Buffers. A buffered solution resists changes in ph when small amounts of acids or bases are added or when dilution occurs.

Buffers A buffered solution resists changes in ph when small amounts of acids or bases are added or when dilution occurs. The buffer consists of a mixture of an acid and its conjugate base. Example: acetic acid/sodium acetate

The Henderson-Hasselbalch equation HA(aq) H + (aq) + A - (aq) a H A HA log a log H A HA log H log A HA log H log a log A HA ph p a ph p a log A HA

For base log log log log log log log log B BH p poh B BH OH B BH OH B BH OH B BH OH b b b b

When A - = HA, ph = pa a H A HA H A A log a log H ph p a Regardless of how complex a solution may be, whenever ph = p a, A - must equal HA.

Application of Henderson- Hasselbalch equation Sodium hypochlorite (NaOCl, the active ingredient of bleach) was dissolved in a solution buffered to ph 6.20. Find the ratio OCl - /HOCl in this solution. p a = 7.53 for HOCl HOCl(aq) + H 2 O(l) H 3 O + (aq) + OCl - (aq)

Problems 1. What is the ph of a buffered prepared by 0.12 mol L -1 lactic acid (CH 3 CH(OH)COOH) and 0.10 mol L -1 solution? a = 1.4 x 10-4. sodium lactate (CH 3 CH(OH)COONa) 2. Find the ph of a buffered solution prepared by 0.040 mol L -1 NH 4 Cl(aq) and 0.030 mol L -1 NH 3 (aq) solution. b = 1.8 x 10-5. 3. What weight of NH 4 Cl (M. W. = 53.5 g mol -1 ) should be added to 100 ml of 0.1 mol L -1 NH 4 OH solution to give a ph of 9? b = 1.8 x 10-5. 4. A solution of equal concentrations of lactic acid and sodium lactate has ph = 3.08. (a) What are the p a and a values of lactic acid? (b) What would be the ph if the acid concentration was twice the salt concentration?

Effect of adding acid to a buffer If we add 100 ml of 0.1 mol L -1 HCl solution to a one liter of 0.1 mol L -1 acetic acid and 0.1 mol L -1 sodium acetate solution ( a = 1.75x10-5 ), what will the ph be? CH 3 COOH(aq) + H 2 O(l) CH 3 COO - (aq) + H 3 O + (aq) p a = 4.76 Initial: CH 3 COO - n = C V = 0.1 x 1.0 = 0.10 mol CH 3 COOH n = C V = 0.1 x 1.0 = 0.10 mol H 3 O + n = C V = 0.1 x 0.1 = 0.01 mol Final: CH 3 COO - n = 0.10 0.01 = 0.09mol CH 3 COOH n = 0.10 + 0.01 = 0.11mol H 3 O + totally reacted Disregarding the change in volume ph 0.09 4.76 log 4.76 0.09 0.11 4.67 The preceding example illustrates that the ph of a buffer does not change very much when a limited amount of a strong acid or base is added.

Effect of adding acid to a buffer If we add 100 ml of 0.1 mol L -1 HCl solution to a one liter of 0.1 mol L -1 acetic acid and 0.1 mol L -1 sodium acetate solution ( a = 1.75x10-5 ), what will the ph be? CH 3 COOH(aq) + H 2 O(l) CH 3 COO - (aq) + H 3 O + (aq) p a = 4.76 Initial: CH 3 COO - n = C V = 0.1 x 1.0 = 0.10 mol CH 3 COOH n = C V = 0.1 x 1.0 = 0.10 mol H 3 O + n = C V = 0.1 x 0.1 = 0.01 mol Final: CH 3 COO - n = 0.10 0.01 = 0.09mol CH 3 COOH n = 0.10 + 0.01 = 0.11mol H 3 O + totally reacted The same volume of HCl added to one liter of water (ph = 7) would lead to a ph = -log 0.01 = 2 solution, a change of 5 ph units. This shows ph how a buffer solution regulates the ph. Disregarding the change in volume 0.09 4.76 log 4.76 0.09 0.11 4.67 The preceding example illustrates that the ph of a buffer does not change very much when a limited amount of a strong acid or base is added.

Buffer capacity Buffer capacity measure how well a solution resists changes in ph when acid or base is added. The greater the buffer capacity, the less the ph changes. Effect of adding 0.01 mol of H + or OH - to a buffer containing HA and A - (total quantity of HA + A - = 1 mol). The minimum change in ph occurs when the initial ph of the buffer equals p a for HA. That is, the maximum buffer capacity occurs when ph = p a. ph p a 1

Problem 1) When it is desired to study reactions of biological interest at phs close to neutrality, the buffer formed by the mixture of H 2 PO 4 and 2 HPO 4 is often used. a) Find the ph of 0.2 mols of H 2 PO 4 and 0.1 mols of 2 HPO 4 dissolved in 1L solution. b) If you add 0.05 mols of nitric acid to a solution of item a, what will the new ph be? Consider that the volume variation is negligible. c) If you add 0.05 mols of NaOH to a solution of item a, what will the new ph be? Consider that the volume variation is negligible. a 1 = 5.9x10-3 ; a 2 = 6.2x10-8 ; a 3 = 4.8x10-13

Problem 2) Two buffers were prepared: a) 0.1 mol L -1 HAc + 0.1 mol L -1 NaAc b) 0.1 mol L -1 HAc + 0.5 mol L -1 NaAc 25 ml of each solution was added in a flask a and b and on them was added 5 ml of 0.1 mol L -1 NaOH solution. 25 ml of each solution was added in a flask a and b and on them was added 5 ml of 0.1 mol L -1 HCl solution. In which flask there was greater variation of ph when NaOH was added? And HCl? a = 1.8 x 10-5

Problem 3) Which of the buffer systems below would be a good choice to prepare a buffer with ph = 5? And ph = 10? Composition pa CH 3 COOH / CH 3 COO - 4.75 HNO 2 / NO - 2 3.37 HClO 2 / ClO - 2 2.00 NH 4+ / NH 3 9.25 (CH 3 ) 3 NH + / (CH 3 ) 3 N 9.81 H 2 PO 4- / HPO 2-4 7.21

Mix Two weak acids: (1) HA and (2) HL. HA >>> HL H 2 O OH - (Aq.) + H + (Aq.) HA H + (Aq.) + A - (Aq.) H + = OH - + A - + L - HL H + (Aq.) + L - (Aq.) Each species contribution: Equilibrium of proton From water: w OH H HA A HA H and HL L HL H

2 HL HA H HL HA H H HL H HA H H HL HA w HL HA w HL HA w HA HA H H C HA A HA c HA HL HL H H C HL L HL c HL Mix

Mix By replacing the expressions of HA and HL in the equilibrium expression of the próton, a fifth order equation is obtained in relation to the H + ion. This is a complex formulation, requiring calculations of successive approximation. A possible simplification is to assume that: C HA HA e C HL HL, thus a first approximation of the hydrogen ion concentration can be obtained: H 2 0 w HA C HA HL C HL H 0 w HA C HA HL C HL HA 1 C HA H H 0 0 HA HL 1 C HL H H 0 0 HL

Mix H 2 w HA HA 1 HL HL 1 H w HA HA 1 HL HL 1 The approximation test is performed, taking the maximum error value in the calculation of the hydrogen ion concentration as 1.0% (HA 1 HA 0 = 1.0%) and restarting the cycle, if the error is greater than this value. It is a calculation of successive approximation.

Acid-base equilibrium in nonaqueous media In order to treat the acid-base equilibrium in non-aqueous or partially aqueous médium, two fundamental characteristics of the solvents must be considered: the autoprotolysis constant and the dieletric constant. The autoprotolysis constant reflects the amphiprotic or aprotic behaviour of the solvent. Protic solvents are polar and have as main characteristic the ability to make a hydrogen bond, for example water. Aprotic solvents do not have this characteristic. Ex: hexane. The dieletric constant reflects its ability to separate the charges, that is, to avoid the formation of pair between oppositely charged ion. Solvents can be divided into 4 different classes. (in order to facilitate the discussion)

First class Polar amphiprotic solvents have a water-like behaviour and include methanol, formic acid, water-alcohol mixtures, etc.. When an HA acid is dissolved in a polar amphiprotic solvent HS, the following equilibrium reaction occurs: HA + HS H 2 S + + A - where H 2 S + is the solvated proton. The formal equilibrium treatment given to this class of solvents is similar to that given to water. a a H 2 S a. a HA A Acid p a (H 2 O) p a (MeOH) p a (EtOH) Acetic 4.75 9.7 10.4 Benzoic 4.20 9.4 10.1 Phenol 9.97 14.2 15.3

Second class It comprises apolar amphiprotic solvents, the typical example is glacial acetic acid. An HA acid, when dissolved in this type of solvent, ionizes and dissociates according to the equations: HA + HS H 2 S + A - i aha a H 2 H 2 S + A - H 2 S + + A - 2 d a a H H S S 2 S A. a A A

As in amphiprotic solvent with low dielectric constant the ion pair formation is inevitable, the ionization constant is the one that expresses exactly the force of the acid or base. The ionization is the acidity / basicity function of the solvent. The dissociation is a function of the dielectric constant. In solvents with a dielectric constant greater than 30-40, the ion pair formation is negligible and the treatment is done as in water. Acid p a Base p a Percloric 4.87 Tribenzylamine 5.38 Sulfuric ( 1 ) 7.24 Pyridine 6.10 Hydrochloric 8.55 2 5-dichloroaniline 9.48

Third class It is the one that includes the aprotic polar, that are better known as aprotic dipolar. They are relatively inert and can be subdivided into protophilic and protophobic. Protophilics: DMSO, DMF, Py - solved cations and can accept hydrogen bonds. Protophobics: ACN, nitromethane - have little capacity to solvate cations and form hydrogen bonds. Acid p a (ACN) p a (DMSO) Acetic 22.3 12.6 Benzoic 20.7 11.1 Salicylic 16.7 6.8

Fourth class Aprotic apolar - inert This class includes benzene, toluene, n-hexane, etc. They have low dielectric constant (usually less than 10) and play no role in neutralization reactions. The equilibrium treatment in non-aqueous medium, as many be noted, is not as simple as in water, and therefore, much care must be taken.

TITRATIONS IN ANALYTICAL CHEMISTRY

Titration A titration is a procedure in which increments of the known reagent solution the tritant are added to analyte until the reaction is complete. Titrant: is usually delivered from a buret. Analyte: is contained in a flask. Each increment of titrant should be completely and quickly consumed by reaction with analyte until the analyte is used up. Common titrations are based on acid-base, oxidation-reduction, complex formation, or precipitation reactions.

Equivalence and end point The equivalence point occurs when the quantity of titrant added is the exact amount necessary for stoichiometric reaction with the analyte. The equivalence point is the ideal result we seek in a titration. What we actually measure is the end point, which is marked by a sudden change in a physical property of the solution. The end point cannot exactly equal the equivalence point. The difference between the end point and the equivalence point is an inescapable titration error. E V V equivalenc e end

Requirements of a reaction employed in a titration Simple reaction with known stoichiometry Quick reaction Present chemical or physical changes (ph, temperature, conductivity), mainly at the point of equivalence.

Methods Visual indicator Color change is caused by the disappearance of analyte or the appearance of excess titrant. Instrumental methods Change in voltage or current between a pair of electrodes Monitoring the absorbance of light by species in the reaction

Primary standard A primary standard is a reagent that is pure enough to be weighed out and used directly to provide a known number of moles. Ex: sodium carbonate

Requirements primary standard High purity. Established methods for confirming purity should be available. Stability toward air. Absence of hydrate water so that the composition of the solid does not change with variations in relative humidity. Ready availability at modest cost. Reasonable solubility in the titration medium. Reasonably large molar mass so that the relative error associated with weighing the standard is minimized.

Standardization 99.9% pure or better It should not decompose under ordinary storage It should be stable when dried by heating or vacum We say that a solution whose concentration is known is a standard solution.

Secondary standard It is a compound which allows to prepare a titrant solution, but its concentration is determined by comparison (standardization) against a primary standard. Ex: NaOH

Standard solutions The ideal standard solution for a titrimetric method will 1. be sufficiently stable so that it is only necessary to determine its concentration once; 2. react rapidly with the analyte so that the time required between additions of reagente is minimized; 3. react more or less completely with the analyte so that satisfactory end points are realized; 4. undergo a selective reaction with the analyte that can be described by a simple balanced equation.

Types Direct titration Titrant is added to analyte until the end point is observed. analyte + titrant product Indirect titration The reagent to be titrated is generated in the solution. Back titration A known excess of a standard reagent is added to the analyte. Then a second standard reagent is used to titrate the excesso of the first reagent. analyte + reagent 1 product + excess reagent 1 excess reagent 1 + reagent 2 product

Neutralization titration Standard solutions of strong acids and strong bases are use extensively for determing analytes that are themselves acids or bases or analytes that can be converted to such species by chemical treatment. The standard solutions employed in neutralization titrations are Strong acids or strong bases because these substances react more completely with na analyte than do their weaker counterparts and thus yield sharper end points. Strong base: NaOH, OH Strong acid: HCl, H 2 SO 4, HClO 4 OBS: HNO 3 is rarely used due to its high oxidizing power.

Acid/Base indicators An acid/base indicator is a weak organic acid or weak organic base whose undissociated form differs in color from its conjugate base or its conjugate acid form. For example, the behaviour of an acid-type indicator, HIn, is described by the equilibrium HIn (acid color) + H 2 O H 3 O + + In - (base color) Here, internal structural changes accompany dissociation and cause the color change. The equilibrium for a base-type indicator, In, is In (base color) + H 2 O InH + (acid color) + OH -

H O In HIn Acid/Base indicators HIn In 3 a H3O x a The human eye is not very sensitive to color diferences in a solution containing a mixture of In - and HIn, particularly when the ratio In - /HIn is greater than about 10 or smaller than about 0.1. HIn In HIn In 10 1 1 10 H O 3 H O 3 10 1 10 a a ph( acid color) log(10 1 ph( basic color) log( 10 a ) a ) p a p 1 a 1 Indicator range: ph pa 1

Acid/Base indicators PHENOLPHTHALEIN transaction range: 8.3-10.0

Acid/Base indicators PHENOL RED RED YELLOW RED transaction range: 6.8-8.4

Acid/Base indicators METHYL ORANGE YELLOW RED transaction range: 3.1-4.4

Sigmoidal curve Curve Titration