Transcript Determination of the Equilibrium Constant, Ksp

Determination of the
Equilibrium
Constant, Ksp, for a
Chemical Reaction
By: Bronson Weston
Background
Information
 Ksp is a particular type of equilibrium
constant called the solubility product
constant.
 Equilibrium is achieved when an
ionic solid dissolves to form a
saturated solution. The equilibrium
exists between the aqueous ions and
the precipitate, an undissolved solid.
 A saturated solution contains the
maximum concentration of ions of the
substance that can dissolve at the
solution's temperature.
 As the concentration of solute, dissolved
ions, increases, so does the rate of
reprecipitation. When the rate of
reprecipitation equals the rate of
dissolution, and there is no more net
dissolution of solid, equilibrium is
reached.
Ksp Equation
 If given the following reaction:
 AnBm(s)  n Am+ (aq) + m Bn-(aq)
 The Ksp of the reaction is:
 Ksp = [Am+]n[Bn-]m
 Ksp = molarity of solution
 solution is saturated, no precipitate
 Ksp< Molarity of Solution
 solution is saturated, precipitate is formed
 Ksp > Molarity of Solution
solution is unsaturated, no precipitate is
formed
Materials
• Calcium nitrate, Ca(NO3)2, 0.0900
M
• Sodium Hydroxide, NaOH, 0.100 M
• 96- Well Microplate
• Beral pipets
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Procedures- Row 1
1. Arrange the Microplate so that you
have 12 wells across from left to
right
2. Put 5 drops of water in
wells #2 through #12
in the first row
3. Put 5 drops of .0900 M Ca(NO3)2 in
well #1 in the first row
4. Add 5 drops of Ca(NO3)2 to well #2
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5. Using an empty Beral pipet, mix the
solution in Well #2 thoroughly by drawing
the solution into the pipet and then
squirting it back several times
6. Calculate the molarity of the solution in
well #2.
5 drops of 0.0900 M = n moles
5 drops
5 drops Ca(NO3)2 + 5 drops water = 10 drops
n moles = 0.0900 M = 0.0450 M
10 drops
2
7. Using the empty pipet, draw the solution
from well #2 and put 5 drops into well #3
8. Put the remaining solution back into well
#2
9. Mix the solution in well #3 with the empty
Beral pipet as before.
10. Continue this serial dilution
procedure, adding 5 drops
of the previous solution to
the 5 drops of water in each
well down the row until you
fill the last one, well #12.
11. After Mixing the solution in well #12,
discard 5 drops
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Calculations
12. Determine the
concentration of
Ca(NO3)2 solution
in each well,
using the method
used in step 6
Well #1
Well #2
Well #3
Well #4
Well #5
Well #6
Well #7
Well #8
Well #9
Well #10
Well #11
Well #12
0.0900 M
0.0450 M
0.0225 M
0.0113 M
0.00563 M
0.00281 M
0.00141 M
7.03 x 10-4 M
3.51 x 10-4 M
1.76 x 10-4 M
8.79 x 10-5 M
4.39 x 10-5 M
More Procedures
13.Place 5 drops of 0.100 M NaOH in
each well, #1- #12
14.Use an empty pipet to mix the
solution in each well
15.Calculate the concentration of each
reactant, Ca+2 and OH-,
and record the data on
a table.
16.Allow three or four minutes for
precipitates to form.
17.Observe the pattern of precipitation
and record, on the table, which
solutions form a precipitate.
I drew this one myself
Well #
Ca+2
OH-
Precipitate
Well #1
0.0450 M
0.0500 M
Yes
Well #2
0.0225 M
0.0500 M
Yes
Well #3
0.0113 M
0.0500 M
No
Well #4
0.00563 M
0.0500 M
No
Well #5
0.00281 M
0.0500 M
No
Well #6
0.00141 M
0.0500 M
No
Well #7
7.03 x 10-4 M
0.0500 M
No
Well #8
3.51 x 10-4 M
0.0500 M
No
Well #9
1.76 x 10-4 M
0.0500 M
No
Well #10
8.79 x 10-5 M
0.0500 M
No
Well #11
4.39 x 10-5 M
0.0500 M
No
Well #12
2.20 x 10-5 M
0.0500 M
No
18.Assume that the first solution, the
most concentrated, that does not
form a precipitate represents the
saturated solution.
19.Calculate the Ksp of Ca(OH)2, using
the concentration of Ca+2 and OHions in the saturated solution
Calculate the limiting reactant:
Ca+2 + 2OH-  Ca(OH)2
0.0113 mol Ca+2 x 1.00 mol Ca(OH)2
1.00 mol Ca+2)
= 0.0113 mol Ca(OH)2
0.0500 mol OH- x 1.00 mol Ca(OH)2
2.00 mol OH=0.0250 mol Ca(OH)2
0.0113 mol Ca(OH)2 < 0.0250 mol Ca(OH)2
Ca+2 is the limiting reactant
Calculate moles of OH- used in the reaction:
0.0113 M Ca+2 x 2.00 M OH1.00 M Ca+2
= 0.0226 M OH-
Calculate the Ksp:
Ksp = [Am+]n[Bn-]m
Ksp = [Ca+2][OH-]2
Ksp = [0.0113 M][0.0226 M]2
Ksp = 5.77 x 10-6
Calculate Percent Error:
[(Experimental Value-Actual Value) / Actual Value] x 100
Actual Value  Ksp = 6.5 x 10-6
Experimental Value  Ksp = 5.77 x 10-6
((5.77 x 10-6) – (6.5 x 10-6)) x 100
6.5 x 10-6
= 11% Error
Explanation
This lab was intended to demonstrate how you
can figure out a salt’s solubility constant. When
a solution is saturated, it is in equilibrium.
Therefore, when a solution is saturated, we
can use the current concentrations of the ions
to determine the Ksp. This is why we used the
concentrations in the 3rd well to determine the
solubility constant. We were able to assume
that the 3rd well was a saturated solution,
because it was the most concentrated solution
that did not form a precipitate.
References
 http://faculty.kutztown.edu/vitz/limsport/LabMan
ual/KSPWeb/KSP.htm
 http://www.jesuitnola.org/upload/clark/aplabs.ht
m#Determination%20of%20the%20Solubility%
20Product%20of%20an%20Ionic%20Compoun
d
 http://mooni.fccj.org/~ethall/2046/ch19/solubilit
y.htm
 Vonderbrink, Sally Ann. Laboratory
Experiments for Advanced Placement
Chemistry Student Edition. Flinn Scientific, Inc.
Batavia, IL. 1995.