Titrations: A Volumetric Analysis Procedure
Titrations are a common laboratory technique used in measuring the amount of a substance in a solution. In this procedure, we measure the volume of reagent needed to react
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Slide1TitrationsVolumetric analysis Procedures in which we measure the volume of reagent needed to react with an analyte Titration Increments of reagent solution (titrant) are added to analyte until reaction is complete. - Usually using a buret Calculate quantity of analyte from the amount of titrant added. Requires large equilibrium constant (Thermodynamic) Requires rapid reaction (kinetic) aA + tT → products A: analyte T: titrant 1 13920131 http:\\asadipour.kmu.ac.ir 41 slides
Slide2TitrationsBuret Evolution Descroizilles (1806) Pour out liquid Gay-Lussac (1824) Blow out liquid Henry (1846) Copper stopcock Mohr (1855) Compression clip Used for 100 years Mohr (1855) Glass stopcock 2 13920131 http:\\asadipour.kmu.ac.ir 41 slides
Slide3Type of Titrations based on Measuring Techniquesi) Volumetric titrimetry: Measuring the volume of a solution of a known concentration (e.g., mol/L ) that is needed to react completely with the analyte. ii) Gravimetric (weight) titrimetry: Measuring the mass of a solution of a known concentration (e.g., mol/kg ) that is needed to react completely with the analyte . iii) Coulometric titrimetry : Measuring total charge (current x time) to complete the redox reaction, then estimating analyte concentration by the moles of electron transferred . 13920131 TMHsiung@2007 3/42 http:\\asadipour.kmu.ac.ir 41 slides
Slide4Type of Titrations based on Chemical Reactionsi) Acid-Base Titrations, example: H + + OH – → H 2 O ii) Precipitation Titrations, example: Ag + (aq) + Cl – (aq) → AgCl (s) ii) Redox Titrations: 5 H 2 O 2 + 2 MnO 4 – + H + → 5 O 2 + 2 Mn 2+ + 8H 2 O iv) Complexometric Titrations, example: EDTA + Ca 2+ → (Ca –EDTA) 2+ 13920131 TMHsiung@2007 4/42 http:\\asadipour.kmu.ac.ir 41 slides
Slide5Type of Titration CurvesType Example. y-axis x-axis Acid-base HCl/NaOH pH V. NaOH Precipitation Ag + /Cl – pAg + V. Ag + Complexation Ca 2+ /EDTA pCa 2+ V. EDTA Redox MnO 4 – /Fe 2+ Potential V. Fe 2+ Type Example y-axis x-axis Spectro- photometric apotransferrin/ Fe 3+ Absorbance V. Fe 3+ Thermo- metric H 3 BO 4 / NaOH Temperature V. NaOH V. = volume 13920131 TMHsiung@2007 5/42 http:\\asadipour.kmu.ac.ir 41 slides
Slide6Expressing concentrationFormality Molarity (V & W) Molality Normality %W/W %W/V %V/V part per thousand (ppt) - X's 1000 parts per million (ppm) - X's 10 6 parts per billion (ppb) - X's 10 9 6 13920131 http:\\asadipour.kmu.ac.ir 41 slides
Slide7relationship between titrantand analyte # Eqg titrant = # Eqg analyte (V*N) titrant =(V*N) analyte # Eqg titrant = (V*N) titrant #moles titrant =(V*M) titrant #moles analyte =(V*M) analyte 7 13920131 http:\\asadipour.kmu.ac.ir 41 slides
Slide8Standardization : The process by which the concentration of a reagent is determined by reaction with a known quantity of a second reagent Primary standard : The reagent which is ready to be weighted and used prepare a solution with known concentration ( standard ). Requirements of primary reagent are: - Known stoichiometric composition and reaction - High purity - Nonhygroscopic - Chemically stable both in solid and solution - High MW or FW Secondary standard : A standard solution which is standardized against a primary standard. TMHsiung@2007 8/42 13920131 http:\\asadipour.kmu.ac.ir 41 slides
Slide9Standardization Required when a non-primary titrant is used - Prepare titrant with approximately the desired concentration - Use it to titrate a primary standard - Determine the concentration of the titrant - titrant known concentration analyte unknown concentration titrant unknown concentration analyte known concentration Titration Standardization 9 13920131 http:\\asadipour.kmu.ac.ir 41 slides
Slide10Standardization of 0.1 M NaOH1-selection the PS (e.g. KHP) 2-wheing the PS 10* 0.1 =mg/204.1 213.8 3-making solution 4-addind suitable indicator 5-titration 9.1ml 6-calculation 9.1*n=213.8/204.1 n= 0.115 10 13920131 http:\\asadipour.kmu.ac.ir 41 slides N1V1=N2V2
Slide11Blank Titration : Titration procedure is carried out without analyte (e.g., a distilled water sample). It is used to correct titration error. Back titration : A titration in which a ( known ) excess reagent is added to a solution to react with the analyte. The excess reagent remaining after its reaction with the analyte, is determined by a titration. 13920131 TMHsiung@2007 11/42 http:\\asadipour.kmu.ac.ir 41 slides
Slide12Example: To standardizing a KMnO4 stock solution, the primary standard of 9.1129 g Na 2 C 2 O 4 is dissolved in 250.0 mL volumetric flask. 10.00 mL of the Na 2 C 2 O 4 solution require 48.36 mL of KMnO 4 to reach the titration end point. What is the molarity (M) of MnO 4 – stock solution? ( FW Na 2 C 2 O 4 134.0 ) Solution: Ans 5C 2 O 4 2 – (aq) + 2MnO 4 – (aq) + 16H + (aq) → 10CO 2(g) + Mn 2+ (aq) + 8H 2 O (l) Standardization 13920131 TMHsiung@2007 12/42 http:\\asadipour.kmu.ac.ir 41 slides
Slide13Example: A 0.2865 g sample of an iron ore is dissolved in acid, and the iron is converted entirely to Fe 2+ . To titrate the resulting solution, 0.02653 L of 0.02250 M KMnO 4 is required. Also a blank titration require 0.00008 L of KMnO 4 solution. What is the % Fe (w/w) in the ore? (AW Fe 55.847) MnO 4 – (aq) + 5Fe 2+ + 8H + (aq) → Mn 2+ (aq) + 5Fe 3+ + 4H 2 O (l) U nknown Analysis with a Blank Correction 13920131 TMHsiung@2007 13/42 http:\\asadipour.kmu.ac.ir 41 slides
Slide14Back Titration1) Add excess of one standard reagent (known concentration) 2) Titrate excess standard reagent to determine how much is left - Add Fe 2+ to determine the amount of MnO 4 - that did not react with oxalic acid - Differences is related to amount of analyte - Useful if better/easier to detect endpoint 14 MnO 4 – (aq) + 5Fe 2+ + 8H + (aq) → Mn 2+ (aq) + 5Fe 3+ + 4H 2 O (l) 13920131 http:\\asadipour.kmu.ac.ir 41 slides
Slide15Back TitrationExample: The arsenic in 1.010 g sample was pretreated to H 3 AsO 4(aq) by suitable treatment. The 40.00 mL of 0.06222 M AgNO 3 was added to the sample solution forming Ag 3 AsO 4(s) : The excess Ag + was titrated with 10.76 mL of 0.1000 M KSCN . The reaction was : Calculate the percent (w/w) As 2 O 3(s) (fw 197.84 g/mol) in the sample. 13920131 TMHsiung@2007 15/42 http:\\asadipour.kmu.ac.ir 41 slides
Slide1613920131TMHsiung@2007 16/42 http:\\asadipour.kmu.ac.ir 41 slides
Slide17In a back titration analysis of HCO3 -, 25 mL of a bicarbonate solution is reacted with 25.00 mL of 0.100 M NaOH. The excess NaOH was titrated with 0.100 M HCl. This required 14.82 mL. What is the concentration of bicarbonate in solution? NaOH + HCO 3 - → Na + + CO 3 - + H 2 O NaOH + HCl → NaCl + H 2 O 13920131 TMHsiung@2007 17/42 http:\\asadipour.kmu.ac.ir 41 slides Back Titration
Slide18Equivalence point Quantity of added titrant is the exact amount necessary for stoichiometric reaction with the analyte - Ideal theoretical result Analyte Oxalic acid (colorless) Titrant (purple) (colorless) (colorless) Equivalence point occurs when 2 moles of MnO 4 - is added to 5 moles of Oxalic acid 18 13920131 http:\\asadipour.kmu.ac.ir 41 slides
Slide19End pointOccurs from the addition of a slight excess of titrant - Marked by a sudden change in the physical property of the solution - Change in color, pH, voltage, current, absorbance of light. - End point ≈ equivalence point Analyte Oxalic acid (colorless) Titrant (purple) (colorless) (colorless) After equivalence point occurs, excess MnO 4 - turns solution purple Endpoint 19 13920131 http:\\asadipour.kmu.ac.ir 41 slides
Slide20Titration Error - Difference between endpoint and equivalence point Corrected by a blank titration 1) repeat procedure without analyte 2) Determine amount of titrant needed to observe change 3) subtract blank volume from titration 20 13920131 http:\\asadipour.kmu.ac.ir 41 slides
Slide21Calculation of ascorbic acid in Vitamin C tablet: (i) Starch is used as an indicator: (ii) starch + I 3 - starch-I 3 - complex ascorbic acid was oxidized with I 3 - : 1 mole ascorbic acid 1 mole I 3 - 21 Standardization : Suppose 29.41 mL of I 3 - solution is required to react with 0.1970 g of pure ascorbic acid, what is the molarity of the I 3 - solution? Analysis of Unknown : A vitamin C tablet containing ascorbic acid plus an inert binder was ground to a powder, and 0.4242g was titrated by 31.63 mL of I 3 - . Find the weight percent of ascorbic acid in the tablet. 13920131 http:\\asadipour.kmu.ac.ir 41 slides
Slide22Titration of a MixtureExample: A solid mixture weighing 1.372 g containing only sodium carbonate ( Na 2 CO 3 , FW 105.99 ) and sodium bicarbonate ( NaHCO 3 , FW 84.01 ) require 29.11 mL of 0.7344 M HCl for complete titration: Find the mass of each component of the mixture. mass moles Total mixture 1.372 Na 2 CO 3 x x/105.99 g/mol NaHCO 3 1.372 - x 1.372 – x/84.01 g/mol 13920131 TMHsiung@2007 22/42 http:\\asadipour.kmu.ac.ir 41 slides
Slide23AnsAns 13920131 TMHsiung@2007 23/42 http:\\asadipour.kmu.ac.ir 41 slides
Slide24Direct and back (indirect) titration ofAspirin 13920131 TMHsiung@2007 24/42 http:\\asadipour.kmu.ac.ir 41 slides
Slide25A titration in which the reaction between the analyte and titrant involves a precipitation. Ag + (aq) + Cl – (aq) AgCl (s) AgCl (s) Ag + (aq) + Cl – (aq) K sp = 1.8 × 10 –10 s = [Ag + ]=[Cl – ] [Ag + ]=[Cl – ]=1.35x10 –5 pAg = 4.89 pCl = 4.89 Precipitation Titrations 13920131 TMHsiung@2007 25/42 http:\\asadipour.kmu.ac.ir 41 slides
Slide26 Titration curve of50.0 mL of 0.0500 M Cl – with 0.100 M Ag + pCl pAg 13920131 TMHsiung@2007 26/42 http:\\asadipour.kmu.ac.ir 41 slides
Slide27Example: For the titration of 50.0 mL of 0.0500 M Cl – with 0.100 M Ag + . The reaction is: Ag + (aq) + Cl – (aq) AgCl (s) K = 1/K sp = 1/(1.8 × 10 –10 ) = 5.6 x 10 9 Find pAg and pCl of Ag + solution added (a) 0 mL (b) 10.0 mL (c) 25.0 mL (d) 35.0 mL Solution: 13920131 TMHsiung@2007 27/42 http:\\asadipour.kmu.ac.ir 41 slides 50*0.05=0.1*V2 In eq point, 25 ml N1V1=N2V2 (a) 0 mL Ag + added (At beginning) [Ag + ] = 0, pAg can not be calculated. [Cl – ] = 0.0500, pCl = 1.30 (b) 10 mL Ag + added (Before V e ) (d) 35 mL Ag + added (After V e ) √K sp =[Ag + ]=[Cl - ]=1.34*10 -5
Slide28Diluting effect of the titration curves25.00 mL 0.1000 M I – titrated with 0.05000 M Ag + 25.00 mL 0.01000 M I – titrated with 0.005000 M Ag + 25.00 mL 0.001000 M I – titrated with 0.0005000 M Ag + 13920131 TMHsiung@2007 28/42 http:\\asadipour.kmu.ac.ir 41 slides
Slide29Ksp effect of the titration curves 25.00 mL 0.1000 M halide (X – ) titrated with 0.05000 M Ag + 13920131 TMHsiung@2007 29/42 http:\\asadipour.kmu.ac.ir 41 slides
Slide3013920131TMHsiung@2007 30/42 http:\\asadipour.kmu.ac.ir 41 slides
Slide31(a)40.00 mL of 0.0502 M KI + 0.0500 M K Cl , titrated with 0.0845 M Ag + (b) 20.00 mL of 0.1004 M KI titrated with 0.0845 M Ag + Titration of a mixture (uncertainty concerned) 13920131 TMHsiung@2007 31/42 http:\\asadipour.kmu.ac.ir 41 slides
Slide32Example: A 25.00 mL solution containing Br – and Cl – was titrated with 0.03333 M AgNO 3 . K sp (AgBr)=5x10 –13 , K sp (AgCl)=1.8x10 –10 . (a) Which analyte is precipitated first? (b) The first end point was observed at 15.55 mL. Find the concentration of the first that precipitated (Br – or Cl – ?). (c) The second end point was observed at 42.23 mL. Find the concentration of the second that precipitated (Br – or Cl – ?). Solution: (a) Ag + (aq) + Br – (aq) AgBr (s) K = 1/K sp (AgBr) = 2x10 12 Ag + (aq) + Cl – (aq) AgCl (s) K = 1/K sp (AgCl) = 5.6x10 9 Ans: AgBr precipitated first 13920131 TMHsiung@2007 32/42 http:\\asadipour.kmu.ac.ir 41 slides
Slide332)Argentometric Titration Define Argentometric Titration: A precipitation titration in which Ag + is the titrant. Argentometric Titration classified by types of End-point detection: – Volhard method: A colored complex ( back titration ) – Fajans method: An adsorbed/colored indicator – Mohr method: A colored precipitate 13920131 TMHsiung@2007 33/42 http:\\asadipour.kmu.ac.ir 41 slides
Slide34Mohr method The Mohr method was first published in 1855 as a method for chloride analysis. In the precipitation of chloride by silver ion, chromate ion (CrO 4 2 ) is used as an indicator in the formation of Ag 2 CrO 4 , a reddish-brown precipitate formed when excess Ag + is present. Ag + + Cl AgCl (s) white precipitate 2Ag + + CrO 4 2 Ag 2 CrO 4 (s) red precipitate TMHsiung@2007 34/42 13920131 http:\\asadipour.kmu.ac.ir 41 slides K sp = 1.8 x 10 -10 (S = 1.34 x 10 -5 M) CrO 4 2 K sp = 1.2 x 10 -12 (s = 6.7 x 10 -5 M)
Slide35 The titrations are performed only in neutral or slightly basic medium to prevent silver hydroxide formation (at pH > 10). 2Ag + + 2OH 2AgOH (s) Ag 2 O (s) + H 2 O precipitate to prevent chromic acid formation (at pH < 7). CrO 4 2 + H 3 O + HCrO 4 + H 2 O 2 CrO 4 2 + 2 H 3 O + H 2 CrO 4 + H 2 O TMHsiung@2007 35/42 13920131 http:\\asadipour.kmu.ac.ir 41 slides
Slide36Volhard METHOD Back titration for determination of Cl - . . First published in 1874. Reactions: Ag + + Cl AgCl (s) K sp = 1.82 x 10 -10 (excess) white precipitate SCN + Ag + AgSCN (s) K sp = 1.1 x 10 -12 titrant white precipitate SCN + Fe 3+ FeSCN 2+ K f = 1.4 x 10 +2 Indicator red complex TMHsiung@2007 36/42 13920131 http:\\asadipour.kmu.ac.ir 41 slides H + , Fe 3+
Slide37 The titration is usually done in acidic pH medium to prevent precipitation of iron hydroxides, Fe(OH) 3 . Fe 3+ +3(OH) - ⇄ Fe(OH) 3 K sp =1*10 -39 If [Fe 3+ ]=0.001 M pH= ????? TMHsiung@2007 37/42 13920131 http:\\asadipour.kmu.ac.ir 41 slides
Slide38 Since S AgSCN <S AgCl , equilibrium will shift to the right causing a negative error for the chloride analysis. Ag + + Cl AgCl (s) K sp = 1.82 x 10 -10 SCN + Ag + AgSCN (s) K sp = 1.1 x 10 -12 SCN - + AgCl AgSCN + Cl - To eliminate this error, AgCl must be filtered or add nitrobenzene before titrating with thiocyanate; nitrobenzene will form an oily layer on the surface of the AgCl precipitate, thus preventing its reaction with thiocyanate. TMHsiung@2007 38/42 13920131 http:\\asadipour.kmu.ac.ir 41 slides
Slide39Before Ve ( Cl– excess ) Greenish yellow solution After Ve ( Ag + excess ) Fajans Method: An adsorbed/colored indicator . Titrating Cl – and adding dichlorofluoroscein as indicator: 13920131 TMHsiung@2007 39/42 http:\\asadipour.kmu.ac.ir 41 slides
Slide40TMHsiung@2007 40/425) Applications of argentometric titrations: 13920131 http:\\asadipour.kmu.ac.ir 41 slides