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Lecture 2110/24/05
Seminar today
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Precipitation Titration:Titration curve
• Before the equivalence point
• At the equivalence point
• After equivalence point
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• Relate moles of titrant to moles of analyte
• X-axis: Volume titrant added
• Y-axis: Concentration of one of the reactants• often as pXpX = -log[X]
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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
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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
mL50V
M) Ag)V0500.0(mL)00.25)(M I1000.0(
VCVC
Ag
Ag-
AgAgII
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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?
04.8pAg
M101.9]I[]Ag[x
)x)(x(103.8
]I][Ag[K
9
17
sp
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0
2
4
6
8
10
12
14
0 10 20 30 40 50 60 70 80 90 100
mL Ag added
pA
g
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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)
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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
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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
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0
2
4
6
8
10
12
14
16
0 10 20 30 40 50 60 70 80 90 100
mL Ag added
pA
g
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After the equivalence pointx-axis
• Volume of Ag+ added• Add more than 50 mL
• Let’s add 75 mL • (this volume is arbitrary other than > 50 mL)
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After the equivalence pointy-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
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0
2
4
6
8
10
12
14
16
0 10 20 30 40 50 60 70 80 90 100
mL Ag added
pA
g
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0.00
2.00
4.00
6.00
8.00
10.00
12.00
14.00
16.00
0 10 20 30 40 50 60 70 80 90 100
mL Ag added
pA
g
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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
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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+
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25.00 mL of 0.100 M halide titrated with 0.0500 M Ag+
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40.0 mL of 0.052 M KI plus 0.05 M KCl titrated with 0.084 M AgNO3
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Problem 7-11The 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.
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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%
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End-point detection for precipitation reactions
• Electrodes• Silver electrode
• Turbidity• Solution becomes cloudy due to
precipitation
• Indicators• Volhard• Fajans
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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)
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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
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Titration of strong acid/strong base
• 50 mL of 0.02 M KOH with 0.1 M HBr
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Titration of a weak acid with strong base
• 0.02 MES [2-(n-morpholino)ethanesulfonic acid] with 0.100 M NaOH.
• pKa = 6.15
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• Titration of 10.0 mL of 0.100 M B (base) with 0.100 M HCl.
• pKb1 = 4.00
• pKb2 = 9.00
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Finding endpoint with pH electrode
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Titration of H6A with NaOH
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Gran Plot
)VV(K10V beaA
HApHb
Advantage is that you can use data before the endpoint to find the endpoint
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Vb never goes to 0 because 10-pH never gets to 0
Also slope doesn’t stay constant as Vb nears 0
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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
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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?