CHM 320 Lecture 9 Chapt 7Chapter 7 TitrationsHomeworkDue Friday, February 17Chapt. 7: 7-2, 7-7, 7-18Chapter 8 ActivityHomework Due Friday, February 17Chapt. 8: 8-1, 8-3, 8-4, 8-5, 8-7, 8-12CHM 320 Lecture 9 Chapt 7Chapter 7: TitrationsTitration - Incremental addition of a reagent solution (titrant) to react with the analyte. From the amount of reagent added, the quantity of analyte can be determined.Equivalence point – when the quantity of added titrant is the exact amount need to stoichiometrically react with the quantity of analyte.End point – the point that is marked by a change in a physical property that signifies that the reaction of the titrant and the analyteis complete. (this is measured experimentally)Indicator – the substance added to the analyte solution that provides the physical change needed to be observed for the end point.CHM 320 Lecture 9 Chapt 7CHM 320 Lecture 9 Chapt 7Requirements for titrant• Known concentration•Stable• Quantitatively added to analyte• Distinct reaction with analyteOften titrants must be standardized to get an accurate value for their concentration. A primary standard is used to standardize the titrant. A primary standard should be: High purity, Stable in air, Absence of hydrate water, Available at moderate cost, Soluble, Large M.W.CHM 320 Lecture 9 Chapt 7Precipitation Titration Curvep-function pX = - log10[X]Precipitation titration curve:four types of calculations• initial point• before equivalence point• equivalence point• after equivalence pointCHM 320 Lecture 9 Chapt 7Determine the volume of titrant need to react (titrate) the analyte# moles titrant = # moles analyte#molesanalyte=(V*M)analyteVtitrant= # molesanalyte/ Mtitrant1ststep in deriving a calibration curve:CHM 320 Lecture 9 Chapt 7EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.0100 M AgNO3.NaBr + AgNO3= AgBr + Na++ NO3-titration curve => pAg (free Ag) vs. vol. AgNO3(titrant) addedWhat volume of AgNO3is needed to titrate Br¯?VBrMBr= (0.050 L)(0.00500 mole/L NaBr)= 2.5 x 10-4moles Br-VAg(equi pt) = (2.5 x 10-4moles)(1L/0.01 moles)= .025 L (or 25 ml AgNO3)CHM 320 Lecture 9 Chapt 7EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.0100 M AgNO3.before equivalence pointafter 5.0 mL of AgNO3addedVNaBr* MNaBr-VAgNO3* MAgNO3MNaBr unreacted= -------------------------------------VNaBr+ VAgNO3(50.00mL*0.00500M) -(5.00mL*0.01000M)MNaBr unreacted= -------------------------------------------------------(50.00 + 5.00)mLMNaBr unreacted= 3.64 X 10-3MCHM 320 Lecture 9 Chapt 7EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.0100 M AgNO3.before equivalence pointafter 5.0 mL of AgNO3addedMNaBr unreacted= 3.64 X 10-3M[Br-]total= [Br-]NaBr unreacted+ [Br-]dissolved AgBr[Br-]total= 3.64 X 10-3M + [Ag+]dissolved AgBrwhere [Br-]dissolved AgBr= [Ag+]dissolved AgBrCHM 320 Lecture 9 Chapt 7EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.0100 M AgNO3.before equivalence pointafter 5.0 mL of AgNO3added[Br-]total= 3.64 X 10-3M + [Ag+]dissolved AgBrwhere [Br-]dissolved AgBr= [Ag+]dissolved AgBrexcept very near the equivalence point,3.64 X 10-3M >> [Ag+]dissolved AgBrThus [Br-]total~ 3.64 X 10-3MCHM 320 Lecture 9 Chapt 7EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.0100 M AgNO3.before equivalence pointafter 5.0 mL of AgNO3addedKsp= [Ag+][Br-] = 5.2 X 10-13[Ag+] = 5.2 x 10-13/ [Br-] = 5.2 x 10-13/ 3.64 x 10-3= 1.43 x 10-10pAg = - log (1.43 x 10-10)= 9.84CHM 320 Lecture 9 Chapt 7EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.0100 M AgNO3.equivalence pointat 25.00 mL of AgNO3addedKsp= [Ag+][Br-] = 5.2 x 10-13becomes when [Ag+] = [Br-][Ag+]2= 5.2 x 10-13M2[Ag+] = 7.21 x 10-7MpAg = -log [Ag+] = - log[7.2 x 10-7] pAg = 6.14CHM 320 Lecture 9 Chapt 7EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.0100 M AgNO3.after equivalence pointAfter the equivalence point there is very little change in the amount of precipitate present (except very close to the equivalence point).Thus, at 25.10 mL of AgNO3addedVAgNO3*MAgNO3-VNaBr*MNaBrMAgNO3 unreacted= -----------------------------------VAgNO3+ VNaBrCHM 320 Lecture 9 Chapt 7EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.0100 M AgNO3.after equivalence pointVAgNO3*MAgNO3-VNaBr*MNaBrMAgNO3 unreacted= -------------------------VAgNO3+ VNaBr(25.10 mL * 0.01000 M) - (50.00 mL * 0.00500 M)MAgNO3 = --------------------------------------------------------------unreacted (25.10 + 50.00)mLMAgNO3 unreacted= 1.33 X 10-5MCHM 320 Lecture 9 Chapt 7EXAMPLE: Derive a curve for the titration of 50.00 mL of 0.00500 M NaBr with 0.0100 M AgNO3.after equivalence pointthus, at 25.10 mL of AgNO3added[Ag+]total= [Ag+]AgNO3 unreacted+ [Ag+]dissolved AgBr[Ag+]total= 1.33 X 10-5M + [Ag+]dissolved AgBr[Ag+]total~ 1.33 X 10-5MpAg = =-log [Ag+] = -log (1.33 x 10-5) = 4.880CHM 320 Lecture 9 Chapt 7Vol of tpAg5.00 9.8425.00 6.1425.10 4.88Precipitation Titration0246810120.00 5.00 10.00 15.00 20.00 25.00 30.00Vol of AgNO3 addedpAgCHM 320 Lecture 9 Chapt 72.600.00252.810.0015383.150.0007143.250.0005633.380.0004173.560.0002743.870.0001354.176.71E-05.000131600025970246810120.00 10.00 20.00 30.00 40.00 50.00Vol of AgNO3 addedpAgCHM 320 Lecture 9 Chapt 7• Complexometric Titration — Determination of water hardness • Gravimetric Methods — Determination of Chloride • Precipitation Titration — Determination of Chloride by Fajans Methods • Acid/Base Titration – Alkalinity (Bicarbonate in natural waters)Real World Uses of