Chemistry 130 Dr. John F. C. Turner 409 Buehler Hall

Chemistry 130 Dr. John F. C. Turner 409 Buehler Hall
Chemistry 130

Course coverage Chemistry 130 Ch. 12: Properties of Solutions
Ch. 13: Chemical Equilibrium Ch. 15: Acid-base equilibria Ch. 16: Solution Equilibria Ch. 17: Thermodynamics Spring Break Ch. 18: Electrochemistry Ch. 19: Nuclear Chemistry Ch. 20 – 22: Descriptive Chemistry of the elements Ch. 23 – 24: Organic, biological, medicinal and materials chemistry Chemistry 130

Course mechanics Chemistry 130 Grading Scale Lecture course: 80%
6 quizzes: ~25% 3 exams: ~25% Final Examination: 30% Laboratory course: % Grade Cut-offs: A: 80%-100% B+: 78%-79.9% B: 70%-77.9% C+: 68%-69.9% C: 60%-67.9% D: 50%-59.9% F: 0%-49.9% There is no curve; grading is on an absolute scale. There are no dropped quizzes or examinations. Chemistry 130

Course mechanics Chemistry 130 Additional Help
The tutorial center, in Buehler Hall, opposite the General Chemistry office on the 5th floor, is staffed from 9.00 a.m. until 5 p.m. daily during the semester. All Chem 130 students are invited to the three review sessions on Monday, Tuesday and Wednesday 7 p.m. – 10 p.m. Buehler 511. Office hours are held on Monday from 2 – 5 p.m. at Buehler 409. Chemistry 130

Course overview and context
The first half of the course will cover: Properties of Solutions Solution Equilibria Chemical Equilibrium Thermodynamics Acid-base equilibria Electrochemistry The key to understanding all of these areas is the study of thermodynamics Chemistry 130

Properties of Solutions Solution Equilibria Chemical Equilibrium Thermodynamics Acid-base equilibria Electrochemistry All of these topics are very important: On a practical level: “Life takes place in solution” - the human body is ~70% water and fluids in the body are solutions Many technologically important materials are solution: Steel Palladium-hydrogen Chemistry 130

Steel Steel is a complicated system, the majority components of which are iron and carbon. Phase diagram for Fe-Fe3C There are many phases and transitions between phases, with different compositions. Transformation temperatures are also different between different phases. Chemistry 130

Palladium-hydrogen Chemistry 130
Palladium will absorb very large quantities of H2 to form a gas-solid solution. This is a very important process for the storage of hydrogen and the so-called 'hydrogen economy' (nothing to do with cold fusion ) H 2 Chemistry 130

Science is also a powerful description of the physical world. It is a fundamentally important cultural construction. It is new and it is always wrong and it is always right simultaneously. The Scientific Method A natural law is a sum of human experience in light of experiment, theory and hypothesis It is predictive and testable in all aspects The material in Chem 130 addresses one fundamental property of the description of the physical world – change and the nature of change Chemistry 130

Chemistry is the unique science that contains all the elements of a mathematically rigorous science. It also contain elements of art, in terms of creativity Chemists change the local structure of the universe, in principle at will. The definition of chemistry: “That branch of physical science and research, which deals with the several elementary substances, or forms of matter, of which all bodies are composed, the laws that regulate the combination of these elements in the formation of compound bodies, and the various phenomena that accompany their exposure to diverse physical conditions.” Chemistry 130

Change is an essential aspect of chemistry and chemical change takes place within those natural laws that are well understood. Chemical change is one of the fundamental mechanisms that limit or enable the compass of our lives. Chemistry 130 is partially about the restrictions and allowances that govern chemical change. Chemistry 130

Many changes are spontaneous; many are not. Some hypothetical changes have never been observed. Why? Chemical change involves a change in the number of particles that we observe: In both of these reaction, we see the reactants disappear and the products appear. What are the rules that govern the destruction and creation, i.e. the chemical interchange of forms of matter? H 2 C= CH 2  Br 2 Br CH 2 CH 2 Br NaOH 𝑎𝑞  HCl 𝑎𝑞  NaCl 𝑎𝑞  H 2 O 𝑙 Chemistry 130

In Chem 120, we focused on heat and heat flows that are observed in the course of a chemical reaction. We described these heat flows as either The internal energy, U, at constant volume or The enthalpy, H, at constant pressure Both of these extensive properties are descriptions of the system only Chemistry 130

However, even though these chemical changes are occurring locally, after we have performed these reactions, there is a little less bromine or sodium hydroxide and a little more 1,2,dibromoethane and sodium chloride in the universe than before. We must consider the universe, from a local perspective (in order not to be too grandiose.....) H 2 C= CH 2  Br 2 Br CH 2 CH 2 Br NaOH 𝑎𝑞  HCl 𝑎𝑞  NaCl 𝑎𝑞  H 2 O 𝑙 Chemistry 130

Chemistry 130

Chemistry 130 Physical Properties of Solutions Dr. John F. C. Turner
409 Buehler Hall Chemistry 130

Solutions, solvents and solutes
Solutions are extremely important chemical systems in which many reactions and physical phenomena take place. Atmospheric chemistry often, but not always, takes place in solution 𝐹𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛 O 2𝑔 ℎ2 O 𝑔 O 2𝑔  O 𝑔  O 3𝑔 𝐷𝑒𝑠𝑡𝑟𝑢𝑐𝑡𝑖𝑜𝑛 O 3𝑔  O 𝑔 2 O 2𝑔 Chemistry 130

Solutions, solvents and solutes
A solution is defined as a homogenous mixture of two or more components The major component is called the solvent The other, minor components are called the solutes Solutions can occur in the solid state, the liquid state or the gaseous state The pure components can be solid, liquid or gas prior to forming the solution Chemistry 130

Example 1: Sodium in liquid ammonia
Solute Solvent Solution low high concentration concentration low conductivity metallic conductivity Na 𝑠 NH 3𝑙 Chemistry 130

Example 2: Potassium permanganate in water
Solute Solvent Solution Very powerful oxidant KMnO 4𝑠 H 2 O 𝑙 Chemistry 130

Energetics of solution
The dissolution of Na in liquid ammonia and KMnO4 in water both involve solids that are strongly bonded. For Na, m.p. /°C: ~97 b.p. /°C: ~883 resulting in a liquid range of 785 K KMnO4 decomposes before it melts The forces between particles in both Na and KmnO4 are strong So why do they dissolve? Chemistry 130

Increasing polarizability
Energetics of solution: Intermolecular forces When considering the solubility or insolubility of a potential solute in a solvent, the first thing to consider are the intermolecular forces There are forces between all atoms in a condensed state i.e. in the liquid or solid state. A perfect gas has no forces between particles and cannot be liquified or solidified; however, all gases can be condensed at some temperature He Ne Ar Kr Xe Rn m.p./ °C [-272.2] b.p. / °C Increasing polarizability Chemistry 130

Energetics of solution: Intermolecular forces
All molecules exert weak attractions on one another due to the mutual attraction of nuclei and electrons. These attractive forces are only significant at very short distances. At such small distances the intermolecular repulsion of the electrons on different atoms is also significant. The electrons contained in all atoms and molecules can be perturbed by an electric field, with greater or lesser ease. This property is called polarizability. The electron cloud around an atom or molecule can give an instantaneous dipole any time that the electrons are not distributed perfectly symmetrically. Such a dipole can induce dipoles in other species nearby. Chemistry 130

The polarizability of a molecule depends on the size of the atoms and how strongly the electrons are bound within the atom. At any instant, the electron density can fluctuate and the distribution can become instantaneously asymmetric. This instantaneous and transient dipole can induce a similar dipole in a neighboring atom or molecule, resulting in a weak attraction Chemistry 130

The attractive forces between an instantaneous dipole and an induced dipole are called London dispersion forces after the physicist Fritz London. These forces are stronger for more polarizable species. Some molecules have a permanent dipole because of differences in electronegativities among the atoms. Such molecules experience dipole-dipole forces. All molecules experience dispersion forces and induced dipoles, and polar molecules also experience dipole-dipole forces. Chemistry 130

The intermolecular forces that we have seen London dispersion forces induced dipole interactions dipole-dipole interactions have a strong effect on the boiling points of liquids. Chemistry 130

Another type of intermolecular force is the hydrogen bonding interaction. Hydrogen bonding is an additional type of bonding interaction that requires a hydrogen atom on one molecule and a source of electron density on another molecule, usually a lone pair. Hydrogen bonding can be intramolecular as well as intermolecular. It can have a profound effect on the properties of a material when it is present. Chemistry 130

Energetics of solution: Hydrogen bonding
Boiling points of covalent hydrides Chemistry 130

The second-row hydrides NH3, H2O, and HF exhibit much higher boiling points that would be expected based on their molecular weights. Strong hydrogen bonding between the molecules is responsible for the large liquid phase range of these compounds. . CH4 has a low boiling point because it has no lone pairs to form strong hydrogen bonds. Chemistry 130

Chemistry 130

(a) CS2 (b) CH3OH (c) CH3CH2OH (d) H2O (d) C6H5NH2 Chemistry 130

In the gas phase, we can adjust the perfect gas equation to account for these intermolecular forces in a partially empirical manner; the resulting equation is termed the Van der Waals equation: which describes the behavior of a non-ideal gas and includes both attractive and repulsive components. We cannot write a similarly simple equation for the liquid state 𝑃𝑉=𝑛𝑅𝑇 𝑃𝑎  𝑛 𝑉  2 𝑉−𝑏𝑛 =𝑛𝑅𝑇 Chemistry 130

If we imagine two materials, A and B, which are non-ideal, then in the pure material, there will be intermolecular forces between both sets of particles In A In B but until they are mixed, there are clearly no forces between A and B When they mix, the forces present in the mixture will be and i.e. some of the A and B particles interact F 𝐴𝐴 F 𝐵𝐵 F 𝐴𝐴 F 𝐵𝐵 F 𝐵𝐴 𝑎𝑛𝑑 𝐹 𝐴𝐵 Chemistry 130

It is the balance of forces that, in part, controls the solubility properties of solvents and solutes. If the forces between particles are exactly and precisely equal, i.e. then there is no change in the enthalpy or the volume and this is termed an ideal solution. An ideal solution is one where the intermolecular forces between solute particles, solvent particles and solute-solvent particles are identical. In this case, F 𝐵𝐴 = F 𝐴𝐵 = F 𝐴𝐴 = F 𝐵𝐵  𝐻 𝑚𝑖𝑥 =0 𝑉 𝑚𝑖𝑥 =0  𝐇 𝑚𝑖𝑥 =𝟎 𝐕 𝑚𝑖𝑥 =𝟎 Chemistry 130

Ideal solutions are extremely rare, if they exist at all. For non-ideal solutions, which are those where the forces between particles are different in the pure materials and the solution, there are three regimes. A. The forces in the pure materials are very much stronger than those is the solution. In this case, the enthalpy of mixing is prohibitively high and endothermic and the solution does not form spontaneously B. The forces in the pure materials are very much weaker than in the solution In this case, the enthalpy of mixing is large and negative and therefore exothermic. The solution forms spontaneously Chemistry 130

The third regime is where the forces in the pure materials are only slightly stronger than in the solution. In this case, the solution forms but the reaction in endothermic but still spontaneous. This is an example that shows us that consideration of enthalpy is not sufficient to explain the formation of the solution. It is analogous to the expansion of a perfect gas; as there are no forces between particles in a perfect gas, then the energy of the system is independent of the separation of the particles. The energy is then independent of the volume and there is no reason, based on the internal energy of the system, for an ideal gas to expand Chemistry 130

Chemistry 130

Chemistry 130 Physical Properties of Solutions Dr. John F. C. Turner
409 Buehler Hall Chemistry 130

Prior to the formation of a solution containing A and B, one of which is the solute and one of which is the solvent, intermolecular forces exist in A and B Forces in A: Forces in B: After mixing and the formation of the solution, the forces that are present are: Forces in solution: F 𝐴𝐴 F 𝐵𝐵 𝐅 𝐵𝐴 , 𝐅 𝐴𝐵 , 𝐅 𝐴𝐴 and 𝐅 𝐵𝐵 Chemistry 130

When a solution is formed, the relative magnitudes of the respective forces is important. Using a Hess cycle, is negative, whereas , the enthalpy of formation of solution from the gas phase, is positive. It is the balance between these two factors that is important  𝐻 𝑣𝑎𝑝 𝐵 𝐴 𝑔  𝐵 𝑙 𝐴 𝑔  𝐵 𝑔  𝐻 𝑣𝑎𝑝 𝐴  𝐻 𝑓,𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛  𝐻 𝑚𝑖𝑥 𝐴 𝑙  𝐵 𝑙 𝐴 𝐵  𝐻 𝑚𝑖𝑥 = 𝐻 𝑣𝑎𝑝 𝐴  𝐻 𝑣𝑎𝑝 𝐵  𝐻 𝑓,𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛  𝐻 𝑣𝑎𝑝 𝐴𝑜𝑟𝐵  𝐻 𝑓,𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 Chemistry 130

is the enthalpy of vaporization for either A or B and is the energy required to form a gas at infinite separation from liquid or solid A or B.. Clearly, we must supply enough energy to overcome the intermolecular forces that are present in either A or B  𝐻 𝑣𝑎𝑝 𝐴𝑜𝑟𝐵 is the enthalpy of formation of the solution from A and B and therefore must be related to the energy released when the intermolecular interactions between A and B occur.  𝐻 𝑓,𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛  𝐻 𝑣𝑎𝑝 𝐵 𝐴 𝑔  𝐵 𝑙 𝐴 𝑔  𝐵 𝑔  𝐻 𝑣𝑎𝑝 𝐴  𝐻 𝑓,𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛  𝐻 𝑚𝑖𝑥 𝐴 𝑙  𝐵 𝑙 𝐴 𝐵  𝐻 𝑚𝑖𝑥 = 𝐻 𝑣𝑎𝑝 𝐴  𝐻 𝑣𝑎𝑝 𝐵  𝐻 𝑓,𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 Chemistry 130

Energetics of ionic solutions
Many ionic solids will dissolve in water to form solutions that will conduct electricity. We write this type of solution formation as In this reaction, solid sodium fluoride is reacting with water and both sodium and fluoride ions become solvated, denoted by '(aq)' as the state. Both water and NaF are polar but it is the hydration of the ions that provides the energetic counterbalance to the large lattice energy of NaF. NaF 𝑠  H 2 O 𝑙  Na 𝑎𝑞 +  Cl 𝑎𝑞 −  𝐻 𝑠𝑜𝑙 NaF =0.91 kJmol −1  𝐻 𝑓 NaF =−576.6 kJmol −1 𝑈 𝑙𝑎𝑡𝑡𝑖𝑐𝑒 NaF =915 kJmol −1 Chemistry 130

The formation of bonds between the metal ion and the water molecules is sometimes so strong that the complex ion can be isolated. Solutions of NiCl2 in water contain the hexaquo ion Most cations in water are solvated by between 4 and 6 water molecules; larger cation by up to 12 water molecules. These bonds are strong and help to offset the lattice energy of an ionic compound, allowing it to dissolve. NiCl 2𝑠  H 2 O 𝑙  Ni 𝑎𝑞 2+ 2 Cl 𝑎𝑞 − Ni  OH 2  Ni  OH 2  Ni  OH 2  Ni  OH 2  Chemistry 130

The Hess cycle for the dissolution of an ionic compound in solution is easy to write down: NiCl 2𝑠  H 2 O 𝑙  Ni 𝑎𝑞 2+ 2 Cl 𝑎𝑞 − Ni 𝑔 2+ 2 Cl 𝑔 −  H 2 O 𝑙  𝐻 ℎ𝑦𝑑 Ni 2+ 2 𝐻 ℎ𝑦𝑑 Cl − 𝑈 𝑙𝑎𝑡𝑡𝑖𝑐𝑒 Ni  OH 2  2 Cl 𝑎𝑞 − NiCl 2𝑠  H 2 O 𝑙 Ni  OH 2  Ni  OH 2  Chemistry 130

Miscibility and partial solubility
Most materials are only partially soluble. Systems that form a homogenous mixture over all compositions are termed miscible Examples of fully miscible materials are Water-EtOH Ne-He Benzene-hexane Cu-Zn Chemistry 130

Most materials are only partially soluble. When a system has attained the maximum possible solubility at a given temperature, we say that it is saturated. In a saturated solution, the solution is in contact with the solid and there is a dynamic equilibrium that exists between the solid solute and the dissolved solute. Although there is no net dissolution of the solute in a saturated system, the solute is precipitating and dissolving at the same rate. There is chemical exchange between the dissolved solute and the solid solute, although the net change is zero – no more will dissolve. Chemistry 130

Solubility also varies with temperature. Almost all substances become increasingly soluble with temperature. When a solution that is saturated at high temperature is allowed to cool, the solute will crystallize as the saturated concentration at low temperature is less than at high temperature. This is a common method for purifying substances, as usually, only the pure material will crystallize. Chemistry 130

Solubility of gases Chemistry 130
The solubility of gases depends on temperature and the gas pressure of the system. The temperature dependence of the concentration of a gas is given by where is the concentration at temperature and is the enthalpy of solution for the gas. The pressure dependence of the solubility is simpler: where is the pressure of the gas and k is a constant. ln 𝑐 2 𝑐 1 =−  𝐻 𝑠𝑜𝑙 𝑅  1 𝑇 2 − 1 𝑇 1  𝑐 1 𝑇 1  𝐻 𝑠𝑜𝑙 𝑐=𝑘 𝑃 𝑔𝑎𝑠 𝑃 𝑔𝑎𝑠 Chemistry 130

Vapor pressure Chemistry 130
All pure liquids are in equilibrium with the vapor; the boiling point of a liquid is the temperature at which the vapor pressure is equal to the external pressure on the liquid. The solvent in a solution also possesses a well-defined vapor pressure that is directly related to the composition of the solution by Raoult's Law. Here, is the pressure of the solvent in the solution is the pressure of the pure solvent. is the mole fraction, which is the ratio of the number of moles of solvent to the total number of moles of all components. 𝑃 𝑠𝑜𝑙𝑣 = 𝑥 𝑠𝑜𝑙𝑣 𝑃 𝑠𝑜𝑙𝑣 o 𝑃 𝑠𝑜𝑙𝑣 𝑃 𝑠𝑜𝑙𝑣 o 𝑥 𝑠𝑜𝑙𝑣 𝑥 𝑖 = 𝑛 𝑖 𝑗 𝑛 𝑗 Chemistry 130

Vapor pressure Chemistry 130
Raoult's Law applies to any volatile component in the system and applies strictly only to ideal solutions. For strongly non-ideal solutions, it can be used when the solution is very dilute. For two liquids, A and B, that have vapor pressures and when pure, the pressure above the liquid will be given by As long as , then one component will be present in greater amount than the other. Repeated condensation of the vapor, followed by boiling will increase this fraction and eventually, the two components can be separated. This is the basis for distillation. 𝑃 𝑠𝑜𝑙𝑣 = 𝑥 𝑠𝑜𝑙𝑣 𝑃 𝑠𝑜𝑙𝑣 o 𝑃 𝐴 o 𝑃 𝐵 o 𝑃= 𝑥 𝐴 𝑃 𝐴 o  𝑥 𝐵 𝑃 𝐵 o 𝑃 𝐴 o ≠ 𝑃 𝐵 o Chemistry 130

Raoult's Law example Chemistry 130
Chloroform and carbon tetrachloride are both miscible and can be considered to be ideal. What is the pressure and the composition of the vapor of a solution consisting of 1 mole of carbon tetrachloride and 2 moles of chloroform? 𝑃 CCl 4 o =16.5kPa 𝑃 HCCl 3 o =28.5kPa Chemistry 130

Chloroform and carbon tetrachloride are both miscible and can be considered to be ideal. What is the pressure and the composition of the vapor of a solution consisting of 1 mole of carbon tetrachloride and 2 moles of chloroform? 1. Calculate the mole fractions: 𝑃 CCl 4 o =16.5kPa 𝑃 HCCl 3 o =28.5kPa 𝑥 𝑖 = 𝑛 𝑖 𝑗 𝑛 𝑗 𝑗 𝑛 𝑗 = 𝑛 HCCl 3  𝑛 CCl 4 =12=3 𝑛 HCCl 3 = 2 3 ; 𝑛 CCl 4 = 1 3 Chemistry 130

Chloroform and carbon tetrachloride are both miscible and can be considered to be ideal. What is the pressure and the composition of the vapor of a solution consisting of 1 mole of carbon tetrachloride and 2 moles of chloroform? 2. Calculate the individual pressures, using Raoult's Law 𝑃 CCl 4 o =16.5kPa 𝑃 HCCl 3 o =28.5kPa 𝑃= 𝑥 𝐴 𝑃 𝐴 o  𝑥 𝐵 𝑃 𝐵 o 𝑃 CCl 4 = 𝑥 CCl 4 𝑃 CCl 4 o = 1 3 ×16.5kPa=5.5kPa 𝑃 HCCl 3 = 𝑥 HCCl 3 𝑃 HCCl 3 o = 2 3 ×28.5kPa=19kPa Chemistry 130

Chloroform and carbon tetrachloride are both miscible and can be considered to be ideal. What is the pressure and the composition of the vapor of a solution consisting of 1 mole of carbon tetrachloride and 2 moles of chloroform? 3. Calculate the total pressure Answer: the pressure of the solution = 24.5 kPa 𝑃 CCl 4 o =16.5kbar 𝑃 HCCl 3 o =28.5kbar 𝑃= 𝑥 𝐴 𝑃 𝐴 o  𝑥 𝐵 𝑃 𝐵 o 𝑃 CCl 4 = 𝑥 CCl 4 𝑃 CCl 4 o =5.5kPa 𝑃 HCCl 3 = 𝑥 HCCl 3 𝑃 HCCl 3 o =19kPa 𝑃 𝑡𝑜𝑡𝑎𝑙 = 𝑃 HCCl 3  𝑃 HCCl 3 =5.519kbar=24.5kPa Chemistry 130

Chloroform and carbon tetrachloride are both miscible and can be considered to be ideal. What is the pressure and the composition of the vapor of a solution consisting of 1 mole of carbon tetrachloride and 2 moles of chloroform? 4. Calculate the composition of the vapor 𝑃 CCl 4 o =16.5kPa 𝑃 HCCl 3 o =28.5kPa 𝑃= 𝑥 𝐴 𝑃 𝐴 o  𝑥 𝐵 𝑃 𝐵 o 𝐼𝑛 𝑡ℎ𝑒 𝑣𝑎𝑝𝑜𝑟 𝑝ℎ𝑎𝑠𝑒 𝑃 𝑡𝑜𝑡𝑎𝑙 =24.5kPa 𝑥 CCl 4 = 𝑃 CCl 4 𝑃 total = = 𝑥 HCCl 3 = 𝑃 HCCl 3 𝑃 total = =0.776 Chemistry 130

In this example, if the vapor were condensed, the resulting liquid would have the composition and further vaporization and condensation cycles would enrich it further in the more volatile component, in this case Eventually, a complete separation would be achieved and we would have accomplished a fractional distillation. 𝑥 CCl 4 = 𝑥 HCCl 3 =0.776 HCCl 3 Chemistry 130

Colligative properties
Just as the vapor pressure of a solution deviates from that of the pure solvent by virtue of the presence of the solvent, so other properties of the solution also are changed when a solute is present. In particular, the freezing point and the boiling points are depressed and elevated respectively. Depression of the freezing point and elevation of the boiling point are known as colligative properties; they depend only on the number of particles of the solute and not their nature. Chemistry 130

In a dilute solution, the depression of the freezing point of the solution is given by and the elevation of the boiling point by In both cases, m is the molality of the solution – the concentration in mol kg-1  𝑇 𝑓 = 𝑇 𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 − 𝑇 𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 =− 𝐾 𝑓 m  𝑇 𝑏 = 𝑇 𝑏 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 − 𝑇 𝑏 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 = 𝐾 𝑏 m 𝐾 𝑓 𝑖𝑠 𝑡ℎ𝑒 𝑐𝑟𝑦𝑜𝑠𝑐𝑜𝑝𝑖𝑐 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝐾 𝑏 𝑖𝑠 𝑡ℎ𝑒 𝑒𝑏𝑢𝑙𝑙𝑖𝑜𝑠𝑐𝑜𝑝𝑖𝑐 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 Chemistry 130

If the ebullioscopic or cryoscopic constant is known, the colligative properties of a solution are a good way of determining the number of particles present in a solution or the formula mass.  𝑇 𝑓 = 𝑇 𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 − 𝑇 𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 =− 𝐾 𝑓 m  𝑇 𝑏 = 𝑇 𝑏 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 − 𝑇 𝑏 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 = 𝐾 𝑏 m 𝐾 𝑓 𝑖𝑠 𝑡ℎ𝑒 𝑐𝑟𝑦𝑜𝑠𝑐𝑜𝑝𝑖𝑐 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝐾 𝑏 𝑖𝑠 𝑡ℎ𝑒 𝑒𝑏𝑢𝑙𝑙𝑖𝑜𝑠𝑐𝑜𝑝𝑖𝑐 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 Chemistry 130

Chemistry 130 Dr. John F. C. Turner 409 Buehler Hall

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Chemistry 130 Dr. John F. C. Turner 409 Buehler Hall

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