Analytical Chemistry

Analytical chemistry

Denise Lowinsohn [email protected] http://www.ufjf.br/nupis

2018 SCHEDULE

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 -Complex-formation titrations: part 2 09/05/18 -Preparing samples for analysis 16/05/18 1º Test (100 points) 23/05/18 -Preparing samples for analysis -Preparing samples for analysis: activity delivery (100 points) 30/05/18 -Application of Statistics to Analytical Chemistry 06/06/18 -Application of Statistics to Analytical Chemistry 13/06/18 Holiday -Application of Statistics to Analytical Chemistry 20/06/18 -Sampling experiment (100 points) 27/06/18 2º Test (100 points) 04/07/18 Oral presentation (100 points) 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. Acid-base equilibrium

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 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

species that accepts derived from protons acid 1 (base 2) (base 1)

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

species that derived from donates base 2 protons (acid 2) (acid 1)

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

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

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). Brønsted-Lowry Theory

- + 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:

species that species that derived from accepts derived from accepts acid 1 protons acid 1 protons (base 1) (base 2) (base 1) (base 2)

- + - + H2O + NH3 ⇌ OH + NH4 HNO2 + H2O ⇌ NO2 + H3O

species that derived from species that derived from + donates H base 2 donates base 2 + (acid 1) (acid 2) H (acid 2) (acid 1) Strength of acids and bases

• 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. 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)

The concentration of At 25oC water in dilute aqueous   solutions is enormous [H3O ].[OH ] when compared with 2  Keq the concentration of hydrogen and [H 2O] hydroxide ions.   2 [H3O ].[OH ]  Keq .[H 2O]   14 [H3O ].[OH ]  Kw 1x10 CONSTANT

Ion product constant for water Lewis Theory

Acid = accepts a pair of electrons

Base = donates a pair of electrons

Lewis acids and bases don’t need contain protons.

Example: 3+ - 2+ Fe (aq) + SCN (aq) ⇌ Fe(SCN) (aq)

Lewis base: Lewis acid: donates a pair of electrons accepts a pair of electrons

The defintion of Lewis is the most general definition of acids and bases. + pH scale  pH = -log[H3O ]

SÖRENSEN introduced in 1909, the In the most Neutral solution: [H O+] = [OH-] concept of pH, a conveniente way of 3 + - -7 -1 solutions, the [H3O ] = [OH ] = 1.0 x 10 mol L expressing acidity – the negative [H+(aq)] is Acid solution: [H O+] > [OH-] very small. logarithm of hydrogen ion concentration. 3 + -7 -1 [H3O ] > 1.0 x 10 mol L and   [OH-] < 1.0 x 10-7 mol L-1 Kw  [H3O ].[OH ]

+ - Alkaline solution: [H3O ] < [OH ] 0 [H O+] < 1.0 x 10-7 mol L-1 and pK w  pH  pOH 14 (25 C) 3 [OH-] > 1.0 x 10-7 mol L-1

•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.) [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

-1 CHNO3 = 0.010 mol L  total amount of HNO3 present in solution + - -1 Concentration at equilibrium: [H3O ]  [NO3 ] = 0.010 mol L  not considering autoionization of H2O 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”.

+ pH = - log[H3O ] + - HNO3(aq)  H (aq) + NO+3 (aq) - [H3O ] = [NO3 ] = CHNO3 Initial 0.010 mol L-1 - - pH = -log(C) = -log 0.010 Equilibrium - 0.010 mol L-1 0.010 mol L-1 pH = 2.0

-1 CHNO3 = 0.010 mol L  total amount of HNO3 present in solution + - -1 Concentration at equilibrium: [H3O ]  [NO3 ] = 0.010 mol L  not considering autoionization of H2O Practicing.....

-1 What is the pH of 0.040 mol L 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

-1 b) 0.15 g de HNO3 (M.W. = 63 g mol ) in 300 mL of solution c) 10.0 mL de HCl 15 mol L-1 in 750 mL of solution 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:

2- - O (aq) + H2O(l)  2OH (aq) - - H (aq) + H2O(l)  H2(g) + OH (aq) 3- - N (aq) + 3H2O(l)  NH3(aq) + 3OH (aq) Calculation: pH of 0.010 mol L-1 strong base solution

NaOH(aq)  Na+(aq) + OH-(aq) Initial 0.010 mol L-1 - - Equilibrium - 0.010 mol L-1 0.010 mol L-1

-1 CNaOH = 0.010 mol L  total amount of NaOH present in solution Concentration at equilibrium: [Na+]  [OH-] = 0.010 mol L-1  not considering autoionization of H2O

pOH = - log[OH-] pKw = pH + pOH + - [Na ] = [OH ] = CNaOH 14.0 = pH + 2.0 pOH = -log(C) = -log 0.010 pH = 12.0 pOH = 2.0 Practicing.....

What is the pH of: a) 0.028 mol L-1 NaOH solution?

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

What is the molar concentration of: a) KOH, pH = 11.89? b) Ca(OH)2, pH = 11.68? 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. Weak acids

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

EQUILIBRIUM CONSTANT EXPRESSION Very small value

    [H3O ][ A ] [H ][ A ] K  or K  a [HA] a [HA]

Note that [H2O] is omitted in the expression of Ka. (H2O is a pure The higher K , the stronger is the acid (in this liquid) a 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. 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

2 C/Ka > 10  simplified calculation (by simplifying the calculation: the error will be less than 5%)

2 C/Ka ≤ 10  systematic calculation (quadratic equation) Use Ka to calculate pH

HOAc ⇌ H+ + OAc- (aq) (aq) (aq) C / Ka > 102  simplified calculation Initial 0.10 - - Equilibrium 0.10-x x x [H  ][OAc  ] x2 K 1.85x105   a [HOAc] 0.10 x

x  [H  ]  [OAc  ] 1.4x103 molL1

pH  log[H  ]  log(1.4x103 )  2.9 Problems

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

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

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

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

EQUILIBRIUM CONSTANT EXPRESSION Very small value

    [H3O ][ A ] [H ][ A ] K  or K  a [HA] a [HA]

  degree of dissociation

- cTOTAL = [HA] + [A ] [H+] = [A-] = c cxc c.c 2 c 2 K    a c  c c(1) 1 Weak acids – degree of dissociation

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

amount of substance of reactant dissociated [A ] [A ]     CTOTAL [HA][A ]

amount of substance of reactant present initially

The higher the degree of dissociation, the stronger is the acid. % Ionization = strength of acid or base

HOAc ⇌ H+ + OAc- (aq) (aq) (aq) C / Ka > 102  simplified Initial 0.10 - - calculation Equilibrium 0.10-x x x [H  ][OAc  ] x2 K 1.85x105   a [HOAc] 0.10 x

x  [H  ]  [OAc  ] 1.4x103 molL1

 [H ]equ. WEAK %ionization x100 1.4% ACID [HOAc]0 Problems

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

-5 of ionization . Ka = 1.8x10

-4 Calculate the percentage of HF molecules (Ka = 6.8x10 ) ionized in: a) 0.10 mol L-1 HF solution b) 0.010 mol L-1 HF solution 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.

µ = ionic strength 1 2 c = ionic concentration   cizi i z = charge of the ion 2 i

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.

AB ⇌ A+ + B-

(aA+.aB-) / (aAB) = Kdissociation DISSOCIATION CONSTANT

ACTIVITY = CONCENTRATION X ACTIVITY COEFFICIENT

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

    .  [A ].[B ] A B .  Kdissociation [AB] AB

 (activity coefficient) varies with concentration!!!!!! Activity - properties

• 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. 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 = activity coeficient of the species X 2 0.509zX  µ = ionic strength of the solution  log X  z = charge on the species X 1 3,3 X  X

αX = effective diameter os the hydrated ion X in nm Problems

Calculate the ionic strength of:

-1 a) 0.1 mol L solution of KNO3 -1 b) 0.1 mol L solution of Na2SO4

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

+ -1 Use activities to calculate the H3O ion concentration in a 0.120 mol L solution of -1 HNO2 that is also 0.05 mol L in NaCl. Find the relative error introduced by neglecting activities in this calculation.

+ - -4 αH3O = 0.9 nm, αNO2 = 0.3 nm e Ka = 5.1x10 Weak bases

•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.

+ - B(aq) + H2O(l) ⇌ BH (aq) + OH (aq)

[BH  ][OH  ] [BH  ] [BH  ] Kb      [B] CTOTAL [B][BH ] 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):

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

-6 Cocaine is an example of weak base with Kb = 2.6 x 10 . Calculate the pH of 0.0372 mol L-1 solution of this base.

CH3 H .N O + CH3 . C N O - OCH3 C + OH H (aq) OCH3 H O C O C H O H O

[BH  ][OH  ] x2 x2 K   2.6x106   b [B] 0.0372 x 0.0372

0.0372 / 2.6 x 10-6 > 102  YES

[OH ]  3.11x104 molL1 pOH  3.51 pH 10.49 Problems

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

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

-1 -9 -4 of 0.05 mol L : pyridine (Kb = 1.7x10 ), methylamine (Kb = 4.4x10 ) or nitrous -4 acid (Ka = 4.0x10 )?

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. Strength of acid or base

pKa = -log Ka or pKb = -log Kb

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 HA(aq) ⇌ H+(aq) + A-(aq) Fraction of HA in the form A-: + - B(aq) + H2O(l) ⇌ BH (aq) + OH (aq)   [A ] [A ] Ka [B]         A   B  cT [A ][HA] [H ] Ka [A ][HA]

Fraction of HA in the form HA: [HA] [HA] [H  ] [BH  ]         HA   BH  cT [A ][HA] [H ] Ka [A ][HA]

    [H ][ A ] [H ][ A ]  Ka[HA] Ka  [HA]  or [A ]   [HA] Ka [H ] 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.

+ -10 + Ka for the ammonium ion, NH4 is 5.70x10 . 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-. [H  ][ A ] HA(aq) ⇌ H+(aq) + A-(aq) K  a [HA] + [HA][OH  ] - - A (aq) + H2O(l) ⇌ HA(aq) + OH (aq) K  b [A ]

H O(l) ⇌ H+(aq) + OH-(aq) 2 Kw  Ka xKb

pKw  pKa  pKb Problems

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

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

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

2.5x104 [OH  ]   7.1x103 mol L1 35x103

pOH  2.15 pH 11.85 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)

+ - -2 st H2SO3(aq) ⇌ H (aq) + HSO3 (aq) Ka1 = 1.7x10 (1 dissociation constant) - + 2- -8 nd HSO3 (aq) ⇌ H (aq) + SO3 (aq) Ka2 = 6.4x10 (2 dissociation constant)

The decrease in the acid dissociation constant form Ka1 to Ka3, tell us that each successive proton is harder to remove. Polyprotic acids

• 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).

3 • Ka1 / Ka2 >10  only Ka1 Polyprotic bases

• Bases that can accept more than one proton  polyprotic bases

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

2- Example: diprotic system – carbonate ion (CO3 )

2- - - -4 CO3 (aq) + H2O(l ) ⇌ OH (aq) + HCO3 (aq) Kb1 = 1.8x10 - - -8 HCO3 (aq) + H2O(l ) ⇌ OH (aq) + H2CO3(aq) Kb2 = 2.3x10

• Kb1 > Kb2 > Kb3 Problems

CARBONIC ACID (H2CO3) Find the pH of 0.0037 mol L-1 carbonic acid solution

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

OXALIC ACID (H2C2O4) 2- -1 Find the pH and C2O4 ion concentration of 0.0200 mol L oxalic acid solution -2 -5 Ka1 = 5.9x10 and Ka2 = 6.4x10

PHOSPHORIC ACID (H3PO4) Find the pH and all chemistry species concentration envolved in the equilibria of 0.0500 mol L-1 phosphoric acid

-3 -8 -13 Dados: Ka1 = 7.5x10 , Ka2 = 6.2x10 and Ka3 = 4.2x10 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 - + 2- HA ⇌ H + A Ka2

 2 MB :cT [H2 A][HA ][A ] sum of analytical concentration

[H 2 A] 0  Free acid cT    1 [HA ] 0 1 2 - 1  HA species cT [A2 ] 2-  2  A species (totally dissociated species) cT Species distribution diagram

  [H ][HA ]  Ka1 Ka1  [HA ]  [H 2 A]  [H 2 A] [H ]

[H  ][ A2 ] K K K K  [A2 ]  [HA ] a2  [H A] a1 a2 a2 [HA ] [H  ] 2 [H  ]2