Introduction to Analytical Chemistry

Transcription

1 Introduction to Analytical Chemistry ANALYTICAL CHEMISTRY: The Science of Chemical Measurements. ANALYTE: The compound or chemical species to be measured, separated or studied TYPES of ANALYTICAL METHODS: 1.) Classical Methods (Earliest Techniques) a) Separations: precipitation, extraction, distillation b) Qualitative: boiling points, melting points, refractive index, color, odor, solubilities c) Quantitative: titrations, gravimetric analysis 2.) Instrumental Methods (~post-1930 s) a) Separations: chromatography, electrophoresis, etc. b) Qualitative or Quantitative: spectroscopy, electrochemical methods, mass spectrometry, NMR, radiochemical methods, etc. 1

2 Classification of volumetric methods: There are four general classes of volumetric methods: 1-Acid-Base: Many compounds, both inorganic and organic, are either acids or bases and can be titrated with a standard solution of a strong base or a strong acid. The reactions involve the combination of hydrogen and hydroxide ions to form water. The end points of these titrations are easy to detect, either by means of indicator or by following the change in ph with a ph meter. The acidity and basicity of many organic acids and bases can be enhanced by titrating in non-aqueous solvent, so the weaker acids and bases can be titrated. 2- Precipitation: In the case of precipitation, the titrant forms an insoluble product with the analyte. An example is the titration of chloride ion with silver nitrate solution. 3- Complexomietric: In complexometric titrations, the titrant is a complexing agent and forms a water-soluble complex with the analyte, a metal ion. The titrant is often a chelating agent. 2

3 4- Reduction-oxidation: These "redox" titration involve the titration of an oxidizing agent with a reducing agent, or vice versa. An oxidizing agent gains electrons and a reducing loses electrons in a reaction between them. Expressions of concentration: Concentrations of solutions: Standard solutions are expressed in terms of molar concentrations or molarity (M). Molar solution is defined as one that contains one mole of substance in each liter of a solution. Molarity of a solution is expressed as moles per liter or as millimoles per milliliters. Moles = (moles/liter) x liters = molarity x liters millimoles = molarity x milliliters M olarity(m ) mmole = M x ml M oleof solute Volumeof solutionin liters The equivalent weight: The equivalent weight is that weight of a substance in grams that will furnish one mole of the reacting unit. Thus, for HCI, the equivalent weight is equal to the formula weight: 3

4 eq.w t.hcl Formulaw t. HC1(g/mol) 1(eq/ mol) The milliequivalent weight is one thousandth of the eq.wt. A normal solution contains one gram eq.wt. of solute in one liter of solution: No. of gram-equivalents N No. gram- equivalents No. ofliters No. milligram- equivalents No. ofmilliliters By rearrangement of these equations we obtain the expression for calculating other quantities: No of gram eq. = N x No. of liters No. of milligram eq. =Nxml = NxV Back titration: In back-titrations, a known number of millimoles of reactant is taken in excess of the analyte. The unreacted portion is titrated for example, in the titration of antiacid tablets with a strong acid such as HC1. 4

5 Lecture 1 (2 hrs)./ /.: (Acid-base) Theoretical bases of neutralization reactions, Electrolyte and the theory of electrolytic Dissociation, Strong and weak electrolytes, Law of mass action, The dissociation of water, Hydrogen ion exponent (ph), Acids and bases, Acid base equilibrium (ph calculations), Solution of strong acids and strong bases, Solution of weak acids and bases, ph of salts Theoretical bases of neutralization reactions Electrolyte and the theory of electrolytic Dissociation: Aqueous solutions of substances differ in their behavior when submitted to an electric current. Some of them allow the current to pass, i.e. they conduct the electric current, these are termed "electrolytes"; while other do not allow the current to pass, i.e. they yield non conducting solutions, and are called "non electrolytes". The first class includes mineral acid, caustic alkalies and salts, while the second class is exemplified by cane sugar, glycerin and ethyl acetate. Pure water is a bad conductor of electricity, but when acid like HCl, a base such as KOH or a salt like Na2SO4 is dissolved 5

6 in water its conductivity is greatly improved. At the same time, it is noticed that the solute decomposes by the passage of the electric current into its components at the cathode and the anode. These components of the electrolyte are called "ions". The whole phenomenon was called by Arrhenius in 1887 "ionisation" or "dissociation" Strong and weak electrolytes: According to the theory of ionisation, the extent of ionisation increases with dilution, and at very great dilution it is practically complete. On the other hand, for each concentration there is a state of equilibrium between the undissociated molecules and the ions; the process being a reversible one. NaCl Na + + Cl K2SO4 2K + + SO4 2 Na2HPO4 2Na + + H + + PO4 3 The balance can be shifted to the right or the left according to conditions. Arrhenius therefore introduced a quantity " ", called "the degree of dissociation" defined as follows 6

7 Number of solute molecules dissociated Total number of solute molecules before dissociation When the ionisation is complete, i.e. when all molecules have dissociated, the value of "a" will be unity or very nearly so. The electrolyte will be then called "strong electrolyte". On the other hand this value for weak electrolytes is very far from unity. Law of mass action The law is concerned with reactions involving equilibrium between the reactants and productants. Since this the case in the dissociation of most electrolytes, this law is the basis for many calculations in neutralimetric analysis. The law reads: "The rate of a chemical reaction is proportional to the active masses of the reacting substances." In dilute solutions, where conditions approach ideal state "active mass" may be expressed by the concentrations of the reacting species, that is gram molecules or gram ions per liter. 7

8 In a reaction: A + B C + D The velocity of the forward reaction [Vf] and of the backward reaction [Vb] can expressed. Vf = [A].[B]. Kf Vb = [C]. [D]. Kb Where Kf and Kb are the proportionality constants called "Velocity constant and brackets indicate concentration, at equilibrium that is when Vf = Vb And Kf [A]. [B] = Kb [C]. [D] K K f b [C][D] [A][B] Since Kf and Kb are both constants, the fraction Kf / Kb must also be constant. Hence: [C][D] K [A][B] Where K denotes the "equilibrium constant" of the reaction (constant at a given temperature). 8

9 The dissociation of water The dissociation of water is reversible and to a very limited extent as illustrated by its very weak conductivity to an electric current, and can be represented by the equation: H2O H + + OH According to the law of mass action: Since the fraction of water ionised is very minute or negligible, the value of [H2O] is equal to (1) can be regarded as equation 1, and the equation can be written therefore: [H + ]. [OH] = Kw.. (2) Kw is known as "The ionic product of water" Under ordinary experimental conditions and at about 25 o C; the value of Kw is taken to be 1 x and it follows that; [H + ]. [OH ] = Furthermore, since the dissociation of water gives rise to equal 9

10 number of hydrogen and hydroxyl ions. Equation (2) could be written: [H + ] 2 = Kw = 1 x (3) and in other words; [H + ] = = 10-7 it follows that, if the [H + ] = [OH ] = 10 7 the solution is described as "neutral", if [H + ] is more than 10 7, that is 10 6, 10 5, etc. the solution is said to be "acidic" and if [H + ] is less than 10 7, that is 10 8, etc., the solution is called "alkaline". Hydrogen ion exponent (ph): The hydrogen ion exponent was introduced as an easy method for representing small changes in the ion concentrations, and was called the "ph". ph is defined as equal to the logarithm of the hydrogen ion concentration with a negative sign. i.e, ph = log [H + ] if, [H + ] = 1 x ph = log = 10 This method of stating hydrogen ion concentration has the advantage that all degree of acidity and alkalinity between 10

11 that of a solution molar (or normal) with respect to hydrogen and hydroxyl ions can be expressed by a series of positive numbers between 0 and 14. A neutral solution is one in which ph = 7, an acid solution is one in which ph is less than 7 and an alkaline solution in which ph is more than 7. N.B. this method of expressing the concentration of hydrogen ions as its negative exponent (ph) has been extended to express other numberically small values as [OH ], Kw etc. Thus poh = log [OH ] pkw = log Kw The ionic product of water (1 x ) could be thus expressed: pkw = ph + poh ph = pkw poh or ph = 14 poh or poh = 14 ph Acids and bases: According to the classical theory of Arrhenius all acids when dissolved in water dissociate giving rise to hydrogen ions as positive ions. Bases, on the other hands, undergo dissociation with the formation of hydroxyl ions [OH ] as the only negative ions. The old definition of both acids and bases was laid, therefore, on that observation. The acidity of a solution or its 11

12 basicity can also be determined by measuring the amount of either hydrogen ions or hydroxyl ions it contains, respectively. The degree of dissociation of the dissolved acid or base can be used to calculate the concentration of the ions present in the solution. According to the degree of dissociation acids can be divided into two groups: A) Strong acids, having a high degree of dissociation and B) Weak acids, which are feebly dissociated. Similarly strong bases have a high degree of ionisation. While weak bases dissociate feebly. Apart from monbasic acids, which dissociate in one stage, polybasic acids dissociate in consecutive stages. Sulphuric acid, for example, dissociate in two stages, in the first stage one hydrogen is almost completely ionised, thus: H2SO4 H + + HSO4 In the second stage, the other hydrogen is only partially ionised. 12

13 Phosphoric acid dissociates in three stages: H3PO4 H + + H2PO4 H2PO4 H + + HPO4 2 HPO4 2 H + + PO4 3 These stages are called the primary, secondary and tertiary dissociations, respectively, the first stage is the most complete while the others are smaller and smaller. The equilibrium, which exists in a dilute solution of an acid like acetic acid (HAc) at constant temperature, is HAc H + + Ac Applying the law of mass action [H ] x [Ac K [HAc] ] Where K is called "dissociation" "ionisation" or "acidity constant" The stronger the acid, the larger the acidity constant. For a completely ionised acid, the acidity constant is assumed to be 1, and the mass action law does not help much in this case. 13

14 In considering acids with more than one replaceable hydrogen, such as sulphuric or phosphoric acids the dissociation takes place on stages and not on one stage. The corresponding mass action expressions are: H HSO H H SO Also PO PO HPO H H H H H HSO SO 2-4 H HPO PO PO [H ][HSO ] 4 K [H SO ] 2 [H ][SO 4 - [HSO ] [H 4 3 [H - 4 ][HPO 2 2- [H ][H PO 2 (H PO ) ] K2 3x ] 3 PO 1 ] K ] K [H ][PO 4 ] K 2 [HPO ] x x10 3.6x In the same case of phosphoric acid, it may be considered that three acids are present. The first, H3PO4, corresponds to a moderately strong acid, the second is a weak acid, while the third is a extremely weak acid. 14

15 Acid base equilibrium ph calculations 1. Solution of strong acids and strong bases Since strong acids and strong bases are considered completely ionised in solvents., the calculation of ph or poh of such reagents a simple matter, the [H + ] or [OH ] is directly related to the concentration of the substance. The following examples are illustrative. Example 1: Calculate the ph value of a solution of a completely ionised 1.0 N solution of acid; or base.? [H + ] = 1M ph = log 1 = 0 (zero) similarly, in a completely ionised 1.0 N solution of base [OH ] = 1 M poh = log 1 = 0 (zero) 15

16 Example 2 Calculate the [H + ] and ph of N hydrochloric acid? [H + ] = N ph = log (9.0 X 10 3 ) = 2.05 Example 3 Calculate the ph values of a solution of sodium hydroxide whose [OH ] is 1.05 x 10 3? poh = (log 1.05 x 10 3 ) = 2.98 ph = = Example 4 Calculate the hydrogen ion concentration of a solution of ph 5.3? ph = log [H + ] 5.3 = log [H + ] [H + ] = 5.01 x 10 6 M 16

17 Example 5 Calculate the hydroxyl ion concentration of a solution of ph 10.75? poh = = 3.25 [OH ] = the antilog of 3.25 [OH ] = 5.62 x 10 4 M 2. Solution of weak acids and bases A) Calculation of ph of solution of weak acids Weak acids do not ionized freely and only a small fraction of the molecules is partially ionized. To calculate the ph of a weak acid HA, the law of mass action is thus applied to its dissociation equilibrium: HA H + + A and K a [H ] [A [HA] ] Where Ka is the ionisation constant or dissociation constant of the acid. If the acid is pure, its ionisation gives equal concentration of H + 17

18 and A ions, and since their activities may be assumed equal in dilute solutions, therefore: [H + ] = [A ] If the total acid concentration is Ca moles/liter (and can be determined by titration with standard base), then the moles of the unionized acid [HA] must be numerically equal to Ca [H + ], so that the above equation becomes Ka 2 [H ] Ca [H ] Now, the acid is weak and slightly ionised, thus [H + ] is very small compared to Ca and can be neglected in the denominator in the above equation. Therefore: K a [H ] C a 2 and [H ] And ph = ½ (pka + pca) (1) K a C a Example: 1 Calculate the ph and [H + ] of 0.10 N acetic acid (pka= 4.76)? ph = ½ (pka + pca) = ½ x ½ ( log 10 1 ) = =

19 Example 2 Calculate the ph and [H + ] of a M solution of phenobarbital (pka= 7.41)? PCa = log (4.5 x 10 3 ) = 2.35 ph = ½ ( ) = 4.88 [H + ] = antilog 4.88 = = 1.32 x 10 5 M B) Calculation of ph of solution of weak bases: The same method discussed in the previous section is readily adaptable to the calculation of the ph of solutions of weak mono equivalent bases; e.g. ammonia to give: ph = pkw ½ (pkb + pcb) (2) Example 1: Calculate the ph, and the [H + ] of 0.13 N ammonia solution (pkb = 4.76)? ph = 14 ½ (4.76) ½ ( log ) = 14 ½ ( ) = log [H + ] = [H + ] = antilog 0.82 x = 6.6 x M 19

20 Example 2: Calculate the ph and the [H + ] of a M solution of cocaine base (pkb = 5.59)? ph = ½ ( ) = 9.99 [H + ] = antilog of 9.99 = 1.03 x Calculation the ph of salts When salt is dissolved in water, the solution is not always neutral in reaction. Interaction occurs with the ions of water and the resulting solution may be neutral acid or alkaline according to the nature of the salt. A) Salts of strong acids or bases An aqueous solution of such salts, for example potassium chloride, is neutral in reaction. Neither the cation nor the anion has the tendency to hold any hydrogen or the hydroxyl ions of water; the related acids and bases being strong electrolyte KCl K + + Cl - K + + OH - KOH Cl - + H + HCl The equilibrium between the hydrogen and hydroxyl ions in water H2O H + + OH 20

21 B) Salts of weak acids or bases (hydrolysis) Salts of weak acids (or bases) react with water to give basic (or acidic) solutions respectively. This phenomenon is known as hydrolysis; it is the reverse of neutralization. A hydrolytic reaction proceeds because of the tendency of the ions of salts of weak acids (or bases) to react with the hydrogen ions (or hydroxyl ions) of water, forming slightly ionized acids (or bases). The reaction of these salts with water does not proceed to completion; it reaches an equilibrium point and thus represented by an equilibrium expression and an equilibrium constant, known as hydrolysis constant, Kh. the extent to which the hydrolytic reactions proceed; is related to the ionisation constants of the acids or bases formed; the lower the ionisation constant, the more is the extent of hydrolysis. C) Salts of weak acids and strong bases: Sodium acetate is an example of such salts. Its hydrolysis in water may be represented by the following equation: H2O + CH3COO CH3COOH + OH Since the sodium ion of sodium acetate does not react with water it is not included in the equation. But acetate ion reacts with water to form the slightly ionised acetic acid and to satisfy 21

22 the acid constant Ka. This requires that some of the water molecules ionise to maintain the ion product of water, Kw, constant; and in doing so, produces more hydroxyl ions. Consequently, when equilibrium is reached there is an increased hydroxyl ion concentration and a decreased hydrogen ion concentration. For this reason, a solution of a salt of a weak acid and a strong base is alkaline in reaction, and: ph = ½ (pkw pca + pka) (3) The degree of hydrolysis, h, of a salt, analogous to the degree of ionisation of a weak acid or base, is the fraction of the salt hydrolysis when equilibrium is established, thus; y h C S Example 1: Calculate the ph, the [H + ], the [OH ], and the degree of hydrolysis of a 0.1 M solution acetate (pka = 4.76)? ph = ½ ( ) = 8.88 [H + ] = 1.3 x 10 9 M [OH + ] / 1.3x10 9 = 7.5x10 6 M h = 7.5x x

23 6 % h 7.5x10 100x(7.5x10 5 ) 0.007% hydrolysed 0.1 In spite of the small degree of hydrolysis. Yet the [OH ] is 75 fold greater in this salt solution than in pure water Example 2: Calculate the ph of a M solution of sodium sulphathiazole (pka = 7.12)? pcs = log 1.65 x 10 1 = 0.78 ph = ½ ( ) = D) Salts of weak bases and strong acids: Consider the case of ammonium chloride; its hydrolytic reaction is: NH4 + + H2O NH4OH + H + For which ph = ½ (pkw + pcs pkb) (4) The degree of hydrolysis, h, of a salt is y/cs and the percentage of the salt hydrolysed is: 100 x (y/cs) 23

24 Example 1: Calculate the ph, the [H + ] and the degree of hydrolysis of a 0.05 N solution of ammonium chloride.? ph = ½ ( ) = 5.27 [H + ] = 5.4 x 10 6 M 6 Degree of hydrolysis = 5.4x x Percentage of hydrolysis = 100 x (1.1 x 10 4 ) = % Example 2: Calculate the ph of a M solution of ephedrine sulphate (pkb = 4.64)? pcs = log 2.5 x 10 2 = 1.58 ph = ½ ( ) = 5.47 E) Salts of weak acids and weak bases: Such salts, for example ammonium acetate, undergo hydrolysis in aqueous solutions. Provided the dissociation of the acid and the base are not widely different, the hydroxyl and hydrogen ions will be produced in approximately equal amounts. And, ph = ½ pkw + ½ pka ½pKb (5) 24

25 Lecture 2 (2 hrs).././. (Acid-base) Buffer solution, Type of buffers, Henderson equation, Properties of buffer mixtures, Buffer capacity Buffer solutions Buffers are mixtures of compounds which by their presence in solution, resist changes in ph caused by addition of small amounts of acid or base; or upon dilution. The resistance to a change in ph is known as buffer action. Types of buffers One type of buffer solution is readily prepared by dissolving a weak acid [HA] and its salt [A ] in water. Both components are necessary. If to such a buffer solution, acid is added, its hydrogen ions will be removed by the anions of the salt component of the buffer H + + A HA Although more of weak acid is thus formed, yet very little difference will occur in the [H + ] of the solution because the very low ionisation of the weak acid under these conditions. 25

26 If base is added to such a buffer solution, the hydroxyl ions will be removed by the weak acid to form strongly ionised salt. Again, little change is found in the [H + ]: OH + HA A + H2O In such a buffer solution, the weak acid [HA] is said to be in reserve: unable to increase the [H + ] of the solution but available to neutralize any base that may be added. Likewise, the salt, is in reserve; unable to contribute to the [OH ] of the solution, but available to neutralize any acid that may be added. Another type of buffer is prepared from a weak base [BOH] and its salt [B + ]. The mechanism of its buffering action is added its hydrogen ions are removed by the base as the fully ionised salt, with almost no change in the [H + ]: H + + BOH B + + H2O If base is added, its hydroxyl ions will be removed by the salt cations, forming more of the very slightly ionised base, with very little or no change in the [H + ]: OH + B + BOH 26

27 The Henderson equation: The ph of buffer solution and the change in ph upon the addition of an acid or base, may be calculated by the use of a certain equation called the Henderson equation. A) Henderson equation for acidic buffer: In an acidic buffer mixture, the acid is less ionised than if it were alone, because of the presence of the highly ionised salt of the acid which provides a high concentration of the common anion. Whatever the relative proportions of the acid and its salt in the solution, their concentrations must always be related to the hydrogen ion concentration of the acid according to the ionisation constant: K a [H ][A [HA] ] Taking logarithms of both sides, and separating the involving the hydrogen ion concentration ion the right-hand side: logk a log[h ] - [A ] log [HA] [A ] pka = ph log [HA] 27

28 [A ] ph = pka + log. (1) [HA] Since HA is a weak acid and only slightly ionised, and its ionisation suppressed by the presence of the common ion of its salt A -, the quantity [HA] is almost equal to the original concentration of the acid and [A - ] equal to the concentration of salt. Equation 1 could read [ Salt ] ph = pka + log..(2) [ Acid ] Example 1: Calculate the ph of a solution containing 0.10 N acetic acid and 0.10 N sodium acetate? According to Henderson equation: 0.10 ph = log If the ph of this solution is compared with the ph of a solution which contains 0.10 N acetic acid only; then ph = ½ (pka + pca) = ½ ( ) =

29 It is seen that the ph of acetic acid solution has been increased almost 2 ph units; i.e. the acidity has been reduced to about one hundredth of its original value by the presence of an equal concentration of a salt with common ion. Example 2 Calculate the ph of the solution produced by adding 10.0 ml of N HCl to 1 liter of solution which is 0.1 N in acetic acid and 0.1 N in sodium acetate (Ka= 1.82 x 10-5 ) [ Salt ] ph = pka + log [ Acid ] Neglecting the volume change from 1000 to 1010 ml, the HCl reacts with acetate ion forming practically undissociated acetic acid. H + + CH3COO CH3COOH [CH3COO ] = = 0.09 [CH3COOH]= = ph = log = = Hence the addition of strong acid, the ph change only by = 0.09 ph unit, whereas, if 10 ml of N-hydrochloric acid were added to liter of pure water (ph =7), the ph would have changed from 7 to log (0.01) = 2, i.e by 5 ph units. This illustrates the action of the acetic acid sodium acetate buffer 29

30 B) Henderson equation for basic buffer: [B ] ph = pkw pkb log [BOH] Now, the base BOH is weak and only slightly ionised. Also, its ionisation is repressed by the relative large concentration of cations B + from the highly ionised salt. Therefore, [BOH] is numerically equal to the initial concentration of the base, and [B + ] is numerically equal to the initial concentration of the salt. Thus, equation becomes: [Salt] ph = pkw pkb log [Base] Example: Calculate the ph of a solution containing 0.07 N ammonia, and 0.28 N ammonium chloride? 0.28 ph = log = If the ph of this solution s compared with the ph of a solution which contains 0.07 N ammonia only; being: 30

31 ph = ½( ) = Therefore, the addition of the ammonium chloride has decreased the ionisation of the base such that the ph was decreased from to Properties of buffer mixtures: a) Effect of temperature: the change in ph value of buffers with temperature is slight in case of acidic buffers (those composed of weak acid and its salt). But the ph of most basic buffers change more markedly with temperature; owing to Kw which appears in the equation of these buffers and which changes significantly with temperature. b) Effect of dilution: it is noted from Henderson equation (for both acidic and basic buffer solutions) that the ph depends on the ratio of the molar concentrations of the two compounds in a buffer solution. Therefore dilution should have no effect on ph because the volume term cancels out; the concentration of each component changes in a proportionate manner upon dilution. 31

32 Buffer Capacity The buffer capacity is defined as the number gram equivalent of strong acid or strong base required to change the ph of one liter of buffer solution by one unit. Buffer capacity can be calculated if the composition of the buffer is known. N.B. 1. Higher the concentrations of the two components of the buffer solution, the higher is the buffer capacity. 2. Also, the buffer capacity is a maximum when the two components are present in equal concentrations, The following illustrates these points. Example: Calculate the buffering capacity of a solution containing 0.1 gm equivalent of sodium acetate and acetic acid? PKa = 4.76 ph = log 1.00 = 4.76 To increase (or decrease) the ph by one unit, the ratio salt/acid will have to alter by a factor of 10: [salt] 5.76 = log [acid] [salt] [salt] log = 1; therefore = 10 [acid] [acid] To reach ph 5.76, then [salt] will have to increase to a 32

33 concentration of about gram equivalent, and the [acid] to decrease to about The buffering capacity of the buffer mixture is therefore; gram equivalent towards strong base, because the decrease in acid is brought about by adding gram equivalent of strong base. Table 1 gives compositions of some buffer systems and the ph ranges in which they are used Table 1 Some Buffer System Solution ph rang Phethalic acid and potassium phethalate Citric acid and sodium citrate Acetic acid and sodium acetate Sodium dihydrogen phosphate and and disodium hydrogen phosphate Ammonia and ammonium chloride Borax and

34 Lecture 3 (2 hrs)./../. (Acid-base) Acid-Base indicators, Oslwald Theory, Transition of indicators, Mixed indicators, Screened indicators, Turbidity indicators, Universal indicators Acid-Base indicators Acid-base indicators are substances whose presence during a titration renders the end-point visible. Thus, at a certain ph very near, or at the equivalence point of the titration the indicator produces in the system a changes which is easily perceptible to the eye, and may consist of: a- Sharp transformation from one colour to another or to colourless. Most of the colour acid base indicators of practical value are organic in nature. As the colour changes of these indicators depend on the change of the ph, they must themselves be acids or bases. The equilibrium between the indicator molecules and their ions may be represented, as follows; Acidic indicator Hln H + + In- (1) Basic indicator InOH OH- + In + (2) Where HIn is the unionised form of the acidic indicator, which gives the acid colour; In - is the ionised form which produces the basic colour; InOH is the unionised form of the basic indicator, which gives 34

35 the basic colour; In + is the ionised form, which produces the acid colour. Oslwald Theory: If the acid indicator is added to an acidic solution, the concentration of the hydrogen ion term on the right-hand side of equation (1) is increased and the ionization of the indicator is repressed by the common ion effect. The indicator is then predominantly in the unionised form of HIn, the acid colour. If on the other hand, the indicator is added to a basic solution, the [H + ] is reduced by reaction of the acid indicator with the base, and reaction (1) proceeds to the right yielding more ionised indicator [In-] and the basic colour predominates. The reverse is true for basic indicators; in basic solution, the reaction (2) proceeds to the left and the basic colour is prominent. Objectives of Oslwald theory: 1. Phenolphthalein indicator has red colour in slightly alkaline solution, when more alkali is added give colourless, while the expected from the theory, the colour should be increased. 2. Slow colour change in some indicators, while ionic reactions are usually instantanious. 3. Some indicators show their colour changes in non aqueous 35

36 media where ionisation is markedly decreased. In other words, the change in colour of indicators is a process of tautomerism, and the degree of ionisation of indicators, as controlled by the ph, is the factor which determined tautomer predominates. The nature of this process may be demonstrated in the case of the nitrophenols. In basic solutions, p-nitrophenol is present chiefly in the yellow ion; while in acid solution it is present as the colourless nitro compound: It should be noted that not all substances which show tautomeric properties can be used as indicators. The tautomeric change must be rapid, and must occupy a rather small range of ph. Therefore consideration of the tautomeric equilibria modifies the Ostwald equation. If the formula HIn represents the normal indicator molecule and the formula HIn represents the molecule formed by rearrangement (the tautomer), then the indicator salts is eventually produced from neutralisation of weak equilibrium: 36

37 HIn HIn H + + In" (3) Considering the two equilibra separately: [HIn"] Keq. [HIn] and [H ][In " ] Ka [HIn"] (4) Multiplying these two equations: [H ][In " ] Kind. [HIn] This is called the indicator constant, and not die ionization constant of the indicator. The equation can also be written in the manner of Henderson equation for buffers: [In " ] ph = pkind. + log [HIn] It is thus clear that any change in the ph, causes a change in the ratio of the logarithm tern: [In " ] [HIn] i.e [basic colour] [acid colour] So that at any ph value, both colours are present. In case of indicators in which the coloured ion is the cation and not the anion it is possible to derive a relation between colour and ph, similar to the above relation: OHIn OHIn" In''- + OH. ( 5 ) 37

38 In case of indicators in which the coloured ion is the cation and not the anion it is possible to derive a relation between colour and ph, similar to the above relation: OH In OH In" In" + + OH [OH ] [In " ] Kind. [OHIn] and poh = pkind log [OHIn] [In " ] Therefore, ph = pkw pkind log [In " ] [OHIn] in the other work, any change in the ph causes change in the ratio: [In " ] [OHIn] i.e [acid colour] [basic colour] Transition range of acid base indicators: The most efficient transition range of acid base indicators, corresponding to the effective buffer interval, is about 2 ph units; i.e pkind. 1., the reason for the width of this colour range may be explained as follows. The ability of the human eye is not overly acute; and in general, the first change in the acid colour of an indicator becomes visible when the ratio (basic colour) 38

39 /(acid colour) becomes 1/10. The ph value at which this colour change is distinguished is given by the equation: 1 ph = pkind + log = pkind 1 10 Conversely, the eye cannot detect a change in the basic colour of the indicator until the ratio (basic colour)/(acid colour) has become 10 /1, or: 10 ph = pkind + log = pkind +1 1 Therefore, when base is added to a solution of an indicator in its acid form, the eye first visualizes a change in colour when ph = pkind. 1, and the colour ceases to change when ph = pkind In other words, the indicator changes from the full acid colour to the full basic colour through a range that extends 2 ph units, this is the effective transition range of the indicator, and may be expressed as follows: ph = pkind. 1 Between the two ratios: 1/10 and 10/1, one observes an intermediate colour. An indicator therefore, does not change colour suddenly at a definite ph, but changes colour gradually over a certain ph range called the transition range of the indicator. 39

40 Table summarizes the details -of some useful acid-base indicators. Structurally, the indicators form three groups: phthaleins (e.g. phenolphthalein); sulphonephthaleins (e.g. phenol red); and azo compounds (e.g. methyl orange). A range of visual indicators of acid-base titrations Indicator pkind Low ph colour High ph colour Experimental colour change range/ph cresol red -1.0 red yellow thymol blue 1.7 red yellow bromo-phenol blue 4.0 yellow blue methyl orange 3.7 red yellow methyl red 5.1 red yellow bromo-thymol blue 7.0 yellow blue phenol red 7.9. yellow red phenolphthalein 9.6 colourless red alizarin yellow R 11.0 yellow orange nitramine 12.0 colourless orange The well-known indicator phenolphthalein is adiprotic acid and is colorless. It dissociates first to a colorless form and then, on losing the second hydrogen to an ion with a conjugated system; a red colour results 40

41 colorless pink colorless ph 8 ph 8 12 ph 12 Benzenoid structure Quinonoid structure Tribasic salt Methyl orange, another widely used indicator, is a base and is yellow in the molecular form. Addition of a hydrogen ion gives a cation which is pink in color. Yellow (azo structure) Red ph 4.4 ph 3.1 N.B. Increasing the concentration of indicators has a serious effect on one-colour indicators such as phenolphthalein. Let us write the equilibrium expression for the dissociation of the colourless acid form of phenolphthalein as [H ][In [HIn] ] K a or as [H + ] = ka[hin]/[in - ] 41

42 Where Ka is the dissociation constant and [In - ] is the minimum detectable concentration of the pink base form, which we may assume to be constant, hence [H + ] = Ka[HIn] From the above equation, it is seen that the ph at which the pink end point colour appears will depend upon the total concentration of the indicator. If more indicator is present, the ph at the end point will be lower. Conversely, if less indicator is present, the end point ph will be shift toward hitter values. In the case of the two colour indicators where the acidic and the basic forms are both coloured, the transition range is independent of the concentration. Thus, if the total concentration of a two colour indicator is increased, the individual concentrations of the acid and the base forms will increase proportionally and the ph transition range should remain unchanged even though the colour intensities are increased. Mixed indicators: These indicators are used when it is necessary to locate the ph of an end point within close limits. This close adjustment of ph may be obtained by using a suitable mixture of two 42

43 indicators, to produce a definite and characteristic colour change within a very narrow range of ph. An example of such mixed indicator is bromocresol green and methyl red; the acidic and basic colours of the mixture are orange and green respectively. Screened indicators: Screened indicator increase /the sharpness of the colour change at the end point of. a titration. A screened indicator is a mixture of an indicator and an inert dye whose colour does not change with ph. The effect of the dye is to decrease the range of wavelengths transmitted by the solution, so that the light transmitted by the two coloured forms of the indicator is not masked by ught of other wavelengths. An example is the socalled "modified methyl orange" k is a mixture of methyl orange with the inert dye xylene-cyanol F.I. This screened indicator is purple-red in acid medium and green in alkaline medium, and gray at its intermediate point. Turbidity indicators: Turbidity indicators are salts of weak organic acids or bases of high molecular weight, which coagulate and settle out of the solution at a definite ph value. It should be noted that not only 43

44 the ph of the solution influences the coagulation of the indicator but also the temperature, the presence of other salts and protective colloids, the speed of the titration, and the presence of non electrolytes as glycerin, alcohol, etc. Nevertheless, these turbidity indicators are useful in titrating weak acids or bases. Universal of multi-range indicators: By suitable mixing of several indicators, the colour change made to extend over a considerable portion of the ph range, such indicators are called "universal" or "multirange" indicators. They are not suitable for titration but indicate roughly the ph of the solution e.g. a mixed indicators containing thymol blue, methyl red, bromothymol blue and ph.ph. give the following colors at various ph values: ph colour red yellow blue orange green 44

45 Lecture 4 (2 hrs)./ /.. (Acid-base) Neutralisation titration curves, Titration curves of strong acid versus strong bases, Titration curves of strong bases versus strong acids, Titration curves of weak acids versus strong bases, Titration curves of weak bases versus strong acids Neutralization curves An insight into the mechanism of neutralization processes is obtained by studying the changes in hydrogen ion concentration during the course of the appropriate titration. The curve obtained by plotting ph as ordinates against the percentage of acid neutralized (or the number of nil of alkali added) as abscissa is known as the neutralization curve. 1- Strong acid versus strong base: In case of strong acid versus strong base, both the titrant and the analyte are completely ionized. An example is the titration of hydrochloric acid with sodium hydroxide. H + + Cl - + Na + + OH - H2O + Na + + Cl - The H + and OH- combine to form H2O, and the other ions (Na + and Cl-) remain unchanged, so the net results of neutralization is 45

46 conversion of the HC1 to a neutral solution of NaCl, To titration curve for 100 ml of 1 M HC1 with 1 M sodium hydroxide solution. It is a simple matter to calculate the ph values at different points in the titration and from them to plot a titration curve, consider, for example, the following points in the titration: a) At the beginning of titration: We have an acid concentration of 100x1 = 100 milliequivalent per 100ml [H + ]=ln ph= -log [H + ] = - log 1 = zero b) During the titration: For 50 ml of base: [H + ] = 50x1/1 50 =3.33xl0-1, or ph=0.48. For 75 ml of base: [H + ] = 25 x 1/175 = 1.43 x 10-1, or ph = For 90 ml of base: [H + ] = 10 x 1/190 = 5.27 x l0-2, or ph = For 99 ml of base: [H + ]=1x1/199 =5.03x l0-3. or ph = For 99.9 ml of base: [H + ]= 0.1 x 1/199.9 = 5.01x10-4. or ph = 3.30 c) At the equivalence point: (The point at which the reaction is theoretically complete). When acid and alkali have been added in exactly equivalent point, the solution contains only NaCI and water. The ph value of the solution is

47 d) Beyond the equivalence point: The solution contains excess alkali: With ml base [OH - ] = 0.1/200 =5.00x10-4 or poh 3.3 and ph =10.7 With 101 ml base [OH - ] = 1/201 =5.00x10-3 or poh 2.3 and ph =11.7 The results show that as the titration proceeds the ph rises slowly, but between the addition of 99.9 and ml of alkali, the ph of the solution rises from 3.3 to 10.7, i.e in vicinity of the equivalence point the rate of change of ph of the solution is very rapid. The appropriate indicator is one that changes colour between ph 33 and ph l

48 Phenolphthalein methyl red and methyl orange are most often used. 2. Titration curves of Strong bases versus Strong acids: The derivation of the titration curves in this case is analogous to that for strong acid versus strong base involving only the calculation of the concentration of excess base or excess acid at any point in the titration. 3. Titration curves of weak acids versus strong bases: In the derivation of titration curves for a solution of a weak 48

49 acid versus a strong base, the ionisation equilibrium of the acid must be taken into account; and four types of calculations must be used corresponding to the four distinct parts of the curve. Consider the titration of ml. of 0.10 N acetic acid (pka = 4.76) with 0.10 N sodium hydroxide. a) ph before adding titrant: the ph of acetic is calculated using equation ph = ½ (pka + Ca) ph = ½ ( ) = 2.88 b) ph during titration: The additions of base procedures a buffer of acetic acid and sodium acetate. The ph of the solution is calculated from the Henderson equation. [Salt] ph = pka + log [Acid] Thus, on adding 25 ml. of base, the molar concentrations of the two components of the buffer will be: 0.1 [acid] = 75 x [salt] = 25 x ph = log = on adding 50 ml. of base, the molar concentrations of acid and salt will be identical; so that the log term in the Henderson 49

50 equation will be omitted, and the ph = Similar calculations will delineate the curve in the entire buffer region, as in the next figure and table c) ph at the equivalent Point: At the equivalence point, the acetic acid is quantitatively converted to sodium acetate whose molar concentration is: [NaAc] = 100 x The ph of the solution is calculated from equation ph = ½ (pkw pcs + pka) ph = ½ ( ) = 8.73 d) ph beyond equivalence point: Beyond the equivalence point, the excess sodium hydroxide present represses the hydrolysis of the acetate ion to such an extent that its concentration becomes negligibly small. On adding 50.1 ml. of base: [NaOH] = [OH ] = poh = 4.3 ph = = x 0.1 = 5.0 x The data from these calculations are given in table and plotted in the next figure comparing this curve with that in the figure for strong acid, it is apparent that both curves are identical in the 50

51 region beyond the equivalence point. But the ph of the solution is higher in the curve for weak acid for all points up to, and including, the equivalence point. This major difference between the two curves results in a decrease in the magnitude of the ph change in the region of the equivalence point. Changes of ph during titration of 100 ml. of 0.1 N Acetic acid with 0.1 N sodium hydroxide NaOH added ph NaOH added ph

52 Effect of concentration: Using dilute reagents (e.g N), the change in ph associated with the equivalence point becomes less, as seen in figure. The initial ph is higher, but in the buffer region the two curves are identical, because the ph of buffer solutions in almost independent of dilution Indicator choice; Feasibility of titration: The above curves clearly indicate the limited choice of indicators for the titration of a weak acid. Thus, methyl orange cannot be used because its transition range does not fall within the vertical part of the curves. Bromothymol blue is not satisfactory, since its colour change requires the addition of 1 or 2 ml. of excess titrant. A suitable indicator should be one whose transition range exists in the basic region of the ph scale, e.g. phenolphthalein. 4. Titration curves of weak bases versus strong acids: The derivation of titration curves of weak bases with strong acids is analogous to that described for the case of weak acids and strong bases, except that with weak base as indicator having an acid transition range is required. E.g. titration of ammonia against HCl) 52

53 53

54 Lecture 5 & 6 (4 hrs)././.. (Acid-base) Application of neutralization reactions, Direct titration methods, Displacement titration, Biphasic titration, Residual titration Application of Neutralization Titrations Neutralization titrations are used to determine the innumerable inorganic, organic and biological species that posses inherent acidic or basic properties. Equally important, however, are the many applications that involve conversion of analyte to an acid or base by suitable chemical treatment followed by titration with standard strong base or acid. Determination of Inorganic Substances 1. Determination of ammonium salts (e.g. NH4Cl) 1. Formol titration: When formaldehyde is added to solution of ammonium salt, hexamethylene tetramine [(CH2)6N4] is produced accompanied by an amount of acid equivalent to the ammonium salt present in the solution. The librated acid can be titrated against standard alkali using phenolphthalein (ph.ph.) as indicator. 4NH4C1 + 6HCHO (CH2)6N4 + 4HCl + 6 H2O 54

55 2. Direct method: In this method a solution of the ammonium salt is treated with a solution of strong base (e.g. sodium hydroxide) and the mixture distilled using the distillation apparatus. Ammonia is quantitatively expelled, and is absorbed in an excess standard acid. The excess of acid is back titrated in the presence of methylorange as indicator. NH4 + + OH NH3 + H2O 3. The indirect method The ammonium salt (other than carbonate or bicarbonate) is boiled with known excess of standard sodium hydroxide solution. The bailing is continued until no more ammonia is evolved. The excess of sodium hydroxide is titrated with standard acid, using methyl orange as indicator. 2. Determination of Carbonate and Bicarbonate in Mixture The analysis of such mixture requires two titrations, one with an alkaline-range indicator, such as ph.ph., and the other with an acidrange indicator, such as methyl-orange. 55

56 Na2CO3 + HC1 NaHCO3 + NaCl ph 8.3 MaHCO3 + HCl CO2 + H2O + NaCl... ph 3.8 A portion of cold solution is slowly titrated with standard hydrochloric acid using ph.ph. as indicator. This volume of acid (V1) corresponds to half the carbonate: CO H + HCO3 - Another sample of equal volume is then titrated with the same standard acid using methyl-orange, as indicator. The volume of acid (V2) corresponds to carbonate + bicarbonate; hence 2 V4 = carbonate and V2-2Vi = bicarbonate. Titration curve for 50 ml 0.1 sodium carbonate versus 0.1 M HCl 56

57 3- Determination of a Mixture of Carbonate and Hydroxide (analysis of commercial caustic soda) Two methods are used for this analysis. In the first method the total alkali (carbonate + hydroxide) is determined by titration with standard acid, using methyl orange as indicator. In a second portion of solution the carbonate is precipitated with a slight excess of barium chloride solution, and, without filtering, the solution is titrated with standard acid using phenolphthalein as indicator. The latter titration gives the hydroxide content, and by subtracting this from the first titration, the volume of acid required for the carbonate is obtained. Na2CO3+ BaCl2 = BaCO3 (insoluble) + 2 NaCl The second method utilizes two indicators. It has been stated that the ph of half-neutralised sodium carbonate, i.e. at the sodium hydrogen-carbonate stage, is about 8.3, but the ph changes comparatively slowly in fee neighborhood of the equivalence point; consequently the indicator color-change with phenolphthalein (ph range ) is not too sharp. This difficulty may be overcome by using a comparison solution containing sodium hydrogen-carbonate of approximately the same concentration as the unknown and the same volume of indicator. 57

58 A simpler method is to employ a mixed indicator composed of thymol blue and cresol red; this mixture is violet at ph 8.4, blue at ph 8.3 and rose at ph 8.2. With this mixed indicator the mixture has a violet colour in alkaline solution and changes to blue in the vicinity of the equivalence point; in making the titration the acid is added slowly until the solution assumes a rose colour. At this stage all the hydroxide has been neutralised and the carbonate converted into hydrogen carbonate. Let the volume of standard acid consumed be v ml. OH +H + = H2O CO H + = HCO3 Another titration is performed with methyl orange, as indicator. Let the volume of acid be V ml. OH +H + = H2O CO H + = HCO3 H2CO3 = H2O + CO2 Then V - 2(V-v) corresponds to the hydroxide, 2(V-v) to the carbonate, and V to the total alkali. To obtain satisfactory results by this method the solution titrated must be cold (as near 0 C as is practicable), and loss of carbon dioxide must be prevented as far as possible by keeping the tip of the burette immersed in 58

59 4. Determination of Boric Acid Boric acid acts as a weak monoprotic acid (Ka= 6.4x10-10 ), it cannot therefore titrated accurately with standard alkali. However, by the addition of certain organic polyhydroxy compounds, such as glycerol, mannitol, sorbitol, or glucose, it acts as a much strong monobasic and can be directly titrated with sodium hydroxide; using phenolphthalein as indicator. NaOH + H3BO3 = NaBO2 + 2 H2O The effect of polyhydroxy compounds has been explained on the basis of the formation of 1:1 and l:2 -mole ratio complexes between the hydrated borate ion and 1,2- or 1,3-diols: 59

60 5- Determination of Borax When borax is dissolved in water, it is hydrolysed into: Na2B4O7 + 7 H2O = 4 H3BO3 + 2 NaOH If the aqueous solution is titrated with standard hydrochloric acid using methyl orange as indicator, it is the NaOH that is actually titrated; boric acid being of no effect on die indicator, and net reaction is: Na2B 4O7 5 H2O + 2 HC1 = 4 H3BO3 + 2 NaCl Na2B 4O7 10 H2O = 2 HCl The residual solution can be titrated for the remaining boric acid with standard sodium hydroxide after adding glycerol and using phenolphthalein as indicator. The reaction would be: Na2B4O7 + 5 H2O + 2 HC1 = 4 H3BO3 + 2 NaCl Na2B4O7.10 H2O = 4 NaOH In other words, when a pure sample of borax containing no free acid is titrated, the volume of standard alkali used would be exactly double the volume of standard acid. If the borax solution is treated directly with glycerol and titrated against standard sodium hydroxide, the solution could then be 60

61 regarded as boric acid which is half neutralised. Glycerol Na2B4O7 + 5 H2O NaH2BO3 + 2 H3BO3 Na2B4O7.10 H2O = 2 NaOH 6- Determination of Mixture of Boric Acid and Borax Solutions of alkali borates may be titrated with standard acid (e.g. HC1) using methyl orange as indicator. They react towards this indicator as if they were solutions of alkali hydroxides. They behave as dinormal bases when titrated with acids. Na2B4O7.10 H2O + 2 HC1 = 4 H3BO3 + 2 NaCl + 5 H2O While the liberated boric acid consumes 4 molecules of NaOH when titrated with alkali (e.g. NaOH), using phenolphthalein as indicator in presence of glycerol (More details in the practical part). 7- Determination of Nitrogen by Kjeldahl's Method Nitrogen is found in a wide variety of substances. Examples include amino acids, proteins, synthetic drugs, fertilizers, explosives soils, potable water supplies, and dyes. The most common method for determining organic nitrogen is the Kjeldahl method, [n this method, the sample is decomposed in hot, concentrated sulphuric acid to 61