Determination of the Solubility Product Constant by Titration
Experiment 10: Determination of the Solubility Product Constant by Titration
(original “Exp11” prepared by Kelemu Woldegiorgis)
With added video link and virtual data
INTRODUCTION
Many systems in chemistry appear to be static when in fact they are in dynamic equili
ium. When a system is in dynamic equili
ium the rate of the forward process is equal to the rate of the reverse process. One of these equili
ium systems is solubility equili
ium. When an ionic salt is placed in water, if the salt is soluble, the salt will dissociate into ions and the solution will contain hydrated positive ions and hydrated negative ions. An example is the formation of Na+ (aq) and Cl- (aq) when NaCl is dissolved in water.
Consulting solubility tables tell us that most hydroxides are insoluble and therefore do not dissociate into ions when in solution. For example, the net ionic equation for the reaction between MgCl2 and KOH is shown below, Equation 2.
However, even compounds considered insoluble will form a small amount of ions in solution. If a salt is “insoluble” what is really happening is that the equili
ium for the formation of hydrated ions from the reaction between the solid salt and water lies very far to the left (toward the solid). The formation of this small amount of ions and their recombination make up a system in dynamic equili
ium.
The equili
ium constant for this process is called the solubility product, Ksp. The equili
ium reaction and equili
ium constant expression for Mg(OH)2 is shown below. (Note that the Mg(OH)2 does not appear in the equili
ium expression since it is a solid. Similar equili
ium expressions can be written for other insoluble salts.
Ksp = [Mg2+][OH_]2
In this experiment you will determine the Ksp of two (essentially) insoluble hydroxides by titrating saturated solutions of the hydroxides with HCl. The experimental set up for the titration is shown in Fig. 1.
Figure 1. Titration set up. Phenolphthalein indicator has a pink color in base.
APPARATUS AND CHEMICALS
50 mL buret, Graduated cylinder, Buret stand with clamp, Erlenmeyer Flasks, Pipets and pipet bulb, 0.10 M HCl, 0.0010M HCl, and Phenolphthalein indicator.
PROCEDURE
A. Magnesium Hydroxide Titration
1. Use a pipet to withdraw 20.00 mL of a saturated solution of Mg(OH)2 and place it in a 150 mL Erlenmeyer flask; add three drops of phenolphthalein indicator.
2. Prepare a 1.0 L of XXXXXXXXXXM HCl solution by diluting 10.0 mL of 0.10 M HCl with distilled water. Fill the buret with the XXXXXXXXXXM HCl solution up to the zero mL mark.
3. Titrate the hydroxide solution until the mixture turns COLORLESS. Record the buret reading at this point.
4. Repeat steps XXXXXXXXXXRemember to re-fill the buret between titrations if needed and to record initial and final buret readings for each titration.
5. Run a third titration if the first two runs are different by more than 2%.
B. Calcium Hydroxide Titration
1. Use a pipet to withdraw 20.00 mL of a saturated solution of Ca(OH)2 and place it in a 150 mL Erlenmeyer flask.
2. Add three drops of phenolphthalein indicator.
3. Fill the buret to the zero mL mark with 0.10 M HCl solution and titrate the hydroxide solution until the mixture turns colorless.
4. Record the buret reading at the end point.
5. Repeat steps XXXXXXXXXXRemember to re-fill the buret between titrations if needed and to record initial and final buret readings for each titration.
6. Run a third titration if the first two runs are different by more than 2%.
View the video, “Calcium Hydroxide titration” (but a different indicator is used). Note what is involved in creating a saturated solution.
Tine Willis •Mar 21, 2015
https:
www.youtube.com/watch?v=PyFoPszHC74
DISPOSAL
Dispose of the remaining HCl solutions and the reaction mixtures in the sink. Run some tap water afterwards.
Experiment 10: Determination of the Solubility Product Constant by Titration
Name: ______________________________ XXXXXXXXXXSection: _______________
Lab Partner(s): _____________________________ Date: ______________
DATA AND CALCULATIONS
A. Titration of Mg(OH)2 solution
Trial 1 XXXXXXXXXXTrial 2 XXXXXXXXXXTrial 3
1. Volume of Mg(OH XXXXXXXXXX___ 10.0 mL __ XXXXXXXXXX__ 10.0 mL ____ __10.0 mL __
2. Initial buret reading XXXXXXXXXX___ 0.0 mL __ XXXXXXXXXX__ 15.1 mL ____ __29.2 mL __
3. Final buret reading ___ 15.1 mL __ XXXXXXXXXX__ 29.2 mL ____ __44.6 mL __
4. Volume of HCl needed _________ XXXXXXXXXX___________ XXXXXXXXXX_________
5. Moles HCl XXXXXXXXXX__________ XXXXXXXXXX___________ XXXXXXXXXX_________
6. Moles of OH- ____________ ____________ ___________
(Show your calculations)
7. Molarity of OH- ____________ __________ _________
(Show your calculations)
8. Average Molarity of OH- (mol/L XXXXXXXXXX_______________________
9. Ksp of Mg(OH)2 : ______________________
(Show your calculations)
B. Titration of Ca(OH)2 solution
Trial 1 XXXXXXXXXXTrial 2 XXXXXXXXXXTrial 3
1. Volume of Ca(OH XXXXXXXXXX___ 10.0 mL __ XXXXXXXXXX__ 10.0 mL ____ __10.0 mL __
2. Initial buret reading XXXXXXXXXX___ 0.0 mL __ XXXXXXXXXX__ 11.2 mL ____ __21.0 mL __
3. Final buret reading ___ 11.2 mL __ XXXXXXXXXX__ 21.0 mL ____ __30.7 mL __
4. Volume of HCl needed _________ XXXXXXXXXX___________ XXXXXXXXXX_________
5. Moles HCl XXXXXXXXXX__________ XXXXXXXXXX___________ XXXXXXXXXX_________
6. Moles of OH- ____________ ____________ ___________
7. Molarity of OH- ____________ __________ _________
(Show your calculations)
8. Average Molarity of OH- (mol/L XXXXXXXXXX_______________________
9. Ksp of Ca(OH)2: ________________
(Show your calculations)
Questions
1. Comment on the Ksp values of Mg(OH)2 and Ca(OH)2 in relation to the locations of the Mg and Ca in the Periodic Table.
2. Considering the trend of the Ksp values you obtained, assign Ksp values to Be(OH)2 and Ba(OH XXXXXXXXXXx 10-4 , 6.92 x 10-22 )
3. Would you be able to determine the solubility product of barium sulfate using this method? Why or why not?
1
Mg(OH)
2
(s)Mg
2+
(aq)+2OH
-
(aq)K
sp
1(3)
Ringstand
Buret
containsHClsolution
Erlenmyerflask
NaCl(s)Na
+
(aq)+Cl
-
(aq)(1)
H
2
O
Mg
2+
(aq)+2OH
-
(aq)Mg(OH)
2
(s)(2)
Experiment 11: Determination of the Equivalent Mass of an Unknown Metal by Electrolysis
(original “Exp12” prepared by Kelemu Woldegiorgis)
With virtual data
An electrochemical cell consists of two half-cells which may use the same electrolyte or different electrolytes. The chemical reactions in the cell may involve the electrolyte, the electrodes or an external substance (as in fuel cells which may use hydrogen gas as a reactant). During an electrochemical reaction involving electrolytes species from one half-cell lose electrons to their electrode (called anode) while species from the other half-cell gain electrons from their electrode (called cathode). A salt
idge is often employed to provide ionic contact between two half-cells (see Fig. 1). A salt
idge allows the flow of ions to maintain a balance in charge between the oxidation and reduction vessels while keeping the contents of each separate. During the operation of the electrochemical cell, positively charged ions migrate to the cathode while negatively charged ions migrate to the anode.
V
e-
e-
e-
e-
Cathode
Anode
Salt
idge
Figure 1. Schematic view of an electrochemical cell with two cell compartments connected
y a salt
idge and external wire. Electrons flow from the anode electrode to the cathode electrode.
An electrochemical cell which uses electrical energy to
ing about a chemical reaction is called an electrolytic cell. Electrolysis is the passage of a direct electric cu
ent through a molten or dissolved ionic substance to result in redox reactions at the anode and cathode electrodes. At the electrode connected to the negative pole or cathode of the battery, electrons will be accepted by one of the species present in the aqueous solution. The species reduced will normally be a metallic cation, the H+ ion, or possibly water itself. The reaction that is
observed is the one that occurs with the least expenditure of electrical energy