Lecture 2110/24/05

Seminar today

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 pXpX = -log[X]

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

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

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

0

2

4

6

8

10

12

14

0 10 20 30 40 50 60 70 80 90 100

mL Ag added

pA

g

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

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

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)

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

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

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

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

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.

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

)VV(K10V beaA

HApHb

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?