Lab-AP-Solubility Product Key

Experiment 18
Pre-Lab Questions
Before beginning work on this experiment, read the directions and answer the following questions:
The ionic compound silver chromate is not very soluble in water. It ionizes according to the following
equation:
Ag2CrO4(s) ! 2Ag+(aq) + CrO42–(aq)
1. Write the solubility product expression for silver chromate.
Ksp = [Ag+]2 [CrO42–]
2. If one has a solution of 0.10 M silver nitrate and it is diluted by a factor of two, what is the new
concentration?
0.10 M × (½) = 0.050 M
3. The dilution of 0.10 M silver nitrate by a factor of two is carried out five times. What is the concentration
now?
0.10 M × (½)5 = 3.1×10–3 M
4. The value of Ksp of silver chromate is reported to be 1.1×10–12. In a saturated solution of silver chromate, the
silver ion concentration is found to be 2.5×10–4 M. What must the chromate ion concentration be? Show your
work.
K sp = [Ag + ]2 [CrO 4 2− ] = 1.1× 10−12
[CrO 4
2−
1.1× 10−12
1.1× 10−12
]=
=
= 1.8 × 10−5 M
+ 2
−4 2
[Ag ]
(2.5 × 10 )
Data Tables
Procedure A: Reacting serially diluted Ca2+ solutions with OH− solution (constant concentration)
#1
#2
#3
#4
#5
#6
#7
#8
[Ca2+]
0.050
0.025
0.013
6.3×10–3
3.1×10–3
1.6×10–3
7.8×10–4
3.9×10–4
[OH–]
0.050
0.050
0.050
0.050
0.050
0.050
0.050
0.050
PPT?
Y
Y
Y
Y
Y
N
N
N
Procedure B: Reacting serially diluted OH−solutions with Ca2+solution (constant concentration)
#1
#2
#3
#4
#5
#6
#7
#8
[Ca2+]
0.050
0.050
0.050
0.050
0.050
0.050
0.050
0.050
[OH–]
0.050
0.025
0.013
6.3×10–3
3.1×10–3
1.6×10–3
7.8×10–4
3.9×10–4
PPT?
Y
Y
Y
N
N
N
N
N
Laboratory Experiments for AP Chemistry
249
Experiment 18
Post-Lab Calculations and Analysis
1. Write the solubility product expression for the compound calcium hydroxide.
Ksp = [Ca2+][OH–]2
2. Identify the well # and ion concentrations for the first well in the calcium ion serial dilution with no
precipitate. Using these concentrations, determine the solubility product, Ksp, of calcium hydroxide.
Well #6, [Ca2+] = 1.6×10–3 M; [OH–] = 0.050 M
Ksp = [Ca2+][OH–]2 = (1.6×10–3)(0.050)2 = 4.0×10–6
3. Identify the well # and ion concentrations for the first well in the hydroxide ion serial dilution with no
precipitate. Using these concentrations, determine the solubility product, Ksp, of calcium hydroxide.
Well #4, [Ca2+] = 0.050 M; [OH–] = 6.3×10–3 M
Ksp = [Ca2+][OH–]2 = (0.050)(6.3×10–3)2 = 2.0×10–6
4. How did the values obtained from the two trials compare with each other?
These two values are relatively close to each other.
5. Look up the accepted value for the solubility product of calcium hydroxide in your textbook and compare it
to your values. Calculate the % error for each of your values.
Accepted Ksp = 5.5×10–6
% Error (Ca
2+
dilution) =
% Error (OH − dilution) =
4.0 × 10−6 − 5.5 × 10−6
5.5 × 10−6
2.0 × 10−6 − 5.5 × 10−6
5.5 × 10−6
× 100% = 27%
× 100% = 64%
6. Why did we make our Ksp calculations using the concentrations in the first well without a precipitate?
We wanted to use concentrations for which Qsp was less than or equal to Ksp. Any well with a precipitate
must have been over this value, while the first cell with no precipitate could have been up to this value.
250
Laboratory Experiments for AP Chemistry
Experiment 18
7. One should obtain a better value for the Ksp of calcium hydroxide if the concentrations of the last well with a
precipitate were averaged with the concentrations of the first well without a precipitate. Why? What is the
justification for doing this?
The correct value is known to be between the concentrations of the last well containing a precipitate and the
first well without one.
8. Using the method in #7, calculate new Ksp values from your data. Are these results better than before?
Ca
2+
dilution: [Ca ]avg =
2+
1
2
(3.1 × 10
−3
K sp = (2.3 × 10 −3 )(0.050)2 = 5.8 × 10 −6
OH − dilution: [OH − ]avg =
1
2
)
−3
−3
M + 1.6 × 10 M = 2.3 × 10 M
(0.013 M + 6.3 × 10
−3
)
M = 9.6 × 10 −3 M
K sp = (0.050)(9.6 × 10 −3 )2 = 7.3 × 10 −6
The first value is somewhat better than before, but the second value is now about 40% too large.
9. In an attempt to obtain more accurate results, suggest two improvements to the design of the lab.
• Use larger volumes to obtain more precise concentrations of reactants.
• Perform a second set of dilutions between the last well containing a precipitate and the first without, with
smaller differences.
Laboratory Experiments for AP Chemistry
251