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NYB-Exp07-Solubility Product

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NYB
LAB EXERCISE 7
Solubility, Solubility Product and the Common Ion Effect
Introduction
The extent to which sparingly soluble salts dissolve in water is indicated either by the solubility or by
the solubility product of the salt. The solubility of a salt is often expressed as a concentration, that is,
as moles of salt per litre of solution. The solubility product of a salt is the product of the ion
concentrations in a saturated solution, each raised to the power of its coefficient in the balanced
equation.
The purpose of this experiment is to determine the solubility and the solubility product of potassium
bitartrate (KHC4H4O6). This will be done both in water, and in one of several solutions of potassium
nitrate (KNO3) which contain a known concentration of potassium (K+) ion. The calculated solubilities
and solubility products will then be compared.
Potassium bitartrate is sparingly soluble in water and dissolves to give potassium ions and bitartrate
ions.
KHC4H4O6 (s)  K+(aq) + HC4H4O6−(aq)
(1)
When potassium bitartrate is dissolved in water, the potassium ion concentration equals that of the
bitartrate ion. In a saturated solution this concentration is equal to the solubility, x, of the potassium
bitartrate which can be expressed as moles of potassium bitartrate per litre of solution.
[K+] = [HC4H4O6−] = x = solubility in distilled water
(2)
−
That is, the only way x moles of K+ and x moles of HC4H4O6 ion can appear in a litre of solution is if
x moles of KHC4H4O6 dissolve.
The solubility product is the product of the concentrations (in moles per litre) of the two ions in the
saturated solution.
Ksp = [K+] [HC4H4O6−]
(3)
If KHC4H4O6 is added to a solution of a potassium salt such as potassium nitrate (KNO3), then the
bitartrate ion concentration in the saturated solution is not equal to the concentration of potassium ion.
+
−
I.e. [K ]  [HC4H4O6 ].
The new solubility of KHC4H4O6 is still numerically equal to the concentration of bitartrate ions in
solution since all the bitartrate ions come from the dissolution of the potassium salt. However, the
total potassium ion concentration is the sum of the concentration from dissolved potassium nitrate plus
the concentration from the potassium bitartrate which is dissolved in the solution.
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[K+]Total = [K+]KNO3 + [K+]KHC4H4O6
(4)
The solubility product for a saturated solution of KHC4H4O6 in a solution of KNO3, can thus be written
as:
Ksp = [K+]Total [HC4H4O6−]
(5)
Since Ksp is a constant at a given temperature, it follows that the two Ksp values, obtained from
equations (3) and (5) respectively, should be the same. The average value of these two is taken at the
end of the experiment to be the solubility product of potassium bitartrate at that particular temperature.
The bitartrate ion (HC4H4O6−) behaves as an acid and therefore its concentration can be determined by
titration with a strong base:
OH−(aq) + HC4H4O6−(aq)  H2O() + C4H4O62−(aq)
(6)
Knowing the concentration of the hydroxide ion, the concentration of the bitartrate ion can be
calculated.
The solubility of potassium bitartrate in water, (and hence the Ksp value), is temperature dependent:
thus all solubility determinations should be done at the same temperature.
Pre-Lab Questions
1. Which is the common ion in this experiment?
2. What is the difference between solubility and the solubility product of a substance?
3. How do you expect the solubility product of KHC4H4O6 to vary between the distilled water
solution and the KNO3 solution?
4. How do you expect the solubility of KHC4H4O6 to vary between the two solutions?
Procedure
Pour about 150 mL of distilled water into a 500 mL Erlenmeyer flask and add an excess
(approximately 2 grams) of potassium bitartrate. Mix the contents of the flask thoroughly for 5
minutes, then let the mixture stand for 5 minutes more. Record the temperature of the solution. Filter
the solution through a dry filter paper in a dry funnel, into a dry 250 mL beaker. The apparatus used
here must be dry to avoid dilution of the saturated solution.
Rinse and fill a clean burette with a standard sodium hydroxide solution. The precise concentration
will be available in the lab.
Pipette 25.00 mL of potassium bitartrate solution into a clean 125 mL Erlenmeyer flask for the
titration. Add 3 drops of phenolphthalein indicator and titrate the bitartrate solution with sodium
hydroxide to a permanent pale pink endpoint. Repeat the titration. If results aren’t within 0.10 mL,
repeat the titration a third and final time.
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Repeat the experiment above replacing the distilled water in the solution by the assigned potassium
nitrate solution. The flask can be rinsed with the assigned potassium nitrate solution (not water!)
before preparing the actual solution. Prior to titrating, note the temperature of the solution; if this is
not within 1C of your first experiment, warm the flask with your hand or cool it under the tap, as
appropriate.
Collect titration results from other students (at least two more groups) who have used potassium
nitrate solutions of different concentrations.
Use the collected results to calculate the solubility and solubility product of potassium bitartrate in
both water and in the potassium nitrate solutions. Plot a graph of the solubility of potassium bitartrate
against the potassium nitrate concentration.
Mauro Di Renzo & Vanier College Chemistry Department
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CHEMISTRY NYB
DATA SHEET EXPERIMENT 7: SOLUBILITY PRODUCT
Student Name: __________________________
Student Name: __________________________
Concentration of assigned KNO3 solution: ________
Concentration of NaOH titrant: ________
Titration
Water
Trial 1
Trial 2
KNO3 solution
Trial 3
Trial 1
Trial 2
Trial 3
Volume of analyte
Initial burette reading
Final burette reading
Volume of NaOH used
Moles of NaOH
Moles of HC4H4O6− in
saturated solution
[HC4H4O6−] in saturated
solution
Solubility of KHC4H4O6
[K+] in saturated solution
Ksp of KHC4H4O6
Average Ksp
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Results from other groups:
[KNO3]
Solubility of KHC4H4O6
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Ksp
Ver. 1.0.4
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