13 Titrimetric Methods; Precipitation Titrimetry

Transcription

1 13 Titrimetric Methods; Precipitation Titrimetry Titrimetric method: analytical procedures in which the amount of analyte is determined from the amount of a standard reagent required to react with the analyte completely. Three types of quantitative titrimetry 1. volumetric (the most widely used). gravimetric 3. coulometric 13A Some Terms Used in Volumetric Titrimetry *Standard solution (standard titrant) 1. accurate known conc. : 4 significant figures. stable 3. stoichiometric reaction : wholenumber ratio 4. rapid and quantitatively complete reaction : 99.9 % *Titration *Direct titration *Backtitration, residualtitration: when the rate of reaction between the analyte and reagent is slow or when the reagent lacks stability. *Equivalence point, theoretical point: the point in a titration when the amount of added standard reagent exactly equivalent to the amount of analyte. *End point: the point in a titration when a physical change occurs that is associated with the condition of chemical equivalence. *Titration error E t : the difference in volume or mass between the equivalence point and the end point. (V ep V eq ) *Indicator: large changes in the relative conc. of analyte or titrant occur in the equivalencepoint region. a. appearance or disappearance of a color. b. change in color c. appearance or disappearance of turbidity. Instruments for detect end point: voltmeters, ammeters, ohmmeters, colorimeters, temperature recorders, refractometers. 7

2 Fig. 131 The titration process *Primary standards: 1. high purity. stability in air 3. absence of hydrate water 4. ready availability at modest cost 5. reasonable solubility in the titration medium 6. reasonably large formula weight *Secondary standard 13B Standard Solutions *Standardization, establishing the conc. of standard soln. 1. Direct method: carefully weighed quantity of a primary standard dissolved diluted to an exactly known volume.. Standardized by titrating a. a weighed quantity of a primary standard b. a weighed quantity of a secondary standard c. a measured volume of another standard soln *Methods for expressing the conc. of standard solutions a. molarity C : no. of moles/l soln b. normality C N : no. of equivalent/l soln. 13C Volumetric calculations mass A (g) amount A (mol) V(L) C molar mass A (g/mol) A secondary standard solution mol ( ) L mass A (g) amount A (mmol) V(mL) C molar mass A (g/mmol) A mmol ( ) ml 73

3 *Calculation of the Molarity of Standard Solutions (SS) Ex Describe the preparation of.000 L of M AgNO 3 ( g/mol) from the primary standardgrade solid. amount AgNO L.000 mol/l 0.00 mol mass AgNO mol g/mol g dissolving g AgNO 3 in water and diluting to.000 L. Ex. 13. Describe how 500 ml of standard 0.00 M Na + solution can be prepared from primary standard Na CO 3 (5.99 g/mol). amount Na CO ml 0.00 mmol/ml ½.50 mmol mass Na CO 3.50 mmol g/mmol 0.65 g dissolving 0.65 g Na CO 3 in water and diluting to 500 ml. Ex How would you prepare 50.0mL portions of SS that are M, M and M in Na + from the soln in Ex.? V concd C concd V dil C dil V + concd Vdil Cdil 50.0mL mmol Na / ml C + concd 0.00mmol Na / ml 5.0mL M M 0.000M 0.00 M Na + soln 5.0 ml.0 ml 5.00 ml Ex Describe how you would prepare.0 L of approximate 0.5 M HClO 4 (0.46 g/mol) from the conc. reagent, which has a specific gravity of 1.67 g/ml and contains 71 % (w/w) HClO 4. CHClO (g/ml)/0.046 (g/mmol) M no. mmol HClO 4 required 000 ml 0.5 mmol/ml 500 mmol vol conc. reagemnt 500 mmol/11.8 (mmol/ml) 4.4 conc. reagent diluted about 4 ml of 71 % HClO 4 to.0 L. *Calculation of Molarities from Standardization Data Ex A 50.00mL of an HCl solution required 9.71 ml of M Ba(OH) to reach an end point with bromocresol green indicator. Calculate the molarity of the HCl. Ba(OH) + HCl BaCl + H O 1 mmol Ba(OH) reacts with mmol of HCl amount Ba(OH) 9.71 ml mmol/ml mmol amount HCl mmol ( ) mmol HCl C HCl M M 50.0 ml soln 74

4 Ex Titration of 0.11 g pure Na C O 4 ( g/mol) required ml KMnO 4. What is the molarity of the KMnO 4 soln? MnO 4 + 5C O H + Mn + + CO + 8H O stoichiometric ratio amount Na C O 4 amount KMnO ( ) C ml mmol KMnO 5 mmol Na C O 0.11g g/ mmol mmol mmol KMnO M *Calculation the Quantity of Analyte from Titration Data Ex A g sample of an iron ore is dissolved in acid. The iron is then reduced to Fe + and titrated with 47. ml of 0.04 M KMnO 4 solution. Calculate the results of this analysis in term of (a) % Fe ( g/mol); and (b) % Fe 3 O 4 (31.54 g/mol). MnO 4 + 5Fe + + 8H + Mn + + 5Fe H O 5 mmol Fe (a). stoichiometric ratio 1mmol KMnO4 amount KMnO ml 0.04 mol/l mmol amount Fe ml 0.04 mol/l mmol mass Fe mmol g/mmol g ( ) g percent Fe + 0% 36.77% g sample (b). stoichiometric ratio: 5 Fe + 1 MnO 4 5 Fe 3 O 4 15 Fe + 3 MnO 4 amount KMnO ml 0.04 mol/l mmol amount Fe 3 O ml 0.04 mol/l 5/ mmol mass Fe 3 O mmol g/mmol g 5 ( ) g percent Fe 3 O 4 3 0% 50.81% g sample + 75

5 Ex.138. A 0.0mL sample of brackish water was made ammonical, and the sulfide it contained was titrated with ml of 0.03 M AgNO 3. The analytical reaction is Ag + + S Ag S(s) Calculate [H S] in the water in ppm. 1 ( ) g 6 [ H S] ppm 66. ppm H S 0.0 ml 1.000g/mL sample Ex The phosphorus in a 4.58g sample of a plant food was converted to PO 4 3 and precipitated as As 3 PO 4 through the addition of ml of M AgNO 3. The excess AgNO 3 was backtitrated with 4.06 ml of M KSCN. Express the results of the analysis in terms of % P O 5. P O 5 + 9H O PO H 3 O + PO Ag + (excess) Ag 3 PO 4 (s) Ag + + SCN AgSCN(s) 76 1mmol PO 5 6 mmol AgNO 3 1 mmol KSCN and 1mmol AgNO 1 ( ) g/mmol P O 6 0%.14% 4.58 g sample % 5 Ex. 13 The CO in a 0.3L sample of gas was converted to CO by passing the gas over iodine pentoxide heated to 150, and I was distilled and collected to 8.5 ml of M Na S O 3, then back titration with.16 ml of M I solution. Calculate the mg of CO (8.01 g/mol) per liter of sample. I O 5 (s) + 5CO(g) 5CO (g) + I (g) [CO: I 5 : 1] I (aq) + S O 3 (aq) I (aq) + S 4 O 6 (aq) [I : S O 3 1 : ] [CO: S O 3 5 : ] ( ) Mass CO 0.3 L mmol mg/l Ex. 11 The organic matter in a 3.776g sample of a mercuric ointment is decomposed with HNO 3. After dilution, the Hg + is titrated with 1.30 ml of a M soln of NH 4 SCN. Calculate the percent Hg (00.59 g/mol) in the ointment. Hg + + SCN Hg(SCN) (aq) stoichiometric ratio 1 mmol Hg + / mmol NH 4 SCN amount NH 4 SCN 1.30 ml mmol/ml.4367 mmol amount Hg mmol 1/ mmol mass Hg mmol g/mmol g ( ) g percent Hg 0% 6.47% g sample 3

6 Ex. 131 A g sample [(NH 4 ) C O 4 and inert ]/ H O, added KOH to alkaline [NH + 4 NH 3 ], distilled into ml of M H SO 4. Then back titration with ml of M NaOH. Calculate (a) the % N ( g/mol) and (b) the % (NH 4 ) C O 4 (14. g/mol) in the sample. ( ) (a) % N 0% g %.85% / 8.58 (b) % (NH4) CO4 0% 0% g % 13D Gravimetric Titrimetry 13 E Titration curves in Titrimetric Methods Fig. 13 Two types of titration curves. (a) Sigmoidal curve (b) Linearsegment curve Table 131 Concentration changes during a titration of ml of 0.00M AgNO 3 with 0.00M KSCM 0.00M KSCN, ml [Ag + ] mmol/l ml of KSCN to cause a tenfold decrease in [Ag + ] 77 pag pscn

7 Fig. 133 Titration curve for the titration of ml of0.00 M AgNO 3 with 0.00 M KSCN. 13F Precititation Titrimetry Silver nitrate titrations : Argentometric methods for : halides, halidelike anion (SCN, CN, CNO ) several divalent anions, mercaptans, fatty acids and several divalent and trivalent inorganic anions. End point: 1. A change in color due to the reagent, the analyte or an indicator.. A change in potential of an electrode that responds to the conc. of one of the reactants. Titration curves: plots of a conc.related variable as a function of reagent volume. 13F1 Precipitation titration curves involving silver ion preequivalence points equivalence point post equivalence points Ex Perform calculations needed to generate a titration curve for ml of M NaCl with 0.00 M AgNO 3. Ag + (aq) + Cl (aq) AgCl(s), K SP [Ag + ][Cl ] 1.8 Initial point: M in Ag, and pag is indeterminate. preequivalence points, At.00 ml original no. mmol Cl no. mol AgNO [Cl ] C 3 NaCl total volume of solution M [Ag + ] K sp , pag log ( )

8 Equivalence point [Ag + ] sp 5 K , pag log ( ) 4.87 post equivalence point, After Addition of 6.0 ml of Reagent + ( ) [Ag ] C AgNO pag log ( ).88 Tab.13 Changes in pag in the titration of Cl with AgNO 3. AgNO 3, ml ml of M NaCl with 0.00M AgNO ml of M NaCl with 0.000M AgNO The shapes of titration curves Fig. 134 Titration curve for A, ml of M NaCl with 0.00 M AgNO 3, and B, ml of M NaCl with M AgNO 3. Fig. 135 Effect of reaction completeness on precipitation titration curve, ml of a M solution of the anion was titrated with 0.00 M AgNO 3. *Factors influencing endpoint sharpness Satisfactory end points require a change of in pfunction within ±0.1 ml of the equivalence point 79

9 a. reagent conc. : conc. sharpness b. reaction completeness : product ppt Ksp sharpness *Chemical indicators for precipitation titration A + R AR(s) analyte A with titrant R, In + R InR indicator In For a color change to be seen, [InR]/[In] must change by a factor of to 0. 13F Titration curves for mixtures of anions Titration of ml of a solution ([I ] M and [Cl ] M) with 0.00 M AgNO [Ag ][I ] [I ] (4.56 )[Cl ] + [Ag ][Cl ] after added 5.00 ml AgNO 3, c Cl M, [I ] M no. mmol I 75.00mL (.43 8 mmol I /ml) I unprecipitated 0% 7.3 % [Ag ] 3.41 pag log(3.41 ) after added ml AgNO 3, c Cl [Cl ] M [Ag ] pag , Fig. 136 Titration curves for ml of a solution M in Cl and M in I or Br. AgI: Ksp AgBr: Ksp AgCl: K sp

10 13F3 Indicators for Argentometric Titrations *AgNO 3 titrations Method Mohr (Direct) Fajans (Direct) Volhard (Residual) Titrant AgNO 3 AgNO 3 AgNO 3 KSCN Indicator Na CrO 4 Fluorescein Fe 3+ End point Ag CrO 4 (s) red ppt AgX:Ag + Fl (s) red ppt Fe(SCN) + red solution Titration ph 6.5 ~.3 7 Acidic 1. Chromate Ion; The Mohr method 1865 K. F. Mohr, a German pharmaceutical chemist sample: Cl, Br, CN AgNO 3 + X AgX(s) + NO 3 white AgNO 3 + CN Ag(CN) (aq) + NO 3 End point: AgNO 3 + CrO 4 Ag CrO 4 (s) + NO 3 yellow red (Ksp M 3 ) not for arsenate, I, SCN Solubility: Ag CrO 4 > AgX *Choice of indicator: if indicator I anion (AgI Ksp M ) AgCl Ksp 1.8 M [Ag + ] ep (Ksp) ½ (1.8 M ) ½ M if indicator I anion conc M AgI ppt formation [Ag + ]min M /0.005 M I M preequivalence point : Ag + AgI (s) *Concentration of indicator [CrO 4 ] equivalence point : [Ag + ] [Cl ] M [CrO 4 ] Ksp/[Ag + ] 1. 1 /( ) M *ph : (7~) ph < 6.5 Ag CrO 4 (s) Ag + + CrO 4 CrO 4 + H + Cr O 7 + H O ph >.3 Ag + + OH AgOH(s) 81

11 . Adsorption indicators: The Fajans method 196, Polish chemist, K. Fajans Advantages: rapid, accurate and reliable AgNO 3 + X AgX(s) + NO 3 End point: Ag + + AgX(s) + Fl AgX:Ag + Fl (s) Yellowgreen red Indicator: Fluorescein anion * avoid ppt coagulation ppt surface a. adding dextrin or polyethylene glycol b. quick titration and avoid excessive stirring c. halide ion conc. : ~ 0.05 M *Fluorescein dyes: weak ionized acids ph [Fl ] weak end point ph 7, (dichlorofluorescein: ph 4) *Photochemical decomposition : ppt black Indicator ph Sample Fluorescein 7 Cl, Br, I, SCN Dichlorofluorescein 4 Cl, Br, I, SCN Eosin Br, I, SCN O O OH COOH fluorescein 3. Iron(III) Ion; The Volhard method 1874 Jacob Volhard, a German chemist, (back titration): Iron(III) ion as indicator sample: halide ion, C O 4, AsO 3 4, SCN nagno 3 (excess) + B n Ag n B(s) + nno 3 KSCN + unreacted AgNO 3 AgSCN(s) + K + + NO 3 End Fe 3+ + SCN Fe(SCN) + [Fe(SCN) K point: 3+ red [Fe ][ SCN *ph : in acidic soln to prevent Fe(III) Fe(OH) 3 (s) *advantage: carbonate, oxalate & arsenate do not interfere. *For Cl ion in blood serum, urine. (sample: HNO 3 digestion) *Quantitativeness of the back titration n SCN + Ag n B(s) n AgSCN(s) + B n + f ] ] 50 a. compare AgSCN and Ag n B : molar solubility AgSCN: M 8

12 b. Calculate Krxn n [B ] K K rxn n [SCN ] (K if AgSCN solubility < Ag n B or Krxn > 1 filter Ag n B ppt or adding liquid nitrobenzene sp sp of of Ag n B AgSCN) Table 133 Typical Argentometric Precipitation Methods Analyte End point Remarks AsO 3 4, Br, I, CNO, SCN Volhard Removal of Ag salt not required CO 3, CrO 4, CN, Cl, C O 4, PO 3 4, S, NCN BH 4 Volhard Removal of Ag salt required before backtitration of excess Ag + Modified Volhard Epoxide Volhard Titration of excess Cl following hydrohalogenation K + Modified Volhard Titration of excess Ag + following BH 4 + 8Ag + + 8OH 8Ag(s) + H BO 3 + 5H O Precipitation of K + with known excess of B(C 6 H 5 ) 4, addition of excess Ag + giving AgB(C 6 H 5 ) 4 (s), and backtitration of the excess Br, Cl Mohr method In neutral solution Br, Cl, I, SeO 3 Adsorption indicator V(OH) + 4, fatty Electroanalytical Direct titration with Ag + acids, mercaptans Zn + Modified Volhard F Modified Volhard Precipitation as ZnHg(SCN) 4, filtration, dissolution in acid, addition of excess Ag +, backtitration of excess Ag + Precipitation as PbClF, filtration, dissolution in acid, addition of excess Ag +, backtitration of excess Ag + Ex. The As in a 9.13g sample of pesticide was converted to AsO 4 3 and precipitated as Ag 3 AsO 4 with ml of M AgNO 3. The excess Ag + was then titrated with 4.75 ml of M KSCN. Calculate the % of As O 3 in the sample. no. mmol AgNO ml mmol no. mmol KSCN 4.75 ml mmol 0.05 no. mmol AgNO 3 consumed by AsO As O 3 AsO AgNO 3 % As O (1/6) / % % n 83

13 *Quantitativeness of AgNO 3 titration Ag + + X AgX(s) Krxn 1/Ksp 1/[Ag + ][X ] a % reaction: at equivalence point, max % Cl : 0.1 % ex: AgCl, Ksp 1.8 ; [Cl 5 ] K 1.34 M K of AgCl 0 5 sp %dissolved 0.13% M of initial Cl 0.0 M b. min. theo. Krxn Krxn Krxn 1/Ksp 0. M sample [X ] 0.1 % upper limit of [X ] (0. M)(0.1 %) 4 M equivalence point : [Ag + ] [X ] 1 8 min theo. K rxn ( M Ag )( M X ) for 0.1 M X, actual Krxn 8 for 0.01 M X, actual Krxn sp 84