UNIT 03: Solubility
BCLN CHEMISTRY 12 - Rev. July, 2015
Project: Determination of K
Potential Credits:
Name: __Om Patel_________
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/15
Goal and Instructions:
To determine the solubility product constant (K ) for PbI by watching a video of an experiment.
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2
Begin by watching this video: Determination of a Solubility Product Constant
Data and Observations:
1. Write the Dissociation equation for Lead (II) Iodide. It appears in the video but you should ensure
you put your charges and subscripts in the correct place.
PbI (s)
2
=
Pb (aq) + 2 I (aq)
2+
-
2. Write the K expression for PbI .
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2
Ksp = [Pb2+][I–]^2
3. When determining if a precipitate will form you compare a Trial K to the actual K . Will you see
a precipitate when the Trial K is larger than or smaller than the actual K ?
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4. Fill in Table 1 by watching the video
Table 1: Data needed to calculate the Trial K at which a precipitate begins to form
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Test
1
2
3
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4
5
6
UNIT 03: Solubility
BCLN CHEMISTRY 12 - Rev. July, 2015
Tube
Volume
of
0.010M
Pb(NO )
10.0
8.0
6.0
4.0
3.0
2.0
Volume
of
water(m
L)
added
0.0
2.0
4.0
6.0
7.0
8.0
Volume
of
0.020M
KI (mL)
10.0
8.0
6.0
4.0
3.0
2.0
Volume
of water
(mL)
added
0.0
2.0
4.0
6.0
7.0
8.0
Precipit
ate or
no
precipit
ate at
room
tempera
ture
Precipi
tate
Precip
itate
no
precipi
tate
no
precipi
tate
Temper
ature at
which
precipit
ate
dissolve
s ( C)
63
3
2
Precip
itate
54
Precipit
a’te
44
0
Calculations:
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35
N/A
N/A
UNIT 03: Solubility
BCLN CHEMISTRY 12 - Rev. July, 2015
1. For each test tube you will need to calculate the Trial K . In order to do this you will need the
diluted concentration of the Pb ion and the I ion. The dilution formula is shown below and the
calculations for the first test tube are done for you as an example. The old concentrations for each
compound are found in Table 1. Hint: The new volume will always be 20.0 mL
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2+
-
Dilution formula (should I take photo for each question)
New() = Old() x (Old Volume/New Volume)
Test Tube 1 - 10.0 mL of each solution
New (Pb2+) = 0.010M x (10.0ml/20.0ml) = 0.0050M
New (I-) = 0.020M x (10.0ml/20.0ml) = 0.010M
Ksp = (0.005M)(0.010M)^2 = 5.0*10^-7
Test Tube 2 - 8.0 mL of each solution
New [Pb ] = 0.010 M x 8 mL / 20 mL = 0.004 M
2+
New [I ] = 0.020 M x 8 mL / 20 mL = 0.008 M
‒
Ksp = (0.004 M) x (0.008 M) = 2.56 x 10
2
-7
Test Tube 3 - 6.0 mL of each solution
New [Pb ] = 0.010 M x 6 mL / 20 mL = 0.003 M
2+
New [I ] = 0.020 M x 6 mL / 20 mL = 0.006 M
‒
Ksp = (0.003 M) x (0.006 M) = 1.08 x 10
2
-7
Test Tube 4- 4.0 mL of each solution
New [Pb2+] = 0.010 M x 4 mL / 20 mL = 0.002 M
New [I‒] = 0.020 M x 4 mL / 20 mL = 0.004 M
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UNIT 03: Solubility
BCLN CHEMISTRY 12 - Rev. July, 2015
Ksp = (0.002 M) x (0.004 M)2 = 3.2 x 10-8
Test Tube 5 - 3.0 mL of each solution
New [Pb ] = 0.010 M x 3 mL / 20 mL = 0.0015 M
2+
New [I ] = 0.020 M x 3 mL / 20 mL = 0.003 M
‒
Ksp = (0.0015 M) x (0.003 M) = 1.35 x 10
2
-8
Test Tube 6 - 2.0 mL of each solution
New [Pb ] = 0.010 M x 2 mL / 20 mL = 0.001 M
2+
New [I ] = 0.020 M x 2 mL / 20 mL = 0.002 M
‒
Ksp = (0.001 M) x (0.002 M) = 4.0 x 10
2
-9
Questions:
1. According to the course formula sheet, what is the actual K of Lead (II) Iodide?
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8.5*10^-9
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UNIT 03: Solubility
BCLN CHEMISTRY 12 - Rev. July, 2015
2. Based on your calculations what is the minimum value for K ?
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3. Based on your calculations what is the maximum value for K ?
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4. Does the actual K fall between your calculated min and max?
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5. Between which two test tubes should the change from precipitate to no precipitate have occurred?
6. Can you think of a reason for this discrepancy?
7. Fill in Table 2 below using your K calculations and the temperature numbers from the video or
Table 1. The solubility is the same as the diluted [Pb ] you calculated.
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2+
Test tube 1
Test tube 2
Test tube 3
Temperature
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Test tube 4
UNIT 03: Solubility
BCLN CHEMISTRY 12 - Rev. July, 2015
Ksp
Solubility
8. What is the trend in solubility as the temperature is increased?
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