Lecture 20 10/19/05

Titrations Analyte + titrant  Reaction

Requirements for titration reaction Large equilibrium constant so reacts to completion Fast kinetics Selective reaction Way to measure if reaction has occurred

Equivalence vs. Endpoint Are they Equal?

Direct vs. back titration Primary standard: Can be weighed and used directly Standard solution When primary standard not available for titrant Get by standardization Unknown titrant used to titrate known primary standard Direct vs. back titration

Kjeldahl analysis Digestion Neutralization Organic C,H,N  (boiling H2SO4)  NH4+ + CO2 + H2O Neutralization NH4+ + OH-  NH3 (g) + H20 Distillation into standard HCl NH3 + H+  NH4+ Titration of unreacted HCl OH- + H+  H20

Kjeldahl (7-12) Used to analyze 256 μL of a solution containing 37.9 mg proteins/mL. Liberated NH3 was collected in 5.00 mL of 0.0336 M HCl. 6.34 mL of 0.010 M NaOH needed to neutralize unreacted acid. Weight % of nitrogen in protein?

Kjeldahl (7-12)

Kjeldahl (7-12)

Kjeldahl (7-12)

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

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]

Titration of 0.25 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

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

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?

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)

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- = 0.002 moles Find new I- concentration [I-]=(0.002 moles)/(0.035L) = 0.0571 M Find concentration of Ag+ [Ag+]=Ksp/ [I-] [Ag+]=8.3x10-17 / 0.0571 = 1.4 x 10-15 pAg+= 14.84

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-]=0.0571 M Find concentration of Ag+ [Ag+]=Ksp/ [I-] [Ag+]=8.3x10-17 / 0.0571 = 1.4 x 10-15 pAg+= 14.84

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)

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+] = 0.0125 M pAg = 1.9

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

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+

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

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

CaCO3(s) [FM 100.087] + 2H+  Ca2+ + CO2(g) + H2O Problem 7-11 The carbonate content of 0.5413g of powdered limestone was measured by suspending the powder in water, adding 10.00 mL of 1.396 M HCl, and heating to dissolve the solid and expel CO2: CaCO3(s) [FM 100.087] + 2H+  Ca2+ + CO2(g) + H2O The excess acid required 39.96 mL of 0.1004M NaOH for complete titration to a phenolphthalein end point. Find the weight % of calcite in the limestone.

Problem 7-11 (solutions) Moles OH- = (39.96 mL)*(0.1004 M) = 4.012 mmol Moles H+ = (10 mL)*(1.396 M) = 13.96 mmol Moles H+ used to titrate CaCO3 = 9.948 mmol Moles CaCO3 = 9.948 mmol H*(1 mol CaCO3 / 2 mol H) Moles CaCO3 = 4.974 mmol Mass CaCO3 = 4.974 mmol *(100.087 g/mol) = 0.498 g Weight % = 0.498 g / 0.5413 * 100 = 92%

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

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)

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

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

Titration of a weak acid with strong base 0.02 MES [2-(n-morpholino)ethanesulfonic acid] with 0.100 M NaOH. pKa = 6.15

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

Finding endpoint with pH electrode

Titration of H6A with NaOH

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

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

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

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 0.035 M NaOH for a complete titration. How many moles of Nitrogen were in the original sample?