SOLUBILITY
In chapter 4 we defined solubility as the amount of a substance that dissolves in a given quantity of solvent at a given temperature to form a saturated solution. We approached solubility in a very simple manner by classifying salts as soluble or insoluble and solubility guidelines were established indicating if a precipitate would form when two solution are combined.
Overview of Solubility
Experimentally, we can consider solubility equilibria to make predictions about the amount of a given salt that will dissolve. First, we need to define what we mean by solubility and we need to consider what we already know about solubility and compare that to what a chemist thinks about when they hear the term solubility.
When chemists discuss solubility they do so in terms of the solubility product constant. Keep in mind that the solubility product constant and molar solubility are two completely different terms. As chemists, we are able to calculate one from the other, but directly comparing the two without a calculation can lead to some improper assumptions. It is very important to recognize the similarities/differences between solubility in terms of the amount (in grams) that dissolve per liter of solution, the molar solubility, and the Ksp.
The Solubility Product Constant
Example Problem: 1.4 x 10-6 grams of ZnCO3 dissolve in 1.000 mL of solution and 2.8 x 10-6 grams of BaCrO4 dissolve in 1.000 mL of solution, which one has the largest Ksp? Solution: They are the same. Click here to see the explanation.
One common mistake students make is to simply look at the Ksp and think that constant tells them everything they need to know about the solubility when two slightly soluble salts are compared. If you are thinking this way, then you certainly aren't thinking like a chemist. A chemist would think through the process shown in the example below.
Ranking the solubility of slightly soluble salts given the Ksp Part I
Ranking the Solubility of slightly soluble salts given the Ksp Part II
Homework:
Read Section 17.4 and the Ksp Lab
Mastering Chemistry Solubility Constant Expression Tutorial
Mastering Chemistry Introduction to Solubility and the Solubility Product Constant Tutorial
Mastering Chemistry The Solubility Product in Medicine Tutorial
*Mastering Chemistry Ksp Pre-Lab Assignment
*Solubility Graded Homework #1
*Solubility Graded Homework #2
Questions 1-18 from the Solubility homework set posted on Carmen.
LAB #1
DEVELOPING A MASTERY OF THE SOLUBILITY PRODUCT CONSTANT LAB
Once you have a handle on saturated aqueous solutions, we can start to analyze and manipulate the solubility of these solutions and determine the criteria for precipitation.
Criteria For Precipitation
With a background on precipitate formation, we can now predict if a precipitate will form when two solutions are combined.
If Two Solutions are Mixed Will a Precipitate Form?
In certain cases, it is beneficial to separate a mixture of ions in solution. By utilizing the Ksp values and performing calculations, we can calculate the concentration needed to give the best separation of ions in solution.
Order of Precipitation, Minimum Concentration Needed to Facilitate Precipitation, and Best Separation
Homework:
Read Section 17.6
Mastering Chemistry Precipitation Tutorial
Mastering Chemistry Selective Precipitation Tutorial
Mastering Chemistry Precipitation Calculation Tutorial
Mastering Chemistry Fractional Precipitation of Metal Carbonates Tutorial
*Solubility Graded Homework #3
Questions 58-69 from the Solubility homework set posted on Carmen.
A common laboratory experiment constructed to illustrate the principles of solubility is Qualitative Analysis. This experiment is designed to answer what is present. Unfortunately, there is no spot test for each individual cation, so a scheme was developed, which groups cations based on their solubility characteristics. Each ion is then isolated and identified.
Applied Qualitative Analysis Scheme
The Group I cations precipitate as chlorides under acidic conditions.
Group I Separations
Why does our HCl need to be cold and dilute?
Group I Analysis
APPLIED QUALITATIVE ANALYSIS LAB
Similar analysis is performed for the Group II and Group III cations and is investigated in detail in your lab manual.
Chemists love to manipulate things. Ksp expressions are equilibrium expressions and as Le Chatlier showed us, equilibria can be manipulated if you apply a stress. There are many factors that influence solubility and the factors we will investigate in this class are: Common Ion, pH, Complex Ion Formation, and Amphoterism. In all of these effects, it is very important to WRITE OUT THE EQUILIBRIUM EXPRESSION so you can properly analyze how the solubility will be influenced.
The Common Ion effect is simply a direct application of Le Chatlier's principle.
Common Ion Effect
Homework:
Mastering Chemistry Solubility of "Insoluble" Salts Tutorial
Mastering Chemistry The Common-Ion Effect Tutorial
Mastering Chemistry Common-Ion Effect on Solubility for Lead Thiocyanate Tutorial
Mastering Chemistry Common-Ion Effect on Solubility for a Metal Hydroxide Tutorial
*Solubility Graded Homework #4-#5
The pH of a solution has a dramatic influence of the solubility. When investigating pH effects, it is important for you to know the strong acids, which will allow you to identify all the neutral anions in solution.
pH Effects
How does adding acid/base influence solubility?
Homework:
Mastering Chemistry The Effect of Acid on Solubility Tutorial
Mastering Chemistry Effect of pH on Solubility Tutorial
Mastering Chemistry Acid Rain: Effect on Solubility of Calcium Carbonate Tutorial
*Solubility Graded Homework #6-#7
Mastering Chemistry Creating a Buffer Solution Tutorial
Mastering Chemistry pH of a Buffer Solution Tutorial
*Solubility Graded Homework #8-#9
Questions 24-32 from the Solubility homework set posted on Carmen
Le Chatlier's principle does a great job of explaining the solubility in terms of the common ion effect and pH effects, but in some instances some puzzling results are obtained. For instance, when concentrated NH3 is added to a saturated solution of Zn(OH)2 its solubility increases. Now we need to determine why this is, and if Le Chatlier's principle is still valid.
Does zinc hydroxide follow the rules we've discussed so far?
It turns out Le Chatlier's principle still applies, but there is something else going on in solution responsible for these results. The concept responsible for the difference in solubility is complex ion formation.
Complex ion formation and coordination complexes
This allows us to investigate zinc hydroxide in a more complex way and it allows us to perform several calculations based on complex ion formation.
Re-analyzing zinc hydroxide
Solubility of zinc hydroxide in 15 M NH3
Determining the concentration of free metal cations in solution
Amphoterism is a special case where a slightly soluble salt has its solubility increased due to pH effects in acidic solution and has its solubility increased due to complex ion formation under basic conditions.
Amphoterism
Solubility of Al(OH)3 in 15 M NH3
Molar Solubility of Al(OH)3 in 15 M NH3 continued
Amphoteric Effects on Solubility
Homework:
Mastering Chemistry Solubility of Zinc Hydroxide in Basic Solution
Mastering Chemistry Cyanide Poisoning
*Solubility Graded Homework #10-#12
Questions 33-46 from the Solubility homework set posted on Carmen
The theory behind the separations of the Group II and Group III encompass all these effects and utilizes the solubility of the sulfide ion in solution.
Group II and Group III Sulfide Solubility
Will FeS precipitate?
At what pH will FeS begin to precipitate?
Homework:
Read Section 17.7 and the Qualitative Analysis Lab
Mastering Chemistry Qualitative Analysis of Metal Cations Tutorial
Mastering Chemistry Selective Precipitation in an Acidic Solution Tutorial
*Solubility Graded Homework #13-#14
Questions 19-23 & 72-84 from the Solubility homework set posted on Carmen
THERMOCHEMISTRY
Review sections 5.2 – 5.7 in the textbook, especially topics including: The 1st Law of Thermodynamics, Enthalpy, and how q and ΔH are related.
In Chapter 5, the term enthalpy was introduced and in this chapter we will use enthalpy and introduce a new concept, entropy, to tell a more complete story of Thermodynamics.
First, we will look into combustion reactions (which are exothermic) and see that as we progress from reactants to products there is a lowering in the potential energy stored in the chemical bonds, and the potential energy is converted to thermal energy (release of heat).
Chapter 19 Thermodynamics
The first law of Thermodynamics provides the means for accounting for energy, but it gives no hint as to why a particular process occurs in a given direction. A process is considered to be spontaneous if it occurs without outside intervention, and the driving force for a spontaneous process is an increase in entropy. We can define entropy as the measure of randomness or disorder and entropy can be expressed mathematically using macro/microstates and we can also compare the entropy of various systems.
Spontaneous Process and Entropy
Mathematical Definition of Entropy
Show PHET tutorial of an ideal gas.
Macro and Micro States
Comparing Entropy of Various Systems
Homework:
Read Sections 19.1 (skip pages 804, 805, and the first two paragraphs of 806), 19.2, and 19.3.
Mastering Chemistry Entropy and Microstates Tutorial
Mastering Chemistry Qualitative Predictions about Entropy Tutorial
*Thermochemistry Quiz Questions #1 - #4
The second law of Thermodynamics states that in any spontaneous process there is always an increase in entropy of the universe.
2nd Law of Thermodynamics
The third law of Thermodynamics states that the entropy of a pure crystal at zero degress kelvin is zero. The change in entropy of a reaction can be calculated from the standard entropy of each substance.
3rd Law of Thermodynamics
Homework:
Read Section 19.4
Mastering Chemistry Entropy of Reaction for Nitrogen Dioxide Formation Tutorial
Mastering Chemistry Standard Entropy of Reaction Tutorial
Mastering Chemistry Entropy and the Second Law of Thermodynamics Tutorial
Mastering Chemistry Third Law of Thermodynamics Tutorial
*Thermochemistry Quiz Question #5 - #6
Gibbs Free Energy is a state function combining enthalpy and entropy in the form of G = H - TS. J. Willard Gibbs developed this equation, which provides a convenient way to use the change in enthalpy and change in entropy to predict whether a given reaction occurring at constant pressure and constant temperature will be spontaneous. The derivation of this equation can be seen below.
Free Energy (G)
We can also consider the various situations for the relative signs of ΔH and ΔS to predict how ΔG will change with temperature. We can use the chart generated in the video below to investigate the effect of temperature on the spontaneity of a chemical reaction.
Predicting the Sign of Delta G
DEVELOPING A MASTERY OF THERMODYNAMIC RELATIONSHIPS LAB
Homework:
Read Section 19.5
Mastering Chemistry Interactive Activity – Temperature Dependence of Entropy Tutorial
Mastering Chemistry Standard Free Energy of Formation Tutorial
Mastering Chemistry Gibbs Free Energy: Temperature Dependence Tutorial
Mastering Chemistry Melting and Boiling Points Tutorial
*Thermochemistry Quiz Question #7 - #9
Read Section 19.6
Mastering Chemistry Enthalpy, Entropy, and Spontaneity Tutorial
Mastering Chemistry Gibbs Free Energy: Spontaneity Tutorial
*Thermochemistry Quiz Question #10 - #11
Free Energy Under Non-standard Conditions
Thermochemistry of the Haber Process
Read Section 19.7
Mastering Chemistry A Molecular View of Thermodynamics Tutorial
Mastering Chemistry Free Energy and Chemical Equilibrium Tutorial
Mastering Chemistry Free Energy and the Reaction Quotient Tutorial
Mastering Chemistry Gibbs Free Energy and Equilibrium Tutorial
Mastering Chemistry Gibbs Free Energy: Equilibrium Constant Tutorial
Mastering Chemistry Isomerization of Glucose to Fructose Tutorial
*Thermochemistry Quiz Question #12 - #23
After completing these tutorials and quiz questions, you should now be prepared to handle the more challenging problems:
*Quiz Questions #24-#34.
ELECTROCHEMISTRY
Review section 4.4 of the textbook, especially the Oxidation numbers rules on page 137.
Read Sections 20.1, 20.3, and 20.4
Mastering Chemistry Oxidation-Reduction Reactions Tutorial
Mastering Chemistry Oxidation States Tutorial
Mastering Chemistry Identifying Oxidizing and Reducing Agents Tutorial
Chemistry is an experimental science where measurements are made. The theories and equations we see in the text are backed up by experimental results. The following example in the video below helped shape Electrochemistry.
Electrochemistry Observations
Electrochemistry is the study of chemical and electrical energy. The two processes we will discuss are the generation of an electrical current from a chemical (redox) reaction (this is a spontaneous process) and the use of an electric current to produce a chemical change (a nonspontaneous process).
Electrochemistry
Now that we have observed various electrochemical processes, we need a convention to describe each voltaic cell.
Silver-Zinc Voltaic Cell Part 1
Silver-Zinc Voltaic Cell Part 2
Silver-Iron Voltaic Cell
When you describe voltaic cells, be sure to keep a few things in mind:
*When a half reaction is reversed, the sign of the standard cell potential is reversed.
*The standard cell potential is an intensive property.
*To run spontaneously, the standard cell potential must be positive.
*A chemically inert conductor is required if none of the substances participating in the half reaction is a conducting solid.
Voltaic Cells
Homework:
Mastering Chemistry Introduction to Galvanic Cells Tutorial
Mastering Chemistry Animation – Analysis of Copper-Zinc Voltaic Cell Tutorial
Mastering Chemistry A Nickel–Aluminum Galvanic Cell Tutorial
Mastering Chemistry A Nickel-Silver Galvanic Cell Tutorial
Mastering Chemistry Cell Potential Tutorial
*Electrochemistry Quiz Questions #1 - #6
Now that we have investigated Voltaic Cells, we need at come up with a convention to balance the reactions occurring in these cells.
Balancing Redox Reactions
Read Section 20.2
Mastering Chemistry Balancing Redox Equations and Stoichiometry Tutorial
Mastering Chemistry Balancing Redox Equations: Half-reaction Method Tutorial
Work can be done when electrons are transferred through a wire. The amount of work depends on the potential difference between the anode and cathode, and we can determine the relationship between the Cell Potential and the Gibbs Free Energy.
Ecell and Delta G (part 1)
Ecell and Delta G (part 2)
All the calculations to this point have been calculated under standard conditions. When the concentrations of the solutions in the anode and cathode compartment are changed, the cell potential will change. This relationship, referred to as the Nerst equation, will allow us to calculate the cell potential under non-standard conditions.
Cell Potential and Concentration
Application of Nerst Equation
Homework:
Read Section 20.5 and 20.6
Mastering Chemistry Cell Potential and Free Energy of a Lithium–Chlorine Cell Tutorial
Mastering Chemistry Interactive Activity – The Relationship of E°cell, Keq, and Gibbs Mastering Chemistry Free Energy Tutorial
Mastering Chemistry Cell Potential and Free Energy Tutorial
Mastering Chemistry Introduction to the Nernst Equation Tutorial
Mastering Chemistry Cell Potential and Equilibrium Tutorial
Mastering Chemistry The Nernst Equation Tutorial
Mastering Chemistry The Nernst Equation and pH Tutorial
*Electrochemistry Quiz Question #7 - #13
DEVELOPING A MASTERY OF ELECTROCHEMICAL RELATIONSHIPS LAB
In a Galvanic/Voltaic Cell, a spontaneous redox reaction produces a current (electricity). In an electrolytic cell, electrical energy is needed to produce a chemical change. During electrolysis, the electrical energy forces a non-spontaneous reaction to occur.
Electrolysis
In an electrolytic process, we can use the stoichiometric relationships of an electrolytic process to calculate various electrolytic properties. In the first example shown below, we will calculate the mass of solid copper that is plated when a current of 10 amps is passed for 30 minutes through a Cu2+(aq) solution.
Stoich of Electrolytic Processes
The next example shows how long a current of 5 amps must be applied to a solution of Ag+ to produce 10.5 grams of Ag(s).
Electrolysis Example
Producing H2(g) has gained widespread attention from its potential application as a source of alternate energy. The following example shows why the electrolysis of water is not an effective way to produce hydrogen gas and shows how much hydrogen gas is liberated during the passage of 2 amps for 30 minutes.
Electrolysis of H2O
We can also calculate the concentration of a particular ion remaining in a solution after a current has been passed through its solution. In this case, the [Cu2+] remaining in 335 mL of solution that was originally 0.215 M copper(II) sulfate after the passage of 2.17 amps for 235 seconds, can be calculated.
Calculating Concentration in Electrolysis
Homework:
Read Section 20.9
Mastering Chemistry Introduction to Electrolysis Tutorial
Mastering Chemistry Introduction to Electroplating Tutorial
Mastering Chemistry Electrolysis of Aqueous Salts Tutorial
Mastering Chemistry Analysis of Electroplating Tutorial
Mastering Chemistry Electrolysis and Current
*Electrochemistry Quiz Question #14 - #15
After completing these tutorials and quiz questions, you should now be prepared to handle the more challenging problems:
*Quiz Questions #16-#23.
TRANSITION METAL COMPLEXES
Transition Metals utilize their valence d-orbitals to form coordination complexes, which have characteristics important to industry, technology, and medicine. Coordination complexes exist in every color of the rainbow and can be found in jewelry, steel, paints, anticancer drugs, and photographic films. Most catalysts contain transition metal complexes and they are commonly used in the pharmaceutical industry. There two areas are vitally important to research chemists and are rapidly growing. A better understanding of the fundamentals of coordination complexes will help you understand current materials and will help chemists improve how these materials function.
In the early days, chemists were fascinated by transition metal complexes due to their color. The color of these complexes depends heavily on the d electron count of the transition metal, and the energy levels in the complex ions.
Transition Metals and Coordination Complexes
Being able to properly interpret the properties of the electromagnetic spectrum will allow us to interpret how our eyes detect color.
The Electromagnetic Spectrum and Color
The discreet nature of the energy levels in molecules allows us to determine the energy it would take to excite an electron from an energy level of lower energy to one of higher energy. If this excitation falls in the visible range of the electromagnetic spectrum, our eye will detect color.
Orbital Energies
Chemists are able to measure the wavelength of light a material absorbs using a UV-Vis Spectrometer, and by measuring which wavelengths of radiation are absorbed and which are not, we can gain valuable information on a molecules electronic excitations, or electronic structure.
UV-Vis Spectroscopy
UV-Vis Spectroxcopy (cont.)
Keep in mind that a transition metal atom has very different properties than when it is the central atom in a coordination complex. This is due to the difference in how the electrons fill the orbitals to achieve the lowest energy configuration possible.
Electron Configurations of Transition Metal Complexes
Coordination Compounds consist of a transition metal complex ion (transition metal center with ligands attached) and counter ions. The relative energies of the d orbitals in these transition metal complexes lead to some unique physical properties, including color and magnetism.
Chemistry of Coordination Complexes
In 1893 Alfred Werner proposed a theory that successfully explaining the difference in color for various transition metal complexes. This theory is still used today, but before we go into the modern day theory, lets take a look into how Werner developed his theory.
Coordination Complexes Before 1893
Modern Day Formulas for Transition Metal Complexes
Once the modern day formulas were introduced, chemists began to investigate how ligands surrounded the transition metal center. The term isomer is given to complexes that have the same overall composition, but different structures. This prompted chemists to introduce two new terms: coordination sphere and coordination number.
Arranging Ligands Around the Transition Metal Center: Introducing Isomers
Once Werner showed coordination complexes could exist as isomers, chemists began to investigate various isomers and in this class we will cover two types of structural isomers (linkage and coordination sphere) and two types of stereo isomers (geometric and optical).
Isomer Overview
Linkage Isomers
Coordination Sphere Isomers
Geometric Isomers
Optical Isomers
At this point we really havn't scratched the surface of why a transition metal wants to form a bond with a ligand.
Complex Formation: The Metal-Ligand Bond
The Chelate Effect
When you are given a transition metal complex you will need to know how to determine its oxidation state, its coordination number, and its geometry.
Transition Metal Complexes: Oxidation States, Coordination Number, and Geometry
When a transition metal complex has a coordination number of 4 it can exist as either a tetrahedral or square planar molecule. There are a set of empirical observations that can allow you to predict whether a complex with a coordination number of 4 will exist as a tetrahedral molecule or a square planar molecule.
Stereochemistry
Determining if an ML4 complex is Td or Square Planar.
Example Problem: Determining if an ML4 complex is Td or Square Planar.
Transition metal complexes bond to a variety of surrounding atoms/molecules which we refer to as ligands. The nature of the ligand that binds to the transition metal center has a large influence on the properties of the transition metal complex.
Ligands
We have discussed that transition metal complexes exhibit a wide variety of colors and have varying magnetic properties. In order to explain the similarities and differences in these properties we need to look into the bonding theories of transition metal complexes.
Bonding Theories of Transition Metal Complexes
Crystal Field Theory
In order to understand the bonding theories of transition metal complexes you need to visualize how the ligands interact with the 5 different d orbitals. You cannot visualize how the d orbitals interact with the ligands if you do not know the shapes of the d orbitals like the back of your hand.
Shapes of d Orbitals
By observing the colors of various transition metal complexes, we can start to see how the nature of the ligand bound to the transition metal center plays a big role in determining the resulting color of a transition metal complex.
Orbital Overlap and Orbital Energies in Crystal Field Theory
Crystal Field Theory
The Spectrochemical Series
High Spin/Low Spin Co Complexes
Octahedral Field Splitting vs. Tetrahedral Field Splitting
Octahedral Field Splitting vs. Square Planar Field Splitting
Tuesday, June 14, 2011
Monday, May 2, 2011
Developing a Mastery of the Solubility Product Constant
Learning Objectives
• Be able to determine the solubility product constant for calcium iodate by titration of the iodate ion in a saturated solution.
• Be able to set up an equilibrium expression for a slightly soluble salt and write out the expression for Ksp in terms of concentrations of products and reactants.
• Given the solubility of a slightly soluble salt, in either mol/L or g/L, calculate the Ksp.
• Given the Ksp of a slightly soluble salt, calculate the molar solubility.
• Given a list of slightly soluble salts and their Ksp values, be able to list the salts in order of increasing solubility.
• Given a list of slightly soluble salts and their Ksp values, be able to determine the concentration of ions in solution.
Why Study Solubility?
Solubility data has many important technological applications in various fields of science. In designing drugs for medicinal purposes, a knowledge of solubility is required to facilitate drug delivery systems, as well as controlling the solubility of the drug in the blood stream. Engineers utilize theory to design water treatment facilities, which remove hazardous chemicals from drinking water. In addition, environmental chemists can explain pollutants by carefully analyzing solubility data. Our solubility analysis begins with analyzing titration data to determine molar solubility and the solubility product constant (Ksp) and will continue in subsequent labs with manipulating the solubility of a slightly soluble salt by analyzing the common ion effect, pH, complex ion formation, and amphoterism.
Where you Start
How do the solubility rules relate to the Ksp tables?
In Chemistry 121 we discussed the solubility rules by simply stating a set of rules indicating if a compound was “soluble” or “insoluble.” In reality, there are varying degrees of solubility from compound to compound. For example PbF2 and SrSO4 are both “soluble” according to the solubility rules, but it can be determined experimentally that 0.466 g of PbF2 will dissolve in 1 L of H2O, while only 0.092 g of SrSO4 will dissolve in 1 L of H2O.
By analyzing solubility equilibria, the chemist can make predictions about the amount of a given compound that will dissolve. The solubility product constant, Ksp, is a measure of just how much of a solid dissolves to form a saturated solution. Your TA will distribute a table of various Ksp values for common ionic compounds. You will refer to it frequently in the solubility unit of this class.
Where you are
How do we go from a balanced equilibrium expression to a Ksp expression?
The solubility product constant, or Ksp, is the constant established between a solid solute and its ions in a saturated solution. For PbF2 the solubility equilibrium expression is written as:
PbF2(s) ⇔ Pb2+(aq) + 2 F−(aq)
From the balanced chemical reaction describing the equilibrium we can write the equation for determining the Ksp, which is expressed as the product of the molar concentrations of the ions [in this case Pb2+(aq) and F−(aq)] raised to the power of their coefficients. The Ksp equation for PbF2(s) is:
Ksp = [Pb2+][F-]2
In this laboratory section one of the objectives is to experimentally determine a Ksp value. If we were determining this value for PbF2(s) we would need the concentrations of Pb2+ and F- in solution. Using stoiciometry, if the moles of Pb2+ are known we can convert it to moles of F-. As a result we only need to know the concentration of either Pb2+(aq) or F-(aq).
In this particular lab you will need to determine the Ksp for Ca(IO3)2(s) by calculating the concentration of either Ca2+(aq) or IO3−(aq). The balanced equilibrium and Ksp expression are shown below:
Ca(IO3)2(s) ⇔ Ca2+(aq) + 2 IO3−(aq)
Ksp = [Ca2+] [IO3−]2
Complete the following example problems in your lab notebook before reading on. Show them to your TA and have them checked for accuracy before moving on. NOTE: You are highly encouraged to work in groups on the Example Problems, but must perform all chemical experiments individually.
Example Problem 1.
Write equations for the solubility equilibrium and for the solubility product constant for each of the following slightly soluble salts.
a) strontium carbonate b) lead (II) bromide c) zinc phosphate d) silver sulfide
In this experiment it is much easier to calculate the concentration of IO3−(aq) and this concentration is determined by a common technique used by chemists called a titration. In a typical titration the concentration of a certain ion [in this case IO3−(aq)] is unknown and it is combined with a reagent solution whose concentration is known. Chemists refer to the known solution as the standard solution [in this case S2O32−]. Notice that the Na2S2O3 standardized solution is listed as having a concentration of 0.05xx in the lab procedure. It really doesn’t matter if we use 0.0510, 0.0599, 0.0500, or 0.0513 as the concentration of the standard solution, but it is very important that we know the EXACT concentration to four decimal places.
The key to a titration is to have some kind of indicator to tell the chemist when the end point is reached. In this experiment we can utilize the fact that I2(aq) turns blue when it is in contact with a starch solution. This presents a small problem because we are trying to calculate [IO3−] not [I2]. We can use a little oxidation – reduction chemistry to come up with an overall balanced equation in a two step process where first the IO3−(aq) is reduced to I2(aq) in step 1. Then the I2(aq) is further reduced by S2O32−(aq) and in the process is used up to form I−(aq) in step 2. The key here is that in the presence of the starch indicator once all the I2(aq) is consumed the solution will turn from blue to clear and tell us where the end point is. The chemical reactions for this process are listed below:
IO3−(aq) + 5 I−(aq) + 6 H+(aq) → 3 I2(aq) + 3 H2O(l) 3 I2(aq) + 6 S2O32−(aq) → 6 I−(aq) + 3 S4O62−(aq) IO3−(aq) + 6 H+(aq) + 6 S2O32−(aq) → I−(aq) + 3 S4O62−(aq) + 3 H2O(l)
In this titration we know the EXACT amount (in mL) of S2O32−(aq) that was used. We also know the concentration of S2O32−(aq) so multiplying the volume times the concentration gives us the number of moles of S2O32−. We then see from the overall equation that for every 6 moles of S2O32−(aq) we have one mole of IO3−(aq). Lastly to get the [IO3−] we need moles and liters. Suppose we titrated 10.00 mL of the solution, take the number of moles of IO3−(aq) obtained and divide it by 0.01000 L. Likewise we can use molar relationships to find the [Ca2+] and plug those values into the Ksp expression and determine the Ksp for Ca(IO3)2(s).
Complete the following example problems in your lab notebook before reading on. Show them to your TA and have them checked for accuracy before moving on.
Example Problem 2.
To determine the solubility product constant (Ksp) at 50oC for Ni(IO3)2, a chemist prepares a saturated solution of the salt at this temperature. They filter and remove a 10.00 mL aliquot of this solution, and add 50.0 mL of distilled water, 2 g of KI and 10 mL of 3 M HCl to the flask. Thiosulfate solution is added to the flask and after 2% starch solution is added it turns blue. The chemist keeps adding thiosulfate solution from the burette until the solution in the flask turns colorless. It takes 14.58 mL of 0.0515 M Na2S2O3 to reach the endpoint of the titration. The chemical reactions governing the titration are shown above.
a) How many moles of S2O32−(aq) are needed to reach the endpoint of the titration?
b) How many moles of IO3−(aq)are present at the endpoint of the titration?
c) If 10.00 mL of a the saturated Ni(IO3)2 solution was used, what is the concentration of IO3−(aq) at the end point?
d) Based on your results from part c), what is the concentration of Ni2+(aq) at the end point?
e) Write out the Ksp expression for Ni(IO3)2.
f) Determine the Ksp of Ni(IO3)2 at 50oC.
Procedure
The Ksp of any slightly soluble salt is determined from a saturated solution. In this experiment you will experimentally determine the Ksp of calcium iodate, and in order to do so, you must first prepare a saturated solution of Ca(IO3)2.
1. To prepare the saturated solution of Ca(IO3)2 measure roughly 20 mL of 1 M Ca(NO3)2 into a 250-mL beaker. Next add about 50 mL of 0.2 M KIO3 to the Ca(NO3)2 solution.
You should observe a white precipitate. Note: These volume measurements do not need to be exact, so you can use a graduated cylinder to add the solutions to the beaker.
We ultimately need to isolate the Ca2+ and IO3– ions in the solution in order to determine their concentrations. To do this we must separate or filter them from the solid Ca(IO3)2. The magnitude of the Ksp is proportional to how much solid Ca(IO3)2 will dissolve.
2. Similar to Figure 2, assemble a ring, clay triangle, and funel with filter in order to perform a gravity filtration. Fold a piece of filter paper in half, then fold it in half again. In order to fit the filter paper snugly inside the funnel, tear off a small part of the corner as shown in the video below.
Open the filter paper so that one side of the funnel is in contact with a single layer of the filter paper and the other side of the funnel is in contact with three layers of the filter paper.It seems to work best if the side with the tear has three layers. The paper should be in smooth contact with the funnel at the top, but not near the point. Squirting in water with a wash bottle sometimes helps to keep the filter paper in place.
3. Take the 250-mL beaker from Step 1 and pour as much supernatant liquid through the filter paper as possible, trying your best not to disturb the precipitate. Use a wash bottle filled with distilled water to wash down the sides of the beaker. Swirl the wash water with the solid and quickly pour the mixture on the filter paper. Rinse any remaining solid onto the filter paper with a small amount of water from your wash bottle.
4. Wash the solid Ca(IO3)2 on the filter paper with three small portions of distilled water.
5. Using a microspatula, transfer roughly 1/3 of the wet precipitate to one 100-mL beaker. Leave the left over precipitate on the filter paper. You will only use the precipitate on the filter paper if you mess up one of your trials.
6. Rinse your 100-mL graduated cylinder with distilled water and use it to add 80 mL of distilled water to the beaker containing the wet precipitate.
7. Acquire a stirring motor from the balance room and a stir bar from your TA.
Place the stir bar into the beaker containing your 80 mL of water and your Ca(IO3)2 precipitate.
8. Place your beaker on the stirring motor and turn the stirrer on to its lowest speed setting and allow it to stir in order to obtain a saturated solution.
While you are waiting for your solution to reach equilibrium complete the following problems in your lab notebook, which must be checked off by your TA before you continue. Note: the following problems were taken from Quiz I given during the past two quarters.
Example Problem 3.
The solubility product constant for Cu(IO3)2 is 1.44 x 10-7. What volume of 0.0520 M S2O32- would be required to titrate a 20.00 mL sample of a saturated solution of Cu(IO3)2?
Example Problem 4.
The solubility product constant for Sr(IO3)2 is 1.14 x 10-7. What volume of 0.0250 M S2O32- would be required to titrate a 20.00 mL sample of a saturated solution of Sr(IO3)2?
Example Problem 5.
A titration was performed to determine the solubility product constant (Ksp) of lead iodide. The titration is governed by the following reactions:
2 I−(aq) + 2 Ce4+(aq) → I2(aq) + 2 Ce3+(aq) I2 (aq) + 2 Ce4+(aq) + 4 Cl−(aq) → 2 ICl2− (aq) + 4 Ce3+(aq) 2 I−(aq) + 4 Ce4+(aq) + 4 Cl−(aq) → 2 ICl2− (aq) + 4 Ce3+(aq)
(a) Write out the Ksp expression for lead iodide.
(b) 25.00 mL of a saturated PbI2 solution was titrated with 0.0524 M Ce(NO3)4. The first step of the reaction sequence was performed and a starch indicator was added to the solution turning it blue. After 1.240 mL of the Ce(NO3)4 was added the end point was established as the blue color indicating the presence of I2 turned clear. From the results of this experiment, calculate the Ksp of lead iodide.
Once you get your TA to check off these example problems, you are finished with Day #1 of this lab. Write your name on the Erlenmeyer Flask containing your saturated solution with a Sharpie, place Parafilm over the top of the flask, and hand it to your TA.
You will start by filtering this solution at the beginning of the next lab period.
9. In order to ensure you have a saturated solution of Ca(IO3)2, the mixtures set aside from Step 8 must not contain any solid. Filter the saturated solution in the same manner as Step 2, but for this step use a clean, dry 125-mL Erlenmeyer flask to catch the filtrate. Any glassware, as well as the filer paper, must be dry in order to calculate an accurate concentration of IO3–.
10. Rinse a 10-mL pipet with two small portions of the saturated Ca(IO3)2 solution. You may discard any rinse solutions down the drain.
11. Pipet 10.00 mL of the saturated Ca(IO3)2 solution into a 250-mL Erlenmeyer flask. With a Sharpie, label this flask TITRATION FLASK.
In this experiment we will utilize the following set of oxidation/reduction reactions to determine the number of moles of IO3– present in 10.00 mL of a saturated Ca(IO3)2 solution.
IO3−(aq) + 5 I−(aq) + 6 H+(aq) → 3 I2(aq) + 3 H2O(l) 3 I2(aq) + 6 S2O32−(aq) → 6 I−(aq) + 3 S4O62−(aq) IO3− + 6 H+(aq) + 6 S2O32−(aq) → I−(aq) + 3 S4O62−(aq) + 3 H2O(l)
At this point, you have a clear solution containing Ca2+(aq) and IO3–(aq). In a titration an indicator is often used and these indicators exhibit vivid color in solution. If we add excess I– to the solution (done in this experiemnt by adding solid KI) and excess acid (in this case 3 M HCl will work just fine), then all of the IO3–(aq) will be oxidized to I2(aq). We can verify that I2(aq) is present due to its brown color in solution.
As the second step of the reaction sequence indicates, whenever thiosulfate (S2O32–) is added to a solution of I2, the I2(aq) will be reduced to I–(aq). Whenever a small amount of I2(aq) is in the presence of a starch indicator, a dark blue complex will form. When the solution turns from blue (or brown) to colorless, we know that all of the I2(aq) has been reduced to I–(aq).
12. To your TITRATION FLASK, add approximately 50 mL of water using a graduated cylinder and 2 g of solid KI.
In addition, add about 10 mL of 3 M HCl and swirl. This is converting all of the IO3–(aq) to I2(aq), which will turn the solution brown.
This is consistent with the following equation:
IO3−(aq) + 5 I−(aq) + 6 H+(aq) → 3 I2(aq) + 3 H2O(l).
13. Now we are ready to perform the titration with standardized 0.05XX M Na2S2O3. Find the 0.05XX M Na2S2O3 on the cart and record its concentration to four decimal places in your lab notebook. Use this solution to rinse a 25-mL buret twice, then perform three titrations.
14. In each of your titrations, fill the 25-mL buret with 0.05XX M Na2S2O3 and record your initial buret reading. Begin your titration and as the titration proceeds you will observe the brown color of the I2 solution lighten to a more yellow color.
When you observe a yellow color stop the titration and prepare a fresh starch solution by adding 0.6 grams of vitex starch to roughly 30 mL of distilled water in a separate 250-mL Erlenmeyer flask.
15. Add about 4 mL of the starch indicator solution to the TITRATION FLASK and in doing so, you should observe a blue (or dark brown) solution.
16. Continue the titration until you reach the endpoint, which is the moment the solution turns colorless.
This end point indicates all of the iodine has been used up, which is consistent with the following equation:
3 I2(aq) + 6 S2O32−(aq) → 6 I−(aq) + 3 S4O62−(aq)
17. Record the final buret reading at the endpoint.
18. In your lab notebook, calculate the number of moles of S2O32−(aq) required in the titration.
At the endpoint of a titration, chemists can convert the moles of a substance that is known to the moles of something that is unknown. Most titrations performed up to this point involved neutralization reactions between acids and bases. At the end point of an acid/base titration, the moles of acid are equal to the moles of base. The chemical reaction in this experiment does not involve a neutralization reaction between an acid and a base. What we are observing is the following overall redox reaction:
IO3−(aq) + 6 H+(aq) + 6 S2O32−(aq) → I−(aq) + 3 S4O62−(aq) + 3 H2O(l)
From this overall chemical equation, we can come up with the most important relationship in this particular experiment. At the endpoint of the titration:
6 mol S2O32−(aq) = 1 mol IO3−(aq)
19. Calculate the molarity of IO3− for your first titration.
20. Refill the buret and perform two more titrations given in Steps 11 – 17.
21. If any of the three titrations do not agree to within 0.2 mL, perform another titration.
22. For each trial, calculate the molarity of IO3−, and calculate an average molarity of IO3− with the three trials that agree within 0.2 mL.
23. Using the stoichiometry of the ions in solution, calculate the concentration of Ca2+ from your average molarity of IO3−.
24. Calculate the Ksp for Ca(IO3)2.
25. Fill out your Report Sheet and have it checked over by your TA before you leave the second lab period.
26. Have your TA sign your notebook and after doing so they will give you a handout with all the formal lab report requirements. If you finish early it would be a good idea to start on your lab report.
27. There is a report question assignment for this lab on Mastering Chemistry. Check on-line and with your TA for the due date. No late assignments will be accepted.
28. The Mastering Chemistry pre-lab assignment for the Qualitative Analysis Lab is posted. Check on-line and with your TA for the due date. No late assignments will be accepted.
Where You Wanna Be
The Ksp values for most compounds are fairly straightforward to understand, but typically what a chemist wants to do is somehow manipulate the solubility of these compounds. What happens to the solubility when we increase the temperature? What happens when we change the pH? What happens when we add a common ion? In the Qualitative Analysis Experiments and in class in the next few weeks we will talk about how each of these scenarios can influence the solubility of a complex. But before you learn how to manipulate the solubility you have to have a thorough understanding of how a Ksp expression is set up and if you are given a molar solubility you should be able to calculate the Ksp. If you are given the Ksp you should also be able to calculate the molar solubility. The homework sets posted on Carmen will help you work through these types of problems. Also, as a hint, in order to study for the first exam example questions #3, #4, and #5 serve as a great benchmark to see if you truly understand the concepts behind this lab. Count on seeing an exam question based on how well you understand the concepts of those questions.
• Be able to determine the solubility product constant for calcium iodate by titration of the iodate ion in a saturated solution.
• Be able to set up an equilibrium expression for a slightly soluble salt and write out the expression for Ksp in terms of concentrations of products and reactants.
• Given the solubility of a slightly soluble salt, in either mol/L or g/L, calculate the Ksp.
• Given the Ksp of a slightly soluble salt, calculate the molar solubility.
• Given a list of slightly soluble salts and their Ksp values, be able to list the salts in order of increasing solubility.
• Given a list of slightly soluble salts and their Ksp values, be able to determine the concentration of ions in solution.
Why Study Solubility?
Solubility data has many important technological applications in various fields of science. In designing drugs for medicinal purposes, a knowledge of solubility is required to facilitate drug delivery systems, as well as controlling the solubility of the drug in the blood stream. Engineers utilize theory to design water treatment facilities, which remove hazardous chemicals from drinking water. In addition, environmental chemists can explain pollutants by carefully analyzing solubility data. Our solubility analysis begins with analyzing titration data to determine molar solubility and the solubility product constant (Ksp) and will continue in subsequent labs with manipulating the solubility of a slightly soluble salt by analyzing the common ion effect, pH, complex ion formation, and amphoterism.
Where you Start
How do the solubility rules relate to the Ksp tables?
In Chemistry 121 we discussed the solubility rules by simply stating a set of rules indicating if a compound was “soluble” or “insoluble.” In reality, there are varying degrees of solubility from compound to compound. For example PbF2 and SrSO4 are both “soluble” according to the solubility rules, but it can be determined experimentally that 0.466 g of PbF2 will dissolve in 1 L of H2O, while only 0.092 g of SrSO4 will dissolve in 1 L of H2O.
By analyzing solubility equilibria, the chemist can make predictions about the amount of a given compound that will dissolve. The solubility product constant, Ksp, is a measure of just how much of a solid dissolves to form a saturated solution. Your TA will distribute a table of various Ksp values for common ionic compounds. You will refer to it frequently in the solubility unit of this class.
Where you are
How do we go from a balanced equilibrium expression to a Ksp expression?
The solubility product constant, or Ksp, is the constant established between a solid solute and its ions in a saturated solution. For PbF2 the solubility equilibrium expression is written as:
From the balanced chemical reaction describing the equilibrium we can write the equation for determining the Ksp, which is expressed as the product of the molar concentrations of the ions [in this case Pb2+(aq) and F−(aq)] raised to the power of their coefficients. The Ksp equation for PbF2(s) is:
In this laboratory section one of the objectives is to experimentally determine a Ksp value. If we were determining this value for PbF2(s) we would need the concentrations of Pb2+ and F- in solution. Using stoiciometry, if the moles of Pb2+ are known we can convert it to moles of F-. As a result we only need to know the concentration of either Pb2+(aq) or F-(aq).
In this particular lab you will need to determine the Ksp for Ca(IO3)2(s) by calculating the concentration of either Ca2+(aq) or IO3−(aq). The balanced equilibrium and Ksp expression are shown below:
Complete the following example problems in your lab notebook before reading on. Show them to your TA and have them checked for accuracy before moving on. NOTE: You are highly encouraged to work in groups on the Example Problems, but must perform all chemical experiments individually.
Example Problem 1.
Write equations for the solubility equilibrium and for the solubility product constant for each of the following slightly soluble salts.
a) strontium carbonate b) lead (II) bromide c) zinc phosphate d) silver sulfide
In this experiment it is much easier to calculate the concentration of IO3−(aq) and this concentration is determined by a common technique used by chemists called a titration. In a typical titration the concentration of a certain ion [in this case IO3−(aq)] is unknown and it is combined with a reagent solution whose concentration is known. Chemists refer to the known solution as the standard solution [in this case S2O32−]. Notice that the Na2S2O3 standardized solution is listed as having a concentration of 0.05xx in the lab procedure. It really doesn’t matter if we use 0.0510, 0.0599, 0.0500, or 0.0513 as the concentration of the standard solution, but it is very important that we know the EXACT concentration to four decimal places.
The key to a titration is to have some kind of indicator to tell the chemist when the end point is reached. In this experiment we can utilize the fact that I2(aq) turns blue when it is in contact with a starch solution. This presents a small problem because we are trying to calculate [IO3−] not [I2]. We can use a little oxidation – reduction chemistry to come up with an overall balanced equation in a two step process where first the IO3−(aq) is reduced to I2(aq) in step 1. Then the I2(aq) is further reduced by S2O32−(aq) and in the process is used up to form I−(aq) in step 2. The key here is that in the presence of the starch indicator once all the I2(aq) is consumed the solution will turn from blue to clear and tell us where the end point is. The chemical reactions for this process are listed below:
In this titration we know the EXACT amount (in mL) of S2O32−(aq) that was used. We also know the concentration of S2O32−(aq) so multiplying the volume times the concentration gives us the number of moles of S2O32−. We then see from the overall equation that for every 6 moles of S2O32−(aq) we have one mole of IO3−(aq). Lastly to get the [IO3−] we need moles and liters. Suppose we titrated 10.00 mL of the solution, take the number of moles of IO3−(aq) obtained and divide it by 0.01000 L. Likewise we can use molar relationships to find the [Ca2+] and plug those values into the Ksp expression and determine the Ksp for Ca(IO3)2(s).
Complete the following example problems in your lab notebook before reading on. Show them to your TA and have them checked for accuracy before moving on.
Example Problem 2.
To determine the solubility product constant (Ksp) at 50oC for Ni(IO3)2, a chemist prepares a saturated solution of the salt at this temperature. They filter and remove a 10.00 mL aliquot of this solution, and add 50.0 mL of distilled water, 2 g of KI and 10 mL of 3 M HCl to the flask. Thiosulfate solution is added to the flask and after 2% starch solution is added it turns blue. The chemist keeps adding thiosulfate solution from the burette until the solution in the flask turns colorless. It takes 14.58 mL of 0.0515 M Na2S2O3 to reach the endpoint of the titration. The chemical reactions governing the titration are shown above.
a) How many moles of S2O32−(aq) are needed to reach the endpoint of the titration?
b) How many moles of IO3−(aq)are present at the endpoint of the titration?
c) If 10.00 mL of a the saturated Ni(IO3)2 solution was used, what is the concentration of IO3−(aq) at the end point?
d) Based on your results from part c), what is the concentration of Ni2+(aq) at the end point?
e) Write out the Ksp expression for Ni(IO3)2.
f) Determine the Ksp of Ni(IO3)2 at 50oC.
Procedure
The Ksp of any slightly soluble salt is determined from a saturated solution. In this experiment you will experimentally determine the Ksp of calcium iodate, and in order to do so, you must first prepare a saturated solution of Ca(IO3)2.
1. To prepare the saturated solution of Ca(IO3)2 measure roughly 20 mL of 1 M Ca(NO3)2 into a 250-mL beaker. Next add about 50 mL of 0.2 M KIO3 to the Ca(NO3)2 solution.
You should observe a white precipitate. Note: These volume measurements do not need to be exact, so you can use a graduated cylinder to add the solutions to the beaker.
We ultimately need to isolate the Ca2+ and IO3– ions in the solution in order to determine their concentrations. To do this we must separate or filter them from the solid Ca(IO3)2. The magnitude of the Ksp is proportional to how much solid Ca(IO3)2 will dissolve.
2. Similar to Figure 2, assemble a ring, clay triangle, and funel with filter in order to perform a gravity filtration. Fold a piece of filter paper in half, then fold it in half again. In order to fit the filter paper snugly inside the funnel, tear off a small part of the corner as shown in the video below.
Open the filter paper so that one side of the funnel is in contact with a single layer of the filter paper and the other side of the funnel is in contact with three layers of the filter paper.It seems to work best if the side with the tear has three layers. The paper should be in smooth contact with the funnel at the top, but not near the point. Squirting in water with a wash bottle sometimes helps to keep the filter paper in place.
3. Take the 250-mL beaker from Step 1 and pour as much supernatant liquid through the filter paper as possible, trying your best not to disturb the precipitate. Use a wash bottle filled with distilled water to wash down the sides of the beaker. Swirl the wash water with the solid and quickly pour the mixture on the filter paper. Rinse any remaining solid onto the filter paper with a small amount of water from your wash bottle.
4. Wash the solid Ca(IO3)2 on the filter paper with three small portions of distilled water.
5. Using a microspatula, transfer roughly 1/3 of the wet precipitate to one 100-mL beaker. Leave the left over precipitate on the filter paper. You will only use the precipitate on the filter paper if you mess up one of your trials.
6. Rinse your 100-mL graduated cylinder with distilled water and use it to add 80 mL of distilled water to the beaker containing the wet precipitate.
7. Acquire a stirring motor from the balance room and a stir bar from your TA.
Place the stir bar into the beaker containing your 80 mL of water and your Ca(IO3)2 precipitate.
8. Place your beaker on the stirring motor and turn the stirrer on to its lowest speed setting and allow it to stir in order to obtain a saturated solution.
While you are waiting for your solution to reach equilibrium complete the following problems in your lab notebook, which must be checked off by your TA before you continue. Note: the following problems were taken from Quiz I given during the past two quarters.
Example Problem 3.
The solubility product constant for Cu(IO3)2 is 1.44 x 10-7. What volume of 0.0520 M S2O32- would be required to titrate a 20.00 mL sample of a saturated solution of Cu(IO3)2?
Example Problem 4.
The solubility product constant for Sr(IO3)2 is 1.14 x 10-7. What volume of 0.0250 M S2O32- would be required to titrate a 20.00 mL sample of a saturated solution of Sr(IO3)2?
Example Problem 5.
A titration was performed to determine the solubility product constant (Ksp) of lead iodide. The titration is governed by the following reactions:
(a) Write out the Ksp expression for lead iodide.
(b) 25.00 mL of a saturated PbI2 solution was titrated with 0.0524 M Ce(NO3)4. The first step of the reaction sequence was performed and a starch indicator was added to the solution turning it blue. After 1.240 mL of the Ce(NO3)4 was added the end point was established as the blue color indicating the presence of I2 turned clear. From the results of this experiment, calculate the Ksp of lead iodide.
Once you get your TA to check off these example problems, you are finished with Day #1 of this lab. Write your name on the Erlenmeyer Flask containing your saturated solution with a Sharpie, place Parafilm over the top of the flask, and hand it to your TA.
You will start by filtering this solution at the beginning of the next lab period.
9. In order to ensure you have a saturated solution of Ca(IO3)2, the mixtures set aside from Step 8 must not contain any solid. Filter the saturated solution in the same manner as Step 2, but for this step use a clean, dry 125-mL Erlenmeyer flask to catch the filtrate. Any glassware, as well as the filer paper, must be dry in order to calculate an accurate concentration of IO3–.
10. Rinse a 10-mL pipet with two small portions of the saturated Ca(IO3)2 solution. You may discard any rinse solutions down the drain.
11. Pipet 10.00 mL of the saturated Ca(IO3)2 solution into a 250-mL Erlenmeyer flask. With a Sharpie, label this flask TITRATION FLASK.
In this experiment we will utilize the following set of oxidation/reduction reactions to determine the number of moles of IO3– present in 10.00 mL of a saturated Ca(IO3)2 solution.
At this point, you have a clear solution containing Ca2+(aq) and IO3–(aq). In a titration an indicator is often used and these indicators exhibit vivid color in solution. If we add excess I– to the solution (done in this experiemnt by adding solid KI) and excess acid (in this case 3 M HCl will work just fine), then all of the IO3–(aq) will be oxidized to I2(aq). We can verify that I2(aq) is present due to its brown color in solution.
As the second step of the reaction sequence indicates, whenever thiosulfate (S2O32–) is added to a solution of I2, the I2(aq) will be reduced to I–(aq). Whenever a small amount of I2(aq) is in the presence of a starch indicator, a dark blue complex will form. When the solution turns from blue (or brown) to colorless, we know that all of the I2(aq) has been reduced to I–(aq).
12. To your TITRATION FLASK, add approximately 50 mL of water using a graduated cylinder and 2 g of solid KI.
In addition, add about 10 mL of 3 M HCl and swirl. This is converting all of the IO3–(aq) to I2(aq), which will turn the solution brown.
This is consistent with the following equation:
13. Now we are ready to perform the titration with standardized 0.05XX M Na2S2O3. Find the 0.05XX M Na2S2O3 on the cart and record its concentration to four decimal places in your lab notebook. Use this solution to rinse a 25-mL buret twice, then perform three titrations.
14. In each of your titrations, fill the 25-mL buret with 0.05XX M Na2S2O3 and record your initial buret reading. Begin your titration and as the titration proceeds you will observe the brown color of the I2 solution lighten to a more yellow color.
When you observe a yellow color stop the titration and prepare a fresh starch solution by adding 0.6 grams of vitex starch to roughly 30 mL of distilled water in a separate 250-mL Erlenmeyer flask.
15. Add about 4 mL of the starch indicator solution to the TITRATION FLASK and in doing so, you should observe a blue (or dark brown) solution.
16. Continue the titration until you reach the endpoint, which is the moment the solution turns colorless.
This end point indicates all of the iodine has been used up, which is consistent with the following equation:
17. Record the final buret reading at the endpoint.
18. In your lab notebook, calculate the number of moles of S2O32−(aq) required in the titration.
At the endpoint of a titration, chemists can convert the moles of a substance that is known to the moles of something that is unknown. Most titrations performed up to this point involved neutralization reactions between acids and bases. At the end point of an acid/base titration, the moles of acid are equal to the moles of base. The chemical reaction in this experiment does not involve a neutralization reaction between an acid and a base. What we are observing is the following overall redox reaction:
IO3−(aq) + 6 H+(aq) + 6 S2O32−(aq) → I−(aq) + 3 S4O62−(aq) + 3 H2O(l)
From this overall chemical equation, we can come up with the most important relationship in this particular experiment. At the endpoint of the titration:
19. Calculate the molarity of IO3− for your first titration.
20. Refill the buret and perform two more titrations given in Steps 11 – 17.
21. If any of the three titrations do not agree to within 0.2 mL, perform another titration.
22. For each trial, calculate the molarity of IO3−, and calculate an average molarity of IO3− with the three trials that agree within 0.2 mL.
23. Using the stoichiometry of the ions in solution, calculate the concentration of Ca2+ from your average molarity of IO3−.
24. Calculate the Ksp for Ca(IO3)2.
25. Fill out your Report Sheet and have it checked over by your TA before you leave the second lab period.
26. Have your TA sign your notebook and after doing so they will give you a handout with all the formal lab report requirements. If you finish early it would be a good idea to start on your lab report.
27. There is a report question assignment for this lab on Mastering Chemistry. Check on-line and with your TA for the due date. No late assignments will be accepted.
28. The Mastering Chemistry pre-lab assignment for the Qualitative Analysis Lab is posted. Check on-line and with your TA for the due date. No late assignments will be accepted.
Where You Wanna Be
The Ksp values for most compounds are fairly straightforward to understand, but typically what a chemist wants to do is somehow manipulate the solubility of these compounds. What happens to the solubility when we increase the temperature? What happens when we change the pH? What happens when we add a common ion? In the Qualitative Analysis Experiments and in class in the next few weeks we will talk about how each of these scenarios can influence the solubility of a complex. But before you learn how to manipulate the solubility you have to have a thorough understanding of how a Ksp expression is set up and if you are given a molar solubility you should be able to calculate the Ksp. If you are given the Ksp you should also be able to calculate the molar solubility. The homework sets posted on Carmen will help you work through these types of problems. Also, as a hint, in order to study for the first exam example questions #3, #4, and #5 serve as a great benchmark to see if you truly understand the concepts behind this lab. Count on seeing an exam question based on how well you understand the concepts of those questions.
Tuesday, April 19, 2011
Applied Qualitative Analysis Lab
Tips for Success in the Qualitative Analysis Lab
Getting Started
Using Plastic Droppers
Separation of the Group I cations
HCl is used to separate silver and lead from the other Group II and Group III cations.
Isolating Group I
This precipitates out AgCl(s) and PbCl2(s) and leaves Cu2+, Bi3+, Ni2+, and Al3+ in solution. Decant the supernatant solution into a clean, dry test tube. Label the test tube containing the solid: Group I ppt and save it for later analysis.
Decanting Group II/III From Group I
Separating the Group II Cations From the Group III Cations
We will utilize the selective solubility of sulfides to separate the Group II and Group III cations. Before this separation can be performed, the initial solution must be placed in a casserole and heated to remove most of the water, resulting in a more concentrated solution of cations.
Placing Group II&III Cations in Casserole and Evaporating
The sulfide source will be thioacetamide, which produces a saturated solution of H2S when heated.
The solution from the casserole should be placed in a test tube and heated for at least 5 minutes. A dark precipitate will occur indicating the Group II cations have precipitated out of the solution.
Precipitating Group II
Decant the solution containing the Group III cations and now you will have three test tubes. One containing the Group I ppt, another containing the Group II precipitate, and a third containing the Group III solution.
Isolating Groups I, II, & III
Isolating and Identifying the Group I Cations
It is fairly simple to separate the Group I cations once they are isolated from the other Group II and Group III cations. The solubility of AgCl does not change appreciably with temperature, but PbCl2 is much more soluble at high temperatures. If water is added to the test tube containing the Group I ppt, and it is heated in a boiling water bath, the PbCl2 will dissolve and the AgCl will remain as a solid in the solution.
Isolating the Group I Cations
By placing the hot supernatant solution in a clean, dry test tube you have now isolated the Pb2+ from the AgCl(s).
To Confirm the presence of Pb2+ add a drop of potassium dichromate, which produces the CrO42- ion in solution. A yellow solid (PbCrO4) confirms the presence of lead in your sample.
The white solid can now be tested to see if it indeed contains silver. Adding concentrated NH3 to AgCl(s) will form a complex ion and dissolve the solid. [Note: If excess PbCl2 is present it will not dissolve in the presence of NH3 and some white solid might still be present.] To confirm silver is present add nitric acid until the solution is acidic and the white precipitate will re-form, confirming the presence of silver in your sample.
Isolating and Identifying the Group II Cations
The dark precipitate is made up of CuS and Bi2S3. The strategy here is to dissolve this solid using concentrated nitric acid to put both ions in solution.
Dissolving the Group II Solid
If you take a look at the formation constant table you will see that Cu2+ forms a complex ion with ammonia, while Bi3+ does not. This will allow us to separate the Cu2+ from Bi3+ by adding concentrated NH3. The copper ions will remain in solution as a complex ion and the bismuth ions will form a precipitate with hydoxide ions produced in the solution from the ammonia.
Separating the Group II Cations
The supernatant liquid can be decanted into a clean, dry test tube, which effectively separates the Cu2+ and the Bi3+. Cu(NH3)42+ is a distinct blue color and if your solution is blue, it confirms the presence of copper in your sample.
It's possible that if the Pb2+ was not properly removed in the Group I separation, that the white precipitate could be formed from Pb2+. To verify the white solid is Bi(OH)3 we will take advantage of some oxidation/reduction chemistry (we will discuss the details late in Chapter 20). Bismuth can be confirmed by adding NaOH and SnCl2. In the process, Bi3+ is reduced to Bi(s), which is a dark black color. The presence of this dark black solid confirms bismuth.
Confirming Bismuth
Isolating and Identifying the Group III Cations
The Group III solution set aside from the initial separations is acidic. The separation of Ni2+ and Al3+ is very similar to the separation of Ni2+ and Bi3+. Looking at the Kf table Ni2+ forms the Ni(NH3)6 complex ion with ammonia, where Al3+ does not form a complex ion with ammonia. There is one catch though. Al3+ is amphoteric. This means that when concentrated NH3 is added to the solution, producing OH- ions, the Al3+ can redissolve to form the Al(OH)4- complex ion. In order to prevent this complex ion from forming, a buffer must be added to keep the pH around 9. If these steps are properly executed, all of the Al3+ will precipitate out as Al(OH)3 and all of the Ni2+ will be present in solution as Ni(NH3)62+.
Group III Separation
The Ni2+ and Al3+ can be separated by decanting the supernatant solution into a clean, dry test tube.
To confirm the presence of nickel in your sample, add the dimethylglyoxime anion, which will bind to nickel to produce a red gel-like solid.
In order to test for aluminum, the Al(OH)3 solid must be dissolved back into solution. This can be done by adding nitric acid.
An aluminon dye is added to the solution to confirm aluminum is present in your sample.
Getting Started
Using Plastic Droppers
Separation of the Group I cations
HCl is used to separate silver and lead from the other Group II and Group III cations.
Isolating Group IThis precipitates out AgCl(s) and PbCl2(s) and leaves Cu2+, Bi3+, Ni2+, and Al3+ in solution. Decant the supernatant solution into a clean, dry test tube. Label the test tube containing the solid: Group I ppt and save it for later analysis.
Decanting Group II/III From Group ISeparating the Group II Cations From the Group III Cations
We will utilize the selective solubility of sulfides to separate the Group II and Group III cations. Before this separation can be performed, the initial solution must be placed in a casserole and heated to remove most of the water, resulting in a more concentrated solution of cations.
Placing Group II&III Cations in Casserole and EvaporatingThe sulfide source will be thioacetamide, which produces a saturated solution of H2S when heated.
The solution from the casserole should be placed in a test tube and heated for at least 5 minutes. A dark precipitate will occur indicating the Group II cations have precipitated out of the solution.
Precipitating Group IIDecant the solution containing the Group III cations and now you will have three test tubes. One containing the Group I ppt, another containing the Group II precipitate, and a third containing the Group III solution.
Isolating Groups I, II, & IIIIsolating and Identifying the Group I Cations
It is fairly simple to separate the Group I cations once they are isolated from the other Group II and Group III cations. The solubility of AgCl does not change appreciably with temperature, but PbCl2 is much more soluble at high temperatures. If water is added to the test tube containing the Group I ppt, and it is heated in a boiling water bath, the PbCl2 will dissolve and the AgCl will remain as a solid in the solution.
Isolating the Group I CationsBy placing the hot supernatant solution in a clean, dry test tube you have now isolated the Pb2+ from the AgCl(s).
To Confirm the presence of Pb2+ add a drop of potassium dichromate, which produces the CrO42- ion in solution. A yellow solid (PbCrO4) confirms the presence of lead in your sample.
The white solid can now be tested to see if it indeed contains silver. Adding concentrated NH3 to AgCl(s) will form a complex ion and dissolve the solid. [Note: If excess PbCl2 is present it will not dissolve in the presence of NH3 and some white solid might still be present.] To confirm silver is present add nitric acid until the solution is acidic and the white precipitate will re-form, confirming the presence of silver in your sample.
Isolating and Identifying the Group II Cations
The dark precipitate is made up of CuS and Bi2S3. The strategy here is to dissolve this solid using concentrated nitric acid to put both ions in solution.
Dissolving the Group II SolidIf you take a look at the formation constant table you will see that Cu2+ forms a complex ion with ammonia, while Bi3+ does not. This will allow us to separate the Cu2+ from Bi3+ by adding concentrated NH3. The copper ions will remain in solution as a complex ion and the bismuth ions will form a precipitate with hydoxide ions produced in the solution from the ammonia.
Separating the Group II CationsThe supernatant liquid can be decanted into a clean, dry test tube, which effectively separates the Cu2+ and the Bi3+. Cu(NH3)42+ is a distinct blue color and if your solution is blue, it confirms the presence of copper in your sample.
It's possible that if the Pb2+ was not properly removed in the Group I separation, that the white precipitate could be formed from Pb2+. To verify the white solid is Bi(OH)3 we will take advantage of some oxidation/reduction chemistry (we will discuss the details late in Chapter 20). Bismuth can be confirmed by adding NaOH and SnCl2. In the process, Bi3+ is reduced to Bi(s), which is a dark black color. The presence of this dark black solid confirms bismuth.
Confirming BismuthIsolating and Identifying the Group III Cations
The Group III solution set aside from the initial separations is acidic. The separation of Ni2+ and Al3+ is very similar to the separation of Ni2+ and Bi3+. Looking at the Kf table Ni2+ forms the Ni(NH3)6 complex ion with ammonia, where Al3+ does not form a complex ion with ammonia. There is one catch though. Al3+ is amphoteric. This means that when concentrated NH3 is added to the solution, producing OH- ions, the Al3+ can redissolve to form the Al(OH)4- complex ion. In order to prevent this complex ion from forming, a buffer must be added to keep the pH around 9. If these steps are properly executed, all of the Al3+ will precipitate out as Al(OH)3 and all of the Ni2+ will be present in solution as Ni(NH3)62+.
Group III SeparationThe Ni2+ and Al3+ can be separated by decanting the supernatant solution into a clean, dry test tube.
To confirm the presence of nickel in your sample, add the dimethylglyoxime anion, which will bind to nickel to produce a red gel-like solid.
In order to test for aluminum, the Al(OH)3 solid must be dissolved back into solution. This can be done by adding nitric acid.
An aluminon dye is added to the solution to confirm aluminum is present in your sample.
Thursday, April 7, 2011
Lecture #5, Wednesday, April 6th
Videos for this lecture will be posted sometime next week.
With the lectures notes from today you should be able to complete the following problems:
Solubility of "Insoluble" Salts
The Common-Ion Effect
Common-Ion Effect on Solubility for Lead Thiocyanate
Common-Ion Effect on Solubility for a Metal Hydroxide
*Solubility Quiz Question #4
*Solubility Quiz Question #5
The Effect of Acid on Solubility
Effect of pH on Solubility
Acid Rain: Effect on Solubility of Calcium Carbonate
*Solubility Quiz Question #6
*Solubility Quiz Question #7
*Solubility Quiz Question #8
*Solubility Quiz Question #9
Questions 24-32 from the Solubility homework set posted on Carmen
If you want to work ahead, we are going to cover problems involving complex ions next lecture. The following problems should be doable after lecture on Friday:
Solubility of Zinc Hydroxide in Basic Solution
Cyanide Poisoning
*Solubility Quiz Question #10
*Solubility Quiz Question #11
*Solubility Quiz Question #12
*Solubility Quiz Question #13
Questions 33-46 from the Solubility homework set posted on Carmen
With the lectures notes from today you should be able to complete the following problems:
Solubility of "Insoluble" Salts
The Common-Ion Effect
Common-Ion Effect on Solubility for Lead Thiocyanate
Common-Ion Effect on Solubility for a Metal Hydroxide
*Solubility Quiz Question #4
*Solubility Quiz Question #5
The Effect of Acid on Solubility
Effect of pH on Solubility
Acid Rain: Effect on Solubility of Calcium Carbonate
*Solubility Quiz Question #6
*Solubility Quiz Question #7
*Solubility Quiz Question #8
*Solubility Quiz Question #9
Questions 24-32 from the Solubility homework set posted on Carmen
If you want to work ahead, we are going to cover problems involving complex ions next lecture. The following problems should be doable after lecture on Friday:
Solubility of Zinc Hydroxide in Basic Solution
Cyanide Poisoning
*Solubility Quiz Question #10
*Solubility Quiz Question #11
*Solubility Quiz Question #12
*Solubility Quiz Question #13
Questions 33-46 from the Solubility homework set posted on Carmen
Lecture #4, Monday, April 4th
I started lecture discussing the Applied Qualitative Analysis Lab and went into detail about the qualitative analysis schemes used to isolate and identify metal cations in solution. We separate cations into groups based on their solubility characteristics.
Applied Qualitative Analysis Scheme
Once the groups are separated, a flow chart is used to organize the separations within each group.
Group I Separations
We are trying to emphasize they reasons why we use certain chemicals and which concentrations are best used to facilitate the separations.
Why does our HCl need to be cold and dilute?
In the qualitative analysis experiment you will need to isolate and identify various cations in solution. The Group I isolation and confirmations are discussed in detail.
Group I Analysis
When you are performing each step in the lab, be sure to consider how the common ion effect, the pH of the solution, complex ion formation, and amphoteric effects influence the solubility.
With the lectures notes from today you should be able to complete the following problems:
Qualitative Analysis of Metal Cations
Questions 19-23 from the Solubility homework set posted on Carmen
*Solubility Quiz Question #14
*Solubility Quiz Question #15
Applied Qualitative Analysis Scheme
Once the groups are separated, a flow chart is used to organize the separations within each group.
Group I Separations
We are trying to emphasize they reasons why we use certain chemicals and which concentrations are best used to facilitate the separations.
Why does our HCl need to be cold and dilute?
In the qualitative analysis experiment you will need to isolate and identify various cations in solution. The Group I isolation and confirmations are discussed in detail.
Group I Analysis
When you are performing each step in the lab, be sure to consider how the common ion effect, the pH of the solution, complex ion formation, and amphoteric effects influence the solubility.
With the lectures notes from today you should be able to complete the following problems:
Qualitative Analysis of Metal Cations
Questions 19-23 from the Solubility homework set posted on Carmen
*Solubility Quiz Question #14
*Solubility Quiz Question #15
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