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Titrations Volumetric analysis  Procedures in which we measure the volume of reagent needed to react with an analyte Titration  Increments of reagent.

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Presentation on theme: "Titrations Volumetric analysis  Procedures in which we measure the volume of reagent needed to react with an analyte Titration  Increments of reagent."— Presentation transcript:

1 Titrations Volumetric 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 http:\\asadipour.kmu.ac.ir 41 slides

2 Titrations Buret 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 http:\\asadipour.kmu.ac.ir 41 slides

3 Type of Titrations based on Measuring Techniques i)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. 3/42http:\\asadipour.kmu.ac.ir 41 slides

4 Type of Titrations based on Chemical Reactions i)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 MnO 4 – + H + → 5 O Mn H 2 O iv)Complexometric Titrations, example: EDTA + Ca 2+ → (Ca–EDTA) 2+ 4/42http:\\asadipour.kmu.ac.ir 41 slides

5 Type of Titration Curves TypeExample.y-axisx-axis Acid-baseHCl/NaOHpHV. NaOH PrecipitationAg + /Cl – pAg + V. Ag + ComplexationCa 2+ /EDTApCa 2+ V. EDTA RedoxMnO 4 – /Fe 2+ PotentialV. Fe 2+ TypeExampley-axisx-axis Spectro- photometric apotransferrin/ Fe 3+ AbsorbanceV. Fe 3+ Thermo- metric H 3 BO 4 / NaOH TemperatureV. NaOH V. = volume 5/42http:\\asadipour.kmu.ac.ir 41 slides

6 Expressing concentration Formality 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 http:\\asadipour.kmu.ac.ir 41 slides

7 relationship between titrant and 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 http:\\asadipour.kmu.ac.ir 41 slides

8 Standardization: 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. 8/ http:\\asadipour.kmu.ac.ir 41 slides

9 Standardization  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 http:\\asadipour.kmu.ac.ir 41 slides

10 Standardization of 0.1 M NaOH 1-selection the PS (e.g. KHP) 2-wheing the PS 10*0.1=mg/ making solution 4-addind suitable indicator 5-titration 9.1ml 6-calculation 9.1*n=213.8/204.1 n= http:\\asadipour.kmu.ac.ir 41 slides N1V1=N2V2

11 Blank 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. 11/42http:\\asadipour.kmu.ac.ir 41 slides

12 Example: To standardizing a KMnO 4 stock solution, the primary standard of g Na 2 C 2 O 4 is dissolved in mL volumetric flask mL of the Na 2 C 2 O 4 solution require 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 ) Solution: Ans 5C 2 O 4 2 – (aq) + 2MnO 4 – (aq) + 16H + (aq) → 10CO 2(g) + Mn 2+ (aq) + 8H 2 O (l) Standardization 12/42http:\\asadipour.kmu.ac.ir 41 slides

13 Example: A g sample of an iron ore is dissolved in acid, and the iron is converted entirely to Fe 2+. To titrate the resulting solution, L of M KMnO 4 is required. Also a blank titration require L of KMnO 4 solution. What is the % Fe (w/w) in the ore? (AW Fe ) MnO 4 – (aq) + 5Fe H + (aq) → Mn 2+ (aq) + 5Fe H 2 O (l) Unknown Analysis with a Blank Correction 13/42http:\\asadipour.kmu.ac.ir 41 slides

14 Back Titration 1) 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 H + (aq) → Mn 2+ (aq) + 5Fe H 2 O (l) http:\\asadipour.kmu.ac.ir 41 slides

15 Back Titration Example: The arsenic in g sample was pretreated to H 3 AsO 4(aq) by suitable treatment. The mL of M AgNO 3 was added to the sample solution forming Ag 3 AsO 4(s) : The excess Ag + was titrated with mL of M KSCN. The reaction was: Calculate the percent (w/w) As 2 O 3(s) (fw g/mol) in the sample. 15/42http:\\asadipour.kmu.ac.ir 41 slides

16 16/42http:\\asadipour.kmu.ac.ir 41 slides

17 In a back titration analysis of HCO 3 -, 25 mL of a bicarbonate solution is reacted with mL of M NaOH. The excess NaOH was titrated with M HCl. This required mL. What is the concentration of bicarbonate in solution? NaOH + HCO 3 - → Na + + CO H 2 O NaOH + HCl → NaCl + H 2 O 17/42http:\\asadipour.kmu.ac.ir 41 slides Back Titration

18 Equivalence 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) Equivalence point occurs when 2 moles of MnO 4 - is added to 5 moles of Oxalic acid http:\\asadipour.kmu.ac.ir 41 slides

19 End point Occurs 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) After equivalence point occurs, excess MnO 4 - turns solution purple  Endpoint http:\\asadipour.kmu.ac.ir 41 slides

20 Titration 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 http:\\asadipour.kmu.ac.ir 41 slides

21 Calculation 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 Standardization: Suppose mL of I 3 - solution is required to react with 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 g was titrated by mL of I 3 -. Find the weight percent of ascorbic acid in the tablet http:\\asadipour.kmu.ac.ir 41 slides

22 Titration of a Mixture Example: A solid mixture weighing g containing only sodium carbonate (Na 2 CO 3, FW ) and sodium bicarbonate (NaHCO 3, FW 84.01) require mL of M HCl for complete titration: Find the mass of each component of the mixture. massmoles Total mixture1.372 Na 2 CO 3 xx/ g/mol NaHCO x1.372 – x/84.01 g/mol 22/42http:\\asadipour.kmu.ac.ir 41 slides

23 Ans 23/42http:\\asadipour.kmu.ac.ir 41 slides

24 Direct and back (indirect) titration of Aspirin 24/42http:\\asadipour.kmu.ac.ir 41 slides

25 A 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 25/42http:\\asadipour.kmu.ac.ir 41 slides

26 Titration curve of 50.0 mL of M Cl – with M Ag + pCl pAg 26/42http:\\asadipour.kmu.ac.ir 41 slides

27 Example: For the titration of 50.0 mL of M Cl – with 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: 27/42http:\\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 – ] = , 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

28 Diluting effect of the titration curves mL M I – titrated with M Ag mL M I – titrated with M Ag mL M I – titrated with M Ag + 28/42http:\\asadipour.kmu.ac.ir 41 slides

29 K sp effect of the titration curves mL M halide (X – ) titrated with M Ag + 29/42http:\\asadipour.kmu.ac.ir 41 slides

30 30/42http:\\asadipour.kmu.ac.ir 41 slides

31 (a)40.00 mL of M KI M KCl, titrated with M Ag + (b)20.00 mL of M KI titrated with M Ag + Titration of a mixture (uncertainty concerned) 31/42http:\\asadipour.kmu.ac.ir 41 slides

32 Example: A mL solution containing Br – and Cl – was titrated with 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 mL. Find the concentration of the first that precipitated (Br – or Cl – ?). (c)The second end point was observed at 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 32/42http:\\asadipour.kmu.ac.ir 41 slides

33 2)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 33/42http:\\asadipour.kmu.ac.ir 41 slides

34 Mohr 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 34/ http:\\asadipour.kmu.ac.ir 41 slides K sp = 1.8 x (S = 1.34 x M) CrO 4 2  K sp = 1.2 x (s = 6.7 x M)

35 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 35/ http:\\asadipour.kmu.ac.ir 41 slides

36 Volhard METHOD Back titration for determination of Cl -.. First published in Reactions: Ag + + Cl   AgCl (s) K sp = 1.82 x (excess) white precipitate SCN  + Ag +  AgSCN (s) K sp = 1.1 x titrant white precipitate SCN  + Fe 3+  FeSCN 2+ K f = 1.4 x Indicator red complex 36/ http:\\asadipour.kmu.ac.ir 41 slides H +, Fe 3+

37 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* If [Fe 3+ ]=0.001 M pH=????? 37/ http:\\asadipour.kmu.ac.ir 41 slides

38 Since S AgSCN

39 Before V e (Cl– excess) Greenish yellow solution After Ve (Ag + excess) Fajans Method: An adsorbed/colored indicator. Titrating Cl – and adding dichlorofluoroscein as indicator: 39/42http:\\asadipour.kmu.ac.ir 41 slides

40 40/42 5)Applications of argentometric titrations: http:\\asadipour.kmu.ac.ir 41 slides


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