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Queen's University
ENCH 213
Diane Beauchemin

-need to be vertical -to read meniscus Titrations -rinse w/ titrant • Principle -tellifburet is not clean:ifliquid sticks on wall • End point -watch out forbubbles • Requirements -make sure buret lowenough that it doesn't • Weight titrations splash • Kjeldahl method -ifput something on top ofburet it will create a vacuum:difficult forliquid to go • Precipitation titrations out -so put spacers to prevent this and use a Principle of a titration coveron top ofburet to prevent titrant • a X + b Y → Products contamination • reacting a known concentration with the unknown concentration to find the latter • several types ◦ precipitation ◦ acid-base ◦ complexometric (EDTA) → add complexing agent that ◦ redox (reduction/oxidation) End point • indicates the end of the titration • an indicator is often us+d (e.g.+base or acid that reacts with acid or base in the case of a neutralisation): B + H → BH Different protonated forms ↑ ↑ have different colourscolour A colour B • select an indicator so that the change of colour occurs at the point (e.g. pH) where neutralisation is complete ◦ end point as close as possible to equivalence point experimental theoretical The end point of a titration is defined as a) the equivalence point of the titration. b) the actual measured volume of titrant required to complete a titration. c) the volume associated with actual stoichiometric quantity of titrant required to complete a titration. → equivalence point Requirements for a good titration • known stoichiometry • good indicator • known and suitable concentration of titrant → so that relatively large amount required • species to be determined in the proper form ◦ if pretreatment required: eliminate excess reagents that may react with titrant Two types of titrant a) approximate concentration: accurate concentration found by standardization (e.g. 0.3 M HCl) b) known mass of standard material in known volume ↑ Primary standard - stable - >99.9% pure - easy to handle (soluble...) - large f.w. → small RSD on weight Weight titrations • based on mass (instead of volume) of titrant • rely on molality (or weight molarity) → no calibration or thorough cleaning required → no temperature correction → more precise and sensitive • ex.: 50 ± 0.05 mL vs 50.0000 ± 0.0001 g buret balance • can use smaller quantities of reagent without losing much precision Precipitation titration • sharpness of end point depends on: ◦ concentrations of analyte and titrant ◦ size of K sp • p function used because concentration varies over several orders of magnitude. As []decreass, so does sharpness ofendpoint -need something concentratd engouh so see a sharp change, but not so concentrated that adding a drop will make a colourchange Effect of K sp • AgI ⇔(s)+ I + - K sp8.3×10 -17 • Ag + I ⇔AgI (s) K=1.2×10 16 1/K sp • least soluble product: ◦ gives sharpest change at the equivalence point ◦ the larger K, the more pronounced the change in concentration near the equivalence point Regions of precipitation titration curve • before the equivalence point ◦ excess analyte ◦ [analyte] = fraction remaining × original ◦ [analyte] x dilution factor ◦ p(titrant) = -log [titrant]; [titrant] → from Ksp • at the equivalence point ◦ [analyte] and [titrant] from Ksp • after the equivalence point ◦ excess titrant ◦ [titrant] = original [titrant] x dilution factor ▪ dilution factor = excess volume of titrant/total volume of solution - + Example: titration of 25.00 mL of 0.1000 M I with 0.05000 M Ag • equivalence volume: V = 25.00 mL x 0.1000 mmol mL /0.05000 mmol mL = 50.00 mL -1 e • After 10.00 mLAg + - ◦ [I] = ((50.00-10.00 mL)/50.00 mL) (0.1000 M) (25.00 mL/(25.00+10.00 mL) = 0.05714 M + - -17 -15 ◦ [Ag ]=K /[I ]=sp3x10 /0.05714=1.4 x 10 5 M ◦ pAg = -log (1.4 x 10 )5 14.84 -15 + - -9 • At equivalence point: [Ag ]=[I]=√K =9.1 x 10 Mspnd pAg = 8.04 + • After 52.00 mLAg • [Ag ]=0.05000 M x (52.00-50.00 mL)/(25.00 +52.00 mL) = 1.3 x 10 M and pAg = 2.89 -3 You have three flasks with 0.03 M Cl, Br, or I. If you titrated each solution with Cu+, what order will -12 -7 the equivalence points appear in the titration curves? K for CuI = 1 x sp , K for CuCl = 1.9 x 10 sp -9 and K
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