1. Lecture 21 10/24/05
Seminar today

2. Precipitation Titration: Titration curve
Before the equivalence point
At the equivalence point
After equivalence point

3. Relate moles of titrant to moles of analyte
X-axis: Volume titrant added
Y-axis: Concentration of one of the reactants
often as pX = -log[X]

4. Titration of 25 mL of 0.1000 M I- with 0.0500 M Ag+
AgI (s)  Ag+ + I-
Ksp = [Ag+ ][I-] = 8.3 x 10-17
1/Ksp = 1/[Ag+ ][I-] = 1.2 x 1016
So Ag+ + I-  AgI (s) goes to completion

5. At the equivalence point (x-axis)
x: (volume of Ag needed to reach equivalence point)
Use stoichiometry to match moles of titrant and moles of analyte

6. At the equivalence point (y-axis)
y: (concentration of Ag)
All of the Ag+ and I- have reacted to form AgI(s)
Where is the dissolved Ag+ coming from?

8. Before the equivalence point x-axis
Volume of Ag+ added
Add less than 50 mL
Let’s add 10 mL
(this volume is arbitrary other than < 50 mL)

9. Before the equivalence point y-axis
Find moles of I-
Moles of I- = original moles I- - moles of Ag+ added
Moles of I- = (0.025L)(0.1 M) – (0.01L)(0.05M)
Moles of I- = moles
Find new I- concentration
[I-]=(0.002 moles)/(0.035L) = M
Find concentration of Ag+
[Ag+]=Ksp/ [I-]
[Ag+]=8.3x10-17 / = 1.4 x 10-15
pAg+= 14.84

10. Before the equivalence point: y-axis (alternate method)
[I-]=(fraction remaining)(original concentration)(dilution factor)
[I-]=((50mL-10mL)/50mL)(0.1 M)(25mL/35mL)
[I-]= M
Find concentration of Ag+
[Ag+]=Ksp/ [I-]
[Ag+]=8.3x10-17 / = 1.4 x 10-15
pAg+= 14.84

12. After the equivalence point x-axis
Volume of Ag+ added
Add more than 50 mL
Let’s add 75 mL
(this volume is arbitrary other than > 50 mL)

13. After the equivalence point y-axis
Dominated by the unreacted Ag+
[Ag+] = (original concentration)(dilution factor)
[Ag+] = (0.05 M)(volume of excess Ag+/ total volume)
[Ag+] = (0.05 M) x ((75mL-50mL) / (75mL + 25ml))
[Ag+] = M
pAg = 1.9

16. Shape For reactions with1:1 stoichiometry:
Equivalence point is point of maximum slope and is an inflection point (second derivative = 0)
For reactions that do not have 1:1 stoichiometry:
Curve is not symmetric near equivalence point
Equivalence point is not the center of the steepest section of the curve
Equivalence point is not an inflection point

17. Outer curve: 25 mL of 0.100 M I- titrated with 0.0500 M Ag+
Middle curve: 25 mL of M I- titrated with M Ag+
Inner curve: 25 mL of M I- titrated with M Ag+

18. 25.00 mL of 0.100 M halide titrated with 0.0500 M Ag+

19. 40.0 mL of 0.052 M KI plus 0.05 M KCl titrated with 0.084 M AgNO3

20. CaCO3(s) [FM 100.087] + 2H+  Ca2+ + CO2(g) + H2O
Problem 7-11
The carbonate content of g of powdered limestone was measured by suspending the powder in water, adding mL of M HCl, and heating to dissolve the solid and expel CO2:
CaCO3(s) [FM ] + 2H+  Ca2+ + CO2(g) + H2O
The excess acid required mL of M NaOH for complete titration to a phenolphthalein end point. Find the weight % of calcite in the limestone.

21. Problem 7-11 (solutions)
Moles OH- = (39.96 mL)*( M) = mmol
Moles H+ = (10 mL)*(1.396 M) = mmol
Moles H+ used to titrate CaCO3 = mmol
Moles CaCO3 = mmol H*(1 mol CaCO3 / 2 mol H)
Moles CaCO3 = mmol
Mass CaCO3 = mmol *( g/mol) = g
Weight % = g / * 100 = 92%

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

23. 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-  FeSCN2+
(red color indicates end point)

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

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

29. Titration of a weak acid with strong base
0.02 MES [2-(n-morpholino)ethanesulfonic acid] with M NaOH.
pKa = 6.15

34. Titration of 10.0 mL of 0.100 M B (base) with 0.100 M HCl.
pKb1 = 4.00
pKb2 = 9.00

36. Finding endpoint with pH electrode

37. Titration of H6A with NaOH

40. Gran Plot
Advantage is that you can use data before the endpoint to find the endpoint

41. Vb never goes to 0 because 10-pH never gets to 0
Also slope doesn’t stay constant as Vb nears 0

42. Indicator
Acid or base chose different protonated forms have different colors
Seek indicator whose color change is near equivalence point
Indicator error
Difference between endpoint (color change) and true equivalence point
If you use too much can participate in reaction

45. Quiz 4
A sample was analyzed using the Kjeldahl procedure. The liberated NH3 was collected in 5.00 mL of 0.05 M HCl, and the remaining acid required 3 mL of M NaOH for a complete titration. How many moles of Nitrogen were in the original sample?