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Lecture 22 10/26/05. Outer curve: 25 mL of 0.100 M I - titrated with 0.0500 M Ag + Middle curve: 25 mL of 0.0100 M I - titrated with 0.00500 M Ag + Inner.

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Presentation on theme: "Lecture 22 10/26/05. Outer curve: 25 mL of 0.100 M I - titrated with 0.0500 M Ag + Middle curve: 25 mL of 0.0100 M I - titrated with 0.00500 M Ag + Inner."— Presentation transcript:

1 Lecture 22 10/26/05

2 Outer curve: 25 mL of 0.100 M I - titrated with 0.0500 M Ag + Middle curve: 25 mL of 0.0100 M I - titrated with 0.00500 M Ag + Inner curve: 25 mL of 0.00100 M I - titrated with 0.000500 M Ag +

3 25.00 mL of 0.100 M halide titrated with 0.0500 M Ag + Unknowns: Different anions Same volumes

4 40.0 mL of 0.052 M KI plus 0.05 M KCl titrated with 0.084 M AgNO 3

5 End-point detection for precipitation reactions Electrodes Silver electrode Turbidity Solution becomes cloudy due to precipitation Indicators Volhard Fajans

6 Volhard (used to titrate Ag + ) As an example: Cl - is the unknown Precipitate with known excess of Ag + Ag + + Cl -  AgCl(s) Isolate AgCl (s), then titrate excess Ag + with standard KSCN in the presence of Fe +3 Ag + + SCN -  AgSCN(s) When all the Ag + is gone: Fe +3 + SCN -  FeSCN 2+ (red color indicates end point) Must be acidic….WHY?

7 Fajans (use adsorption indicator) Anionic dyes which are attracted to positively charged particles produced after the equivalence point Adsorption of dye produces color change Fluorescein is common dye Signals end-point

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9 Titration of strong acid/strong base 50 mL of 0.02 M KOH with 0.1 M HBr

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12 50 mL of 0.02 M KOH with 0.1 M HBr Construct curve from 4 points Initial pH X = 0 mL, Y = ? Equivalence point X = V eq, Y = 7 Before Equivalence point X = ?, Y = ? After Equivalence point X = ?, Y = ?


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