution, find the pH of final solution.

10.00 mL of 0.6 mol L-1 nitric acid solution added to 0.2 g of NaOH (M.W. = 40 g

mol-1), find the pH of final solution.

To obtain a pH = 2.0, which volume of 2 mol L-1 perchloric acid should be added

to 0.4 g of NaOH (M.W. = 40 g mol-1)?

50.00 ml of 0.6 mol L-1 hydrochloric acid was added to a certain volume of 3 mol

L-1 NaOH and a pH = 12.0 was obtained. What is the initial volume of the NaOH

solution?

Polyprotic acids

• Many acids have more than one acidic proton polyprotic acids

• Series of acid dissociation steps, each characterized by it own acid

dissociation constant.

Example: diprotic system - sulfurous acid (H2SO3)

H2SO3(aq) ⇌ H+(aq) + HSO3-(aq) Ka1 = 1.7x10-2 (1st dissociation constant)

HSO3-(aq) ⇌ H+(aq) + SO3

2-(aq) Ka2 = 6.4x10-8 (2nd dissociation constant)

The decrease in the acid dissociation constant form Ka1 to Ka3, tell us that each

successive proton is harder to remove.

• The calculation of the pH solutions of the most polyprotic acids can be

simplified.

• For most of the polyprotic acids Ka1 is sufficiently larger than Ka2, to allow the

calculation of the concentration of the hydronium ion [H3O+], ignoring the

second ionization. The error in calculating the pH through this approximation is

minimal for most cases.

• Most polyprotic acids behave as weak monoprotic acid (Ka Ka1).

• Ka1 / Ka2 >103 only Ka1

Polyprotic acids

• Bases that can accept more than one proton polyprotic bases

• Series of base dissociation steps, each characterized by it own base

dissociation constant

Example: diprotic system – carbonate ion (CO32-)

CO32-(aq) + H2O(l ) ⇌ OH-(aq) + HCO3

-(aq) Kb1 = 1.8x10-4

HCO3-(aq) + H2O(l ) ⇌ OH-(aq) + H2CO3(aq) Kb2 = 2.3x10-8

• Kb1 > Kb2 > Kb3

Polyprotic bases

Problems

CARBONIC ACID (H2CO3)

Find the pH of 0.0037 mol L-1 carbonic acid solution

Ka1 = 4.3x10-7 and Ka2 = 5.6x10-11

OXALIC ACID (H2C2O4)

Find the pH and C2O42- ion concentration of 0.0200 mol L-1 oxalic acid solution

Ka1 = 5.9x10-2 and Ka2 = 6.4x10-5

PHOSPHORIC ACID (H3PO4)

Find the pH and all chemistry species concentration envolved in the equilibria of

0.0500 mol L-1 phosphoric acid

Dados: Ka1 = 7.5x10-3, Ka2 = 6.2x10-8 and Ka3 = 4.2x10-13

Species distribution diagram

The solution composition of a polyprotic acid can be calculated as a function of

pH, based on the α value and the analytical concentration.

H2A ⇌ H+ + HA- Ka1

HA- ⇌ H+ + A2- Ka2

ion concentrat analytical of sum][][][: 2

2 AHAAHcMB T

Tc

AH ][ 20 Free acid

Tc

HA ][1

HA- species

Tc

A ][ 2

2

A2- species (totally dissociated species)

1210

][

]][[

2

1AH

HAHKa

][

]][[ 2

2

HA

AHKa

][][][ 1

2

H

KAHHA a

2

212

22

][][

][][][

H

KKAH

H

KHAA aaa

2

212

122

2

2

][][

][][][

][][][

H

KKAH

H

KAHAHc

AHAAHc

aaaT

T

Tc

AH ][ 20

Tc

HA ][1

Tc

A ][ 2

2


2

212

122

2

2

][][

][][][

][][][

H

KKAH

H

KAHAHc

AHAAHc

aaaT

T

Tc

AH ][ 20

Tc

HA ][1

Tc

A ][ 2

2

2

212

122

20

][][

][][][

][

H

KKAH

H

KAHAH

AH

aaa


2

212

122

2

2

][][

][][][

][][][

H

KKAH

H

KAHAHc

AHAAHc

aaaT

T

Tc

AH ][ 20

Tc

HA ][1

Tc

A ][ 2

2

2

211

2

212

122

20

][][1

1

][][

][][][

][

H

KK

H

K

H

KKAH

H

KAHAH

AH

aaaaaa


2

212

122

2

2

][][

][][][

][][][

H

KKAH

H

KAHAHc

AHAAHc

aaaT

T

Tc

AH ][ 20

Tc

HA ][1

Tc

A ][ 2

2

2

2112

2

2

211

2

212

122

20

][][1

1

][

][

][][1

1

][][

][][][

][

H

KK

H

Kx

H

H

H

KK

H

K

H

KKAH

H

KAHAH

AH

aaaaaaaaa


2

212

122

2

2

][][

][][][

][][][

H

KKAH

H

KAHAHc

AHAAHc

aaaT

T

Tc

AH ][ 20

Tc

HA ][1

Tc

A ][ 2

2

211

2

2

0

2

2112

2

2

211

2

212

122

20

][][

][

][][1

1

][

][

][][1

1

][][

][][][

][

aaa

aaaaaaaaa

KKKHH

H

H

KK

H

Kx

H

H

H

KK

H

K

H

KKAH

H

KAHAH

AH

D = H+]n + [H+](n-1)Ka1 + [H+](n-2)Ka1Ka2 + . . . + Ka1Ka2. . .Kan

For any system:


cT = [H4Y] + [H3Y-] + [H2Y

2-] + [HY3-] + [Y4-]

D = [H+]4 + Ka1[H+]3 + Ka1Ka2[H

+]2 + Ka1Ka2Ka3[H+] + Ka1Ka2Ka3Ka4

D

KKKK

c

Y

D

HKKK

c

HY

D

HKK

c

YH

D

HK

c

YH

D

H

c

YH

aaaa

T

aaa

T

aa

T

a

T

T

o

4321

4

4

321

3

3

2

21

2

22

3

131

4

4

][

][][

][][

][][

][][

= f (Ka, H+)

0 + 1 + 2 + 3 + 4 = 1



pH of salts

HYDROLYSIS: interaction between species negatively or positively charged with the

solvent.

DEGREE OF HYDROLYSIS: fraction of each mole of hydrolysed anion or cation at

equilibrium.

HYDROLYSIS REACTIONS result in interactions that release H3O+ (aq.) or OH-

(aq.) ions.

CH3 C

O

OH

O

CCH3

O-

+ OH-(aq.)+ H2O

Alkaline hydrolysis - anionic

H H

H

N..+ H2O + H3O

+ (aq.)

Acid hidrolysisis - cationic

N

H

HH +

H

Applications

-Industrial: water treatment.

-In quantitative analyses: predictions about the behaviour of ions with maximum

oxidation state (+3 or +4) in which the maintenance of high acidity can avoid or

take advantage of the hydrolysis for a certain purpose.

-Control of solubility of some salts.

Types

4 types:

-salts of strong acid and strong base neutral solution

Ex: NaCl, NaNO3, Na2SO4, KCl, KNO3, K2SO4

-salts of weak acid and strong base

Ex: CH3COONa, Na2CO3, K2CO3, Na3PO4

-salts of strong acid and weak base

Ex: NH4Cl, CuSO4, NH4NO3, AlCl3, CaCl2

-salts of weak acid and weak base

Ex: CH3COONH4, AlPO4, (NH4)2CO3

not always neutral solution

Salts of weak acid (HA) and

strong bases (MOH)

The final solution is alkaline.

A-(aq) + H2O(l) ⇌ OH-(aq) + HA(aq)

][

]][[

A

HAOHK

h

Ex: CH3COONa


strong bases (MOH)



][

]][[

A

HAOHK

h

Ex: CH3COONa

2H2O(l) ⇌ OH-(aq) + H3O+(aq) ]][[ 3

OHOHKw

HA(aq) + H2O(l) ⇌ H3O+(aq) + A-(aq)

][

]][[ 3

HA

AOHKa


strong bases (MOH)



][

]][[

A

HAOHK

h

Ex: CH3COONa

2H2O(l) ⇌ OH-(aq) + H3O+(aq) ]][[ 3

OHOHKw


][

]][[ 3

HA

AOHKa

a

w

h K

KK

So:

Degree of hydrolysis

Consider x the degree of hydrolysis:

cx-c ][A

cx ][OH [HA]

][][c

-

-

AHA

)1(][

]][[ 2

x

cx

A

HAOHKh

This expression allows the calculation

of the degree of hydrolysis


ProblemsFind the Kh.

•Ions: Na+ and HPO42-

•Na+ from strong base NaOH no influence at pH

•HPO42-(aq) ⇌ PO4

3-(aq) + H+(aq) Ka = 4.2x10-13

•HPO42-(aq) + H2O(l) ⇌ H2PO4

-(aq) + OH-(aq)

•Kh > Ka hydrolysis predominates basic solution

Find the pH of 0.63 mol L-1 NaCH3CO2(aq) solution. Ka acetic acid = 1.8x10-5

Find the degree of hydrolysis of 1 mol L-1 NaCN solution. Ka hydrogen cyanide

= 4.9x10-10

150 mL of 0.02 mol L-1 sodium acetate was diluted until 0.500 L. What is the

concentration of the acetic acid at equilibrium?

024.0102.4

10113

14

x

x

K

KK

a

wh

Salts of strong acid (HA) and

weak base (MOH)

M+(aq) + 2H2O(l) ⇌ MOH(aq) + H3O+(aq)

][

]][[ 3

M

OHMOHKh

MOH(aq) ⇌ M+(aq) + OH-(aq)

]][[ 3

OHOHKw

So:

b

w

h K

KK

2H2O(l) ⇌ OH-(aq) + H3O+(aq)

][

]][[

MOH

OHMKb

Ex: NH4Cl

The final solution is acidic.

pH calculation

Find the pH of 0.15 mol L-1 NH4Cl solution.

•ions: NH4+ and Cl-

•Cl- from strong acid HCl no influence at pH

•NH4+(aq) + OH-(aq) ⇌ NH3(aq) + H2O(l) Kb = 1.8x10-5

•NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O

+(aq)10

5

14

106.5108.1

101

xx

x

K

KK

b

wh

x

xx

15.0106.5

210

16102.9 Lmolxx

pH = 5.04

Salts of weak acid (HA) and weak

base (MOH)

It involves both anionic and cationic hydrolysis

M+(aq) + 2H2O(l) ⇌ MOH(aq) + H3O+(aq) A-(aq) + H2O(l) ⇌ OH-(aq) + HA(aq)

][

]][[

A

HAOHK

h][

]][[ 3

M

OHMOHKh

2H2O(l) ⇌ OH-(aq) + H3O+(aq) ]][[ 3

OHOHKw


][

]][[ 3

HA

AOHKa


][

]][[

MOH

OHMKb


base (MOH)

It involves both anionic and cationic hydrolysis

M+(aq) + 2H2O(l) ⇌ MOH(aq) + H3O+(aq) A-(aq) + H2O(l) ⇌ OH-(aq) + HA(aq)

][

]][[

A

HAOHK

h][

]][[ 3

M

OHMOHKh

2H2O(l) ⇌ OH-(aq) + H3O+(aq) ]][[ 3

OHOHKw


][

]][[ 3

HA

AOHKa


][

]][[

MOH

OHMKb

Ka = Kb neutral solution

Ka < Kb basic solution

Ka > Kb acid solution

MA(aq) ⇌ M+(aq) + A-(aq)

Hydrolysis

M+(aq) + 2H2O(l) ⇌ MOH(aq) + H3O+(aq)


+

M+(aq) + A-(aq) + H2O(l) ⇌ MOH(aq) + HA(aq) Kh

]][[

]][[

AM

HAMOHKh

][

]][[ 3

HA

AOHKa

][

]][[

MOH

OHMKb

]][][][[

]][][][[

3

3

OHMAOH

MOHHAOHOH

xKK

K

ba

w

Ex: CH3COONH4


base (MOH)

pH calculation

Find the pH of 0.1000 mol L-1 CH3COONH4 solution.

Ka CH3COOH = 1.75 x 10-5 e Kb NH3 = 1.78 x 10-5.

NH4+(aq) + CH3COO-(aq) + H2O(l) ⇌ NH4OH(aq) + CH3COOH(aq) Kh

I C C - -

E C-x C-x x x

2

2

2

2

34

34

)()(]][[

]][[

C

x

xC

x

xKK

K

COOCHNH

COOHCHOHNHK

ba

wh

C = 0.100 mol L-1

C >>> x

[CH3COO-] >>> [CH3COOH]

pH calculation




I C C - -

E C-x C-x x x

2

2

2

2

34

34

)()(]][[

]][[

C

x

xC

x

xKK

K

COOCHNH

COOHCHOHNHK

ba

wh

C = 0.100 mol L-1

C >>> x

[CH3COO-] >>> [CH3COOH]

2

22

323 ][][

a

aK

COHx

x

COHK

2

22

323 ][][

a

aK

COHx

x

COHK

aa

ba

aw xKKOHxKK

xKKOH ][][ 3

2

2

3

Kw = Ka x Kb

2

2

3

2

2

34

34 ][

)(]][[

]][[

aba

wh

K

OH

C

x

xKK

K

COOCHNH

COOHCHOHNHK




I C C - -

E C-x C-x x x

C = 0.100 mol L-1

pH calculation

2

22

323 ][][

a

aK

COHx

x

COHK

aa

ba

aw xKKOHxKK

xKKOH ][][ 3

2

2

3

Kw = Ka x Kb

2

2

3

2

2

34

34 ][

)(]][[

]][[

aba

wh

K

OH

C

x

xKK

K

COOCHNH

COOHCHOHNHK




I C C - -

E C-x C-x x x

C = 0.100 mol L-1

pH calculation

[H3O+] = 1.75 x 10-5 mol L-1 pH = 4.75

• The pH depends on the dissociation constants of the

weak base and the weak acid which give rise to the salt.

• pH is independent of salt concentration.

pH calculation

Salts derived from polyprotic acids

Polyprotic acids give rise to two or more anions :

Diprotic acid H2A: HA- and A2-

Tripotic acid H3A: H2A-, HA2- and A3-

Consider an hypothetical acid, designed H2A:

1) Find the pH and concentration of Na2A solution (derived from H2A and

strong base)

2) Find the pH and concentration of NaHA solution (derived from H2A and

strong base)

Case 1 – Na2A – behaviour as weak base

Na2A(aq) 2Na+(aq) + A2-(aq)A2-(aq) + H2O(l) ⇌ HA-(aq) + OH-(aq) Kh (A

2-)

HA-(aq) + H2O(l) ⇌ H2A(aq) + OH-(aq) Kh (HA-)

A2-(aq) + H2O(l) ⇌ HA-(aq) + OH-(aq) Kh (A2-)

I C - - -

E C-x - x x

2

2

2

2 ][

]][[b

a

wh K

K

K

xC

x

A

HAOHK


I x - - x

E x-y - y x+y

1

1

2

2

][

]][[b

a

wh K

K

K

yx

y

HA

AHOHK

[OH-] total

Example

Find the pH and the degree of hydrolysis of 0.200 mol L-1 sodium carbonate solution

CO32-(aq) + H2O(l) ⇌ HCO3

-(aq) + OH-(aq) Kh (CO32-)

I 0,200 - - -

E 0,200-x - x x

4

2

2

1078.1200.0

xK

K

x

xK

a

wh

HCO3-(aq) + H2O(l) ⇌ H2CO3(aq) + OH-(aq) Kh (HCO3

-)

I 5.97x10-3 - - 5.97x10-3

E 5.97x10-3-y - y 5.97x10-3+y

8

1

3

3

1032.21097.5

)1097.5(

x

K

K

yx

yxyK

a

wh

Ka1 = 4.3x10-7

Ka2 = 5.6x10-11

0.200/Kh > 100 YES! x negligible!

x = 5.97x10-3 mol L-1

x/Kh > 100 YES!

y negligible!

y = 2.32x10-8 mol L-1

pH = 11.8

degree = 3%

Case 2 – NaHA

• Salts of intermediate species of polyprotic acids are amphiprotic species,

because they exhibit weak Bronsted & Lowry acid or base behaviour.

• The hypothetical anion HA- is an intermediate compound of the dissociation of

the polyprotic weak acid H2A or of the hydrolysis of the conjugated weak base

and weak A2-.


HA-(aq) + H2O(l) ⇌ H3O+(aq) + A2-(aq)

][

]][[ 2

32

HA

AOHKa

HA-(aq) + H2O(l) ⇌ H3O+(aq) + A2-(aq)

][

]][[ 2

32

HA

AOHKa

[A2-] = [H3O+]

Part of the formed H3O+ may associate with the anion HA- forming the polyprotic

acid H2A.

[A2-] = [H3O+] + [H2A] (I)

][

]][[ 2

32

HA

AOHKa

][

]][[

2

31

AH

HAOHKa

])[(

][][

]][[][

][

][

1

212

3

1

33

3

2

HAK

HAKKOH

K

HAOHOH

OH

HAK

a

aa

a

a

HA-(aq) + H2O(l) ⇌ H3O+(aq) + A2-(aq)

][

]][[ 2

32

HA

AOHKa

[A2-] = [H3O+]

[A2-] = [H3O+] + [H2A] (I)

][

]][[ 2

32

HA

AOHKa

][

]][[

2

31

AH

HAOHKa

If [HA-] >>> Ka1

21

2

3 ][ aa KKOH

213 ][ aa KKOH

)(2

121 aa pKpKpH

])[(

][][

]][[][

][

][

1

212

3

1

33

3

2

HAK

HAKKOH

K

HAOHOH

OH

HAK

a

aa

a

a

The pH is independent of salt concentration.

Part of the formed H3O+ may associate with the anion HA- forming the polyprotic

acid H2A.

Problem

Find the pH of (a) 0.20 mol L-1 NaH2PO4(aq); (b) 0.20 mol L-1 Na2HPO4(aq)

Ka1 = 7.6x10-3; Ka2 = 6.2x10-8; Ka3 = 2.1x10-13

Documents

Denise Lowinsohn [email protected] ...3 was prepared from 0.85 mL of the concentrated acid in 250 mL of distilled water. What is the pH of this prepared solution? The concentrated