Volumetric Titrimetry PDF

Modified by Dr. Mohammed Rasheed Chapter 13 Volumetric Titrimetry Principles of volumetric analysis Volumetry : the volume of standard reagent needed to react with analyte is measured Titrimetry : increments of the titrant are added to the analyte until their reaction is complete. ❖ Titr...

Modified by Dr. Mohammed Rasheed Chapter 13 Volumetric Titrimetry Principles of volumetric analysis Volumetry : the volume of standard reagent needed to react with analyte is measured Titrimetry : increments of the titrant are added to the analyte until their reaction is complete. ❖ Titrant vs Analyte ❖ x mole = y mole ✓ Depending on molar ratio between the anlyte and titrant in the balanced equations  Titration: A procedure for determining the amount of unknown substance (the analyte) by quantitative reaction with a measured volume of a solution of precisely known concentration (the titrant). Titrant: The substance that quantitatively reacts with the analyte in a titration. The titrant is usually a standard solution added carefully to the analyte until the reaction is complete. The amount of analyte is calculated from the volume of titrant required for complete reaction. Standard Solutions (Titrant) ❖ Standard solutions play a key role in titrimetric methods. Desirable Properties of Standard Solutions: accurate known conc.: 4 significant figures Sufficiently stable when preparing stoichiometric reaction: whole-number ratio React rapidly with analyte (should be very fast reaction). React completely with analyte: 99.9 % Endure a selective reaction with analyte  Equivalence point: Point in a titration at which equivalent amounts of titrant is the exact amount necessary for stoichiometric reaction with the analyte. End point: The point in a titration when a physical change occurs that is associated with the condition of chemical equivalence. It represents the experimental estimate of the equivalence point in a titration Vep = actual volume at end point, Veq = theoretical volume of equivalence point titration error : eT = Vep – Veq  Indicator : A substance that undergoes a sharp, easily observable physical change when conditions in its solutions change, for example, acid- base indicator and redox indicator. OR large changes in the relative conc. of analyte or titrant occur in the equivalence-point region. a. Appearance or disappearance of a color. b. Change in color c. Appearance or disappearance of turbidity. Instruments for detect end point: voltmeters, ohmmeters, colorimeter, temperature recorders, refractometers.  Direct titration : Titrant is added to the analyte until the reaction is complete. Back titration or residual-titration : Adding a known excess of reagent to the analyte, then, a second reagent is used to titrate the excess of the first reagent. ❖ This is used when the rate of reaction between the analyte and rea gent is slow or when the reagent lacks stability. A Excess T1 T1 T2 Ex. PO43– + 3Ag+ → Ag3PO4 (s) Ag+ + SCN – → AgSCN (s) Evolution of the buret Buret, buret stand, clamp, white porcelain base, wide-mouth Erlenmeyer flask ❖ Normally, the buret is filled titrant solution to within 1 or 2 mL of the zero position at the top. The initial volume of the buret is read to the nearest 0.01 mL.  Standard solution: A solution of precisely known concentration. Primary standard : An ultra-pure (99.9% purity) compound that serves as the reference material for a titrimetric method of analysis. Secondary standard : A compound whose purity has been established by chemical analysis and that serves as the reference material for a titrimetric method of analysis Standardization : ❖ A process in which the concentration of a solution is determined by using the solution to titrate a known amount of another reagent.  Primary standard: A primary standard is a highly purified compound that serves as a reference material in all volumetric and mass titrimetric properties. The accuracy depends on the properties of a compound and the important properties are: 1. High purity 2. Atmospheric stability 3. Absence of hydrate water 4. Readily available at a modest cost 5. Reasonable solubility in the titration medium 6. Reasonably large molar mass Compounds that meet or even approach these criteria are few, and only a few primary standards are available.  Steps in a titrimetry Weighing 99.9% pure primary standard reagent Find mole of uknown xx.xxxx g V’ ? Dissolve in a volumetric flask N’ ? Transfer pipet Preparation of primary standard solution Erlenmeyer Flask volume Find Volume ml Calculation Molarity indicator Beaker Preparation of Titration unknown unknown sample sample solution solution Buret ❖ Remember mole is calculated by two ways 1- Mass(g)/MM(g/mol) 2- M(mol/L) x V(L) Ex1: Describe the preparation of 2.000 L of 0.0500 M AgNO3 (169.87 g/mol) from the primary standard-grade solid Ans: dissolve 16.9 g in 2.00 L soln Ex2: Describe how 500 ml of 0.01 M Na+ solution can be prepared from primary Na2CO3 standard solution (105.99 g/mol) Ans: 2.5 x10-3 mole Na2CO3; mass Na2CO3 = 0.256 g Dissolve 0.256 g in water and diluted to 500 ml. Ex2: Describe how you prepare 50.0 ml portion of standard solutions that are 0.00500 M, 0.00200 M and 0.00100 M from the stock solution in example 2 Ans: M1V1 = M2V2; 25.0 ml; 10.0 ml, 5.00 ml Ex. 13-4. Describe how you would prepare 2.0 L of approximate 0.25 M HClO 4 (100.46 g/mol) from the conc. reagent, which has a specific gravity of 1.67 and contains 71 % (w/w) HClO4. CHClO4 = 1.67 g/mL× 0.71 = 1.19 g of the acid/ml 1.19g/(100.46 g/mol) = 0.012 mole; 0.012/ml then the conc is 11.8 mole/L (Or conc. In M = (Sp.grav x % (m/m) / molar mass) x 10 Now M1V1=M2V2 11.8 x V1 = 0.25 x 2 ; V1= 0.042 L; i.e diluted about 42 mL of 71 % HClO4 to 2.0 L. Ex. 13-5. A 50.00-mL of an HCl solution required 29.71 mL of 0.01963 M Ba(OH)2 to reach an end point with bromocresol green indicator. Calculate the molarity of the HCl. Ba(OH)2 + 2HCl → BaCl2 + 2H2O ❖ From eqn 1mole Ba(OH)2 react with 2 mole HCl ❖ From titration: 5.83 x 10-4 Ba(OH)2 will react with x mole HCl ❖ Mole HCl = 1.166 x 10-3 mole; M HCl = 1.166 x 10-3/0.05 L Ans: 0.0233 M HCl Ex 13.6: Titration of 0.2121 g of pure Na2C2O4 (134.00 g/mol) required 43.31 mL of KMnO4. What is the molar concentration of the KMnO4 solution? The chemical reaction is ❖From eqn 5 mole Na2C2O4 react with 2 mole MnO4- ❖From the titration 0.2121 g Na2C2O4 = 1.58 x 10-3 mole then 5 mole Na2C2O4 → 2 mole MnO4- 1.58 x 10-3 mole → x mole Mole of the unknown (MnO4-) = 6.33 x 10-4 mole Molarity = 6.33 x 10-4 mole / 0.04331 L = 0.0146 M Ex 13.7: 0.804 g sample of iron ore was dissolved in acid. The iron is then reduced to Fe+2 and titrated with 47.2 mL of 0.02242 M of KMnO4. Calculate the results of this analysis in term of a) %Fe (55.846 g/mol) b) % Fe3O4 (231.54 g/mol). 5Fe2++ MnO−4+ 8H+ → 5Fe3++ Mn2++ 4H2O ❖ From titration; Mole Fe2+ = 5.3 x 10-3 mole ❖ Mass Fe2+ = 0.2956 g; then % Fe2+ = 36.77% ❖ Mole Fe3O4 = 1/3 x mole Fe = 1.767 x 10-3 mole ❖ Mass Fe3O4 = 0.41; % Fe3O4 = 50.8% Ex 13.8: A 100.0-mL sample of brackish water was made ammoniacal, and the sulfide it contained was titrated with 16.47 mL of 0.02310 M AgNO3. The analytical reaction is 2Ag++S2−→Ag2S(s) Calculate the concentration of H2S )34.08 g/mole) the water in ppm. ❖From titration; Mole Ag+ = 3.8x10-4 mole ❖Mole S2- = 1.9 x 10-4 mole ❖Mass H2S = 6.5 x 10-3 g; mass in mg = 6.5 mg ❖Ppm H2S = 6.5 mg / 0.1L = 65 ppm Ex 13.9: The phosphate in a 4.258 g sample of plant food was converted to PO43- and precipitated as Ag3PO4 through the addition of 50.00 ml of 0.0820 M AgNO3. The excess AgNO3 was back titrated with 4.06 mL of 0.0625 M KSCN. Express the result of analysis in terms of the percentage of % P2O5 ❖From eqn3 (titn) Mole SCN- = 2.54 x 10-4 mole ❖Reaction 2; total mole of AgNO3 = 4.1 x 10-3 mole ❖Mole Ag+ reacted with PO43- = 4.1 x 10-3 mole - 2.54 x 10-4 mole = 3.85x10-3 mole ❖From molar ratio, Mole PO43- = 3.85 x 10-3 mole x 2/6 = 1.28 10-3 mole ❖Mole P2O5 = Mole PO43-/2 = 6.42 x 10-4 ❖Mass P2O5 = 6.42 x 10-4 x 141.9 g/mole =0.091 g ❖P2O5 Mass% = 2.14% Titration Curves Two types of titration curves routinely encountered in titrimetric met hods; they are sigmoidal curve and linear segment curve.  Titration curves: plots of a conc.-related variable as a function of reagent volume. Fig. 13-3 Titration curve for the titration of 50.00 mL of 0.1000 M AgNO3 with 0.1000 M KSCN. Precipitation Titrations A titration in which the reaction between the analyte and titrant involves a precipitation. Silver nitrate titrations : Argentometric methods for : halides, halide-like anion (SCN-, CN-, CNO-) several End point: 1. A change in color due to the reagent, the analyte or an indicator. 2. A change in potential of an electrode that responds to the conc. of one of the reactants. Titration curve: Guidance in precipitation titration calculation ❖ Find Ve (volume of titrant at equivalence point) ❖ Find y-axis values: - At beginning - Before Ve - At Ve - After Ve Example 13-13: For the titration of 50.0 mL of 0.0500 M Cl– with 0.100 M Ag+ (Ksp for AgCl= 1.8×10–10). The reaction is: Ag+(aq) + Cl–(aq)  AgCl(s) K here is the inverse of Ksp Find pAg and pCl of Ag+ solution added ❖ Ag+ is the titrant and Cl- is a) 0 mL b) 10.0 mL c) 25.0 mL d) 26.0 mL the analyte a) At 0 ml addition of Ag+ we only have Cl- with a concentration of 0.0500 M; Then PCl = -log Cl- = 1.3; PAg+ is indeterminate b) At 10 ml addition Mole Cl- = 2.5 x 10-3; mole Ag + added = 1 x 10-3; after the titration reaction Mole Cl- = 1.5 x 10-3; M [Cl-] = 1.5 x 10-3/ (0.06L (total volume after mixing) = 0.025 mole Ag + = 0; however, AgCl ppt will be formed and partially soluble depending on Ksp ; AgCl(s)  Ag+(aq) + Cl–(aq) Ksp = 1.8×10–10 (common ion) Ksp = 1.8×10–10 = (X) 0.025 ; then x = [Ag+] = 7.2 x10-9; P Ag + = 8.14; PCl- = 1.6 c) At 25 ml addition; mole Cl- = 2.5 x 10-3; mole Ag + added = 2.5 x 10-3; after the titration reaction; 0ml of both (Equivalence point); but AgCl ppt; Ksp = 1.8×10–10 = X2; then x = [Ag+] = [Cl-] = 1.34 x10-5; PAg + = PCl- = 4.87 d) After 26 ml addition; mole Ag + added = 2.6 x 10-3 mole; M [Ag+]= 2.6 x 10-3 mole/76 x10-3 L (Total Volume) =1.3 x 10-3 M Ksp ; AgCl(s)  Ag+(aq) + Cl–(aq) Ksp = 1.8×10–10 (common ion) Then [Cl-]= 1.3 x 10-7 PAg + = 2.88; PCl- = 6.9 Before equivalence point at equivalence point after equivalence point  Dilution effect of the titration curves 25.00 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+ ❖ As the concentrations of the anlayte and titrant increase, it is easier to detect the end point.  Ksp effect of the titration curves At very small Ksp 25.00 mL 0.1000 M halide (X–) [Ag+] is very small titrated with 0.05000 M Ag+ before the end point ❖As Ksp smaller it is easier to detect the end point. The change in PAg at the eqi velence point is greater, and the re action between the analyte and Ag+ b ecomes more complete (K=1/Ksp) Indicators for Argentometric Titrations Conc. of CrO42- needed to form red ppt ❖ The chromate ion required to initiate the formation of silver chromate At equiv. point Ksp = 1.8×10–10 = X2; then x = [Ag+] = [Cl-] = 1.34 x10-5 ❖ pH should be between 6.5-10.3 2. Adsorption indicators: The Fajans method The indicator is an organic compound that tends to adsorbed onto the surface of the solid in a precipitation Titration. Ideally, the adsorption occur near the equivalence point. ❖ Advantages: rapid, accurate and reliable AgNO3 + X- ⇔ AgX(s) + NO3- End point: Ag+ + AgX(s) + Fl- ⇔ AgX:Ag+ · · Fl-(s) Yellow-green red Indicator: Fluorescein anion 3. Iron(III) Ion; The Volhard method (back titration): Iron(III) ion as indicator sample: halide ion, C2O4, AsO4, SCN- nAgNO3 (excess) + Bn- ⇔ AgnB(s) + nNO3- KSCN + unreacted AgNO3 ⇔ AgSCN(s) + K++ NO3-  2AsO3 + 6AgNO3 → 2Ag3AsO3 + 6NO2 + 3O2 Answer: Total mole AgNO3 = 1.0075 x 10-3 Mole KSCN = 2.052 x 10-4 mole Mole AgNO3 consumed by AsO43- = 0.0008023 mole As2O3= 2AsO43- = 6 AgNO3 Mole As2O3 = (mole AgNO3)/6 % As2O3 = 0.299% Titration curve of a mixture of anions Example: A 25.00 mL solution containing Br– and Cl– was titrated with 0.03333 M AgNO3. Ksp(AgBr)=5x10–13, Ksp(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 in the original solution (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–?). (a) Ag+(aq) + Br–(aq)  AgBr(s) K = 1/Ksp(AgBr) = 2x1012 Ag+(aq) + Cl–(aq)  AgCl(s) K = 1/Ksp(AgCl) = 5.6x109 Ans: AgBr precipitated first (b) The first end point is for Br- (mole Br-= mole Ag+) Mole Ag+ = 5.1 x 10-4 mole = mole Br- Conc. Br-= 5.1 x 10-4 /0.025 L (volume in original soln) = 0.0207 M (c) The second end point is for Cl- (After all Br- reacted with the Ag+ from the first end point); (mole Cl-= mole Ag+) Volume to react with Cl- = 42.23-15.55ml = 26.68ml Mole Ag+ = 8.89 x 10-4 mole = mole Cl- Conc. Cl-= 8.89 x 10-4 /0.025 L (volume in original soln) = 0.0355 M