Analytical Chemistry Lecture 2: Titrimetric Methods PDF

Analytical Chemistry For medical-lab Lecture 2 Titrimetric Methods of Analysis Jimma, Ethiopia 1 2. Titrimetric Methods of Analysis Titration is determining analytes concentration by titrant. Achieved by add...

Analytical Chemistry For medical-lab Lecture 2 Titrimetric Methods of Analysis Jimma, Ethiopia 1 2. Titrimetric Methods of Analysis Titration is determining analytes concentration by titrant. Achieved by adding known conc (C1), volume (𝑉1 ) to analyte of (𝑉2). The addition of titrant is stopped when there is color change (end point) or nearly equivalent point. 𝑁1 𝑉1 = 𝑁2 𝑉2 Equivalent point: when the amount of added standard reagent is equivalent to the amount of analyte. Or where equivalent amounts of analyte and titrant react stoichiometrically.  End point: when a physical change that is associated with the condition of chemical of equivalent occurs.  The two most widely used end points involves:  Changes in color due to reagent or indicators and  Change in potential of an electrode The difference b/n equ’t pt & end pt is titration error Titration error = 𝑉𝐸𝑞𝑢 – 𝑉𝑒𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 Back-titration: excess standard solution is used to consume analyte & determined by titration with second standard sol’n.  Often required when rate of rxn b/n analyte and reagent is slow. or when the standard solution lacks of the stability. Titrimetric methods are classified into four groups.  Acid–base titrations: based on acidic/basic rxn.  Complexometric titrations: by metal–ligand complexing rxn.  Redox titrations: due to oxidizing or reducing agent.  Precipitation titrations: based on precipitate formation. Titration is usually performed by adding standard solution from burette to the analyte beaker, conical or Erlenmeyer flask.  Standard solutions are prepared from compd of high purity.  Standard solutions are two kinds: 1. Primary standard solutions and 2. Secondary standard solution. Acid – base indicators Indicator: is a compound often added to the analyte solution in order to give an observable physical change (at the end point or near the equivalent point) Acid-base indicators: weak organic acid/bases that show color change over different pH range Examples: HIn + H2O In - + H3O+ red Color blue Color 3/15/2023 Study hard! 6 Types of acid – base indicators 1. Phthalein indicator is basic characteristics Are colorless in acidic medium, pink in alkali. Suitable for weak acid & strong base titration eg: phenolphthalein: has pH range 8.3-10.0 2. Sulfolphthalein indicator Two transitions are observed The transitions occur is acid, neutral, moderately basic Color is stable towards strong alkali Common range (6.4 – 8.00), eg: Neutral red 3. Azo indicators Transition usually occurs on the acid side. Used for weak base & strong acid titration. Example: methyl orange (3.1 – 4.4) & methyl red (4.4 - 6.2). 4. Universal indicator is a mixture of indicators which can indicate wide pH value. ranged b/n 1 – 14 by showing successive color changer. When universal indicator show Red the solution is acidic Green = neutral Blue = alkaline Acid-base titration is determination of acid or base concentration based on neutralization reaction. Eg: NaOH + HCl  NaCl + H2O this reaction is takes place by 1:1 stoichiometric ratio Thus conc determination is performed by M1V1 = M2V2 Ex’l: unknown solution of 10 mL NaOH is titrated by 0.1M of HCl, when a color change is appeared volume of HCl is 20 mL determine the conc of NaOH. In this example the equivalent volume is 20 mL, this can lead to the determination of titration curve. Titration curve is a plot that obtained from pH vs volume added or instrument reading vs volume added. Titration Curve  Is a graph as a function of pH vs volume of titrant added.  There are two types the titration curves sigmoidal curve linear segment curve pH vs volume added instrument reading vs volume added Titration Curve determination 1. Strong acid (analyte) Vs strong base (titrant) titration i. Initially: no adding of titrant, pH is from analyte conc. ii. Before the equ’t point: pH calculated from excess acid. 𝑚.𝑚𝑜𝑙𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑒𝑚𝑎𝑖𝑛𝑖𝑛𝑔 𝑎𝑐𝑖𝑑 [𝐻 +] = 𝑡𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 iii. At the equivalent point [H+] = [OH-] = 10-7; pH is from pH of water thus, pH = 7 iv. After the equivalence point, pH is from excess base. 𝑚𝑜𝑙𝑒 𝑒𝑥𝑐𝑒𝑠 𝑏𝑎𝑠𝑒 [𝑂𝐻 −] = , pOH obtained. then pH= 14 - pOH 𝑡𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 Example: consider the titration of 50.0 mL of 0.100M HCl with 0.200M of NaOH. For the reaction of a strong base with a strong acid the only equilibrium reaction of importance is; H3O+ (aq) + OH– (aq) 2H2O (l) To calculate the volume of NaOH needed to reach the equivalence point. At the equivalence point we know from reaction above that : Moles HCl = Moles NaOH MaVa = MbVb  The volume of NaOH needed to reach the equivalence point, is therefore, Vb = MaVa/Mb = 0.100M x 50.0mL/0.200M = 25 mL of NaOH Before the equivalence point: HCl is present in excess and the pH is determined by the concentration of excess HCl. Initially the solution is 0.100 M in HCl, which, since HCl is a strong acid, means that the pH is pH = –log[H3O+] = –log[HCl] = –log(0.100) = 1.00 After adding 10.0 mL of NaOH: the concentration of excess HCl is moles excess HCl MaVa − MbVb [HCl] = = total volume 𝑉𝑎+𝑉𝑏 (0.100Mx0.05L) −(0.200Mx0.01L) = 0.05L+0.01L = 0.050M Therefore the pH of the solution is pH = –log[H3O+] = –log[HCl] = –log(0.050) = 1.30 At the equivalence point: the moles of HCl and the moles of NaOH are equal. Since neither the acid nor the base is in excess, the pH is determined by the dissociation of water. Kw = 1.00 x10–14 = [H3O+][OH–] = [H3O+]2 [H3O+] = 1.00 x10–7 M After equivalence point: pH is determined by the concentration of excess OH–. Up on adding 30.0 mL of titrant the conc of OH– is  moles excess NaOH MbVb−MaVa [OH-] = = total volume Va+Vb (0.200Mx0.03L) −(0.100Mx0.05L) = = 0 0125M 0.05L+0.03L Exercise:  Generate the hypothetical titration curve for the titration of 50 mL of 0.050M of HCl with 0.1M of NaOH.  At different NaOH addition, let A) pH= 1.3 B) pH after addition of 10.0mL= 1.6 C) pH=7 after addition of 25mL D) pH=10.12 after addition of 25.10mL  summery of strong acid with strong base and vs Titration Titration of strong acid Titration of strong base with Stage with strong base strong acid Initial pH -log Ca 14+ log Cb Pre-equ’t -log (CaVa) – (CbVb) 14 + log (VbCb) – (VaCa) pH (Va + Vb) (Va+ Vb) Equ’t pH 7.00 7.00 -log (VaCa) – (VbCb) Post Equ’t 14 + log Va + Vb (VbCb) – (VaCa) (Va+ Vb) pH Titration curve of Strong acid and strong base. ? Assignment: show weak acid and weak base equations Stage of Titration of Weak acid Titration of Weak bases Titration with Strong base. with strong acid. Initial pH 𝟏 𝟏 𝟏 pKa - log Ca pKw - pKb + log Cb 𝟐 𝟐 𝟐 Pre- 𝐕𝐚𝐂𝐚 𝐕𝐛𝐂𝐛 equivalen pKa – log −𝟏 pKw – pKb + log 𝐕𝐛𝐂𝐛 𝐕𝐛+𝐕𝐛 ce pH 𝟏 𝟏 𝟏 𝐕𝐚𝐂𝐚 𝟏 𝟏 𝟏 𝐕𝐛𝐂𝒃 Equivalen 𝟐 𝐩𝐊𝐚 + pKw+ 𝟐 𝟐 log 𝐕𝐛𝐂𝐛 −𝟏 pKw - pKb - log 𝟐 𝟐 𝟐 𝐕𝐛+𝐕𝐛 ce pH Post 𝑽𝒃𝑪𝒃 −(𝑽𝒂𝑪𝒂) 𝑽𝒂𝑪𝒂 −(𝑽𝒃𝑪𝑏 ) pKw + log -log equivalen (𝑽𝒂+𝑽𝒃) (𝑽𝒂+𝑽𝒃) ce pH 19 Precipitation titration It involves pption rxn and formation of precipitate Insoluble compound is produced from analyte & titrant. its application is limited due to:  Lack of suitable indicators  Too slow reaction rate in dilute solution  suitable indicator availability to locate the end point The most important pption rxn involves AgNO3 called Argentometric titration. It is used to determine concentration of halides. This titration has 3 most common end point detection methods. End point detection methods 1. Mohr method: Formation of a colored precipitate Characterized by formation of a second color ppt. It is direct method using CrO4-2 as indicators Example: chloride determination by AgNO3 in the presence of chromate. Analytical reaction: Ag+ + Cl-  AgCl (ppt) Indicator reaction: Ag+ + 𝐶𝑟𝑂4−2  AgCrO4 ( red ppt) Condition Don.t use excess or too small 𝐶𝑟𝑂4−2 because Excess 𝐶𝑟𝑂4−2 : form color change before equivalent point.  Too small 𝐶𝑟𝑂4−2 : consume more Ag+ color formed after eq’t pt pH should be b/n 6 – 10 because: In extremely basic – Ag2O (ppt) is formed In extremely acidic - 𝐶𝑟𝑂4−2  H𝐶𝑟𝑂4− or H2CrO4 is formed 2. Formation of a colored complex or Volhard method Characterized by the formation of a colored complex This is an indirect method or back titration method. It uses excess AgNO 3 which is back titrated with thiocyanate using ferric alum (Fe+3) indicators Analytic reaction: SCN- + Ag+  AgSCN (white ppt) Indicator rxn: Fe+3 + 6SCN-  (Fe(SCN)6)-3 blood red. Back titration: is the process of determining the amount of a titrant needed to react with a first titrant that is in stoichiometric excess of the amount needed to react with analyte 3. Adsorption indicator method (Fajan’s method) Characterized by adsorption of indicators on ppted surface. It is direct method Before the equivalent point: AgCl (ppt) and Cl -. After the equivalent point: AgCl (ppt) and Ag+. Ag+ react with fluorescen ion to adsorb on AgCl surface. then it change color (it impact red color to it). Representative Examples of Precipitation Titrations 24 Example: calculate the titration curve for the titration of 50.0 mL of 0.0500 M Cl– with 0.100 M Ag+. The reaction in this case is Ag+(aq) + Cl–(aq) AgCl(s) Ksp AgCl= 1.8 x10–10 Pre-equivalence point data, The equilibrium constant for the reaction is K = (Ksp)–1 = (1.8 x10–10)–1 = 5.6 x 109 The stoichiometry of the reaction requires that Moles Ag+ = moles Cl– , MAgVAg = MClVCl Solving for the volume of Ag+ MClVCl 0.05Mx50mL VAg+ = = = 25mL MAg 0.1M 25  25.0 mL of Ag+ makes equivalence point. Initially: there is only [Cl–] = 0.05 M, pCl (at 0 mL) = 1.3 Before: the equivalence point Cl– is in excess. The conc. of unreacted Cl– after adding 10.0 mL of Ag+ is MClVCl−MAgVAg [Cl–] = moles excess Cl − = total volume VCl+VAg 0.05Mx50ml −0.1Mx10ml = 60mL , [Cl–] = 2.5 x10-2M pCl (at 10 mL) = –log[Cl–] = –log(2.50x10–2) = 1.60 concentration for Ag+ is calculated from Ksp of AgCl Ksp = [Ag+][Cl–] = 1.8 x10–10 [Ag+] = Ksp = 1.8 x10–10 = 7.2x10-9 [Cl–] 2.50x10–2 26  Then pAg+ = –log[Ag+] = –log(7.2x10–9) = 8.14 At equivalence pt [Ag+] = [Cl–]. Using Ksp expression: Ksp = [Ag+][Cl–] = [x] 2 = 1.8 x 10–10 [Ag+] = [Cl–] = 1.3x10–5 M, then pAg = pCl = 4.89.  pCl- (at 25 mL) = –log[Cl–] = –log(1.3x10–5) = 4.89  pAg+= –log[Ag+]= –log(1.3x10–5) = 4.89 After equivalence pt: excess Ag+ is exist. The conc. of Ag+ after adding 35.0 mL of titrant is; moles excess Ag + MAgVAg −MClVCl [Ag+] = = total volume VCl+VAg (0.1Mx35ml) − (0.05Mx50mL) [Ag+] = = 1.18x10-2M 50mL+35mL pAg+ = –log[Ag+] = –log(1.18x10 -2) = 1.93 27 The concentration of Cl– is: [Cl–] = 𝐾𝑠𝑝 = 1.8 x10–10 = 1.5x10-8 M [Ag+ ] 1.18x10−2 then pCl (at 35 mL)= –log[Cl–] = –log(1.5x10-8) = 7.82 Pption titration curve Volume of pCl AgNO3 (mL) 0.00 1.30 10.00 1.60 25.00 4.89 35.00 7.82 28  Complexometric Titration is the titration in which the reaction between the analyte and titrant is a complexation reaction.  important in many areas eg. black-and-white photography. colored complexes can absorb UV-radiations as result it analyzed by spectrophotometric method.  Some complex are sparingly soluble and used in the:  Gravimetric analysis,  Precipitation analysis. 29 Complexes are widely used: To extract cations from one solvent to anothers,  To dissolve insoluble precipitates.  The most widely used complex forming reagents are;  complexing agents useful in:  Precipitating metals,  Binding metals to prevent interference,  Extracting metals from one solvent to anothers,  Forming complexes for spectropic analysis.  Most common complexometric titration reagent is EDTA.  EDTA Ethylene diamine tetra acetic acid (EDTA), is an amino carboxylic acid. EDTA, which is a Lewis acid, has six binding sites four of them are carboxylate groups and the two are amino groups, Thus EDTA is a Hexadentate ligands. HOOC– CH2 CH2—COOH :N—CH2—CH2—N: HOOC—CH2 CH2--COOH 3/15/2023 Study hard! 31 Complexes of EDTA and metal ions All metal–EDTA complexes have a 1:1 stoichiometry Example: Ag+ and Al3+ complexes are formed by the reactions of: Ag+ + Y4- AgY3- Al3+ + Y4- AlY- In general, we can write the reaction of EDTA anion with metal ion Mn+ Mn+ + Y4- MY (n-4)+ EDTA is the remarkable reagents b/c of:  It forms chelates with all cations except alkali metals. Study hard! 32  Cd2+ (aq) + Y4– (aq) CdY2– (aq) where Y4– is a shorthand notation for EDTA. The formation constant for this reaction. 2 If M = Cd [𝐶𝑑𝑌 ] Kf = 2 4 𝐶𝑑 [𝑌 ] Example : calculate cadmium concentration before the and after equivalent point, if 50.0 mL of Cd2+ is titrated with 0.010M EDTA at a pH of 10 and in the presence of 0.0100M NH3.  At equ’t pt Moles EDTA = moles Cd2+ MEDTAVEDTA = MCdVCd Solving the volume of EDTA MCdVCd 0.005Mx50mL VEDTA = = = 25 ml MEDTA 0.01M Initially: only [Cd] = 0.0005 M is there, thus pCd = 3.30 Before equ’t pt: (adding 5.0 mL) Cd2+ is excess moles excess Cd 2 + M Cd xV Cd −MEDTA VEDTA [Cd] = Total Volume = Total Volume (0.0005Mx50ml) −(0.01Mx5ml) = = 3.64x10-3M 50ml+5ml pCd = 3.49 At equivalence pt [Cd] = [EDTA]. Cd2+ un-complexed we must account for the presence of NH 3; fraction of uncomplexed Cd2+ (αCd2+ ) = [Cd2+] / CCd = 0.0881 thus [Cd2+] = αCd2+ × CCd = (0.0881)(1.87×10−9 M) = 1.64×10−10 M [Cd2+ ] = 1.64×10−10 M, then pCd = pEDTA = 9.78. After equ’t pt: adding 50.0 mL of EDTA M EDTA x V EDTA − M Cd x V Cd [EDTA] = Total Volume 0.01Mx30 ml −(0.0005Mx50ml) = = 6.25x10-4M 30ml+50ml [Cd2+ ]=4.64×10−16 M, then pCd = 15.33. Redox titration Redox reaction is takes place due to electron transfer. The potential is calculated by Nernst equation. EAoxi/red is the standard state potential for half rxn. At the equivalent point the potential is the easiest to calculate using the Nernst equation for titrants half reaction. 36  Ex’l: Determine titration curve for the titration of 50.0 mL of 0.100 M Fe 2+ with 0.100 M Ce 4+ in a matrix of 1 M HClO 4. The reaction in this case is Fe2+ (aq) + Ce4+ (aq) Ce3+ (aq) + Fe3+ (aq) Keq = 6x1015 In 1 M HClO4, the formal potential for the reduction of Fe3+ to Fe2+ is +0.767 V, and for reduction Ce4+ to Ce3+ is +1.70 V. The first task is to calculate the volume of Ce 4+ needed to reach the equivalence point. From the stoichiometry of the reaction we know; Moles Fe2+ = moles Ce4+ , MFeVFe = MCeVCe  Solving for the volume of Ce4+  To determine the titration curve E(V) should be calculated. Initially: no addition of Ce4+ thus no redox rxn & E = 0 V At equivalent point: potential is calculated by Because [Fe2+] = [Ce4+] and [Ce3+] = [Fe3+] at the eq’t point, the log term has a value of zero & eq’t point potential is E (V) = + 1.23 V  Before equivalent point: potential is calculated by E (V) = + 0.731 V  After equivalent point: potential is calculated by E (V) = + 1.66 V  Redox titration curve is plotted from E(v) vs volume added. volume of E (V) Ce4+ (mL) 0.00 0.00 10.00 0.73 50.00 1.23 60.00 1.67