ch. 13: solutions - san diego miramar...
TRANSCRIPT
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Ch. 13: Solutions
Dr. Namphol Sinkaset Chem 201: General Chemistry II
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I. Chapter Outline
I. Introduction II. Types of Solutions and Solubility III. Energetics of Solution Formation IV. Solution Equilibrium V. Solution Concentration VI. Colligative Properties
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I. Solutions Everywhere
• solution: homogeneous mixture of 2 or more substances or components
• solute: substance present in smaller amount • solvent: substance present in greater amount
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II. Types of Solutions
• Aqueous solutions are most common. • solubility: amount of a substance that will
dissolve in given amount of solvent • Solubility depends on IM forces and entropy.
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III. Why Do Solutions Form?
• What is the reason most things happen?*
• Is the PE lowered when the barrier is removed?
• No! So why does this gas/gas solution form?
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III. Entropy • In addition to lowering PE, nature’s
desire to disperse energy is a reason why we observe things happening.
• entropy: measure of energy of randomization or energy dispersal in a system
• KE originally confined now spread out over a larger volume.
• Entropy also explains heat flow. • Much more on entropy later…
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III. What About IM Forces? • In most solutions, PE is a factor. • In solution, new particle-particle interactions form. • IM forces can help or hinder solution formation.
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III. Solution Interactions
• Remember that “like dissolves like.” • If liquids are soluble in all proportions, then
they are said to be miscible.
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III. When Will a Solution Form?
• If solvent molecules and solute molecules interact more strongly with their own kind, a solution tends not to form.
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III. Sample Problem 13.1
• Explain the trends in solubilities in terms of intermolecular forces.
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III. Energetics of Solution Formation
• Although not a reaction, we see energy changes when a solution forms.
• The energy change observed depends on relative strengths of IM forces.
• The enthalpy of solution, ΔHsoln, can be exo or endo and can be estimated using a Hess’s Law type of calculation.
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III. Steps of Solution Formation
1) Separating solute into its constituent particles, ΔHsolute.
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III. Steps of Solution Formation
2) Separating solvent particles to make room for solute particles, ΔHsolvent.
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III. Steps of Solution Formation
3) Mixing solute particles with solvent particles, ΔHmix.
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III. Steps of Solution Formation
• Hess’s Law allows us to break up a process into a series of steps.
• Therefore, the 3-step path for solution formation yields the following equation.*
ΔHsoln = ΔHsolute + ΔHsolvent + ΔHmix
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III. Hess’s Law Applied
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III. Aqueous Ionic Solutions
• Many solutions are ionic, water-based. • For these, ΔHsolvent and ΔHmix can be combined
into ΔHhydration. • heat of hydration: enthalpy change when 1
mole of gaseous solute ions are dissolved in water
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III. Heat of Hydration
• Why is ΔHhydration always very exo for ionic compounds?*
• Compare relative strengths of ΔHsolvent and ΔHmix.
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III. KF Aqueous Solution
• For ionic compounds, ΔHsolute = -ΔHlattice. • Thus, whether or not formation of an aqueous
ionic solution is exo or endo depends on the relative values of ΔHsolute and ΔHhydration.
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IV. Solution Equilibrium
• Initially, when solid is placed in a solvent, solid particles rapidly go into solution.
• Eventually, there will be excess solute particles in solution, and they will begin to redeposit.
• At dynamic equilibrium, the rates of dissolution and deposition are equal.
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IV. Dissolution of NaCl
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IV. Solution Terminology
• Equilibrium between solid and solvated particles can only occur if solution is saturated.
• If there’s less than this equilibrium amount, the solution is unsaturated.
• Some solutions can be made to be supersaturated.
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IV. Solubility of Solids
• In general, the solubility of solids increases with temperature.
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IV. Solubility of Gases
• Gas solubility in water depends on temperature and external pressure.
• Gas solubility decreases with temp. • Gas solubility increases with higher
external pressure.
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IV. Henry’s Law
• Dependence of gas solubility on external pressure is described by Henry’s Law.
• Henry’s constant (kH) usually in M/atm.
𝑆𝑆𝑔𝑔𝑔𝑔𝑔𝑔 = 𝑘𝑘𝐻𝐻𝑃𝑃𝑔𝑔𝑔𝑔𝑔𝑔
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IV. Henry’s Constants
• Why is the value for ammonia (NH3) so high?*
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V. Solution Concentration
• One of the most important aspects about a solution is its concentration.
• There are many different units of concentration, and we need to be able to convert between all of them.
• Which unit we use depends on what type of problem we are solving.
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V. Units of Concentration
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V. Sample Problems 13.2 1) How many grams of KNO3 are needed to
make 200.0 mL of a 0.025 M solution? 2) Calculate the weight fraction of NaCl in a
solution consisting of 12.5 g NaCl and 75.0 g H2O.
3) Calculate the number of grams of CH3OH needed to make a 0.250 m solution w/ 2000 g of H2O.
4) What’s the molality of a 37.0% HCl solution? 5) A 40.0% HBr solution has a density of 1.38
g/mL. What is molarity?
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VI. Colligative Properties
• Salt is added to ice in an ice cream maker. • Icy roads are salted in winter. • Antifreeze is mixed with water. • These are practical uses of colligative
properties. • colligative properties: properties that
depend on number of dissolved particles, not the type particle.
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VI. Colligative Properties
• We will look at 3 colligative properties: Vapor pressure lowering Freezing point depression Boiling point elevation Osmotic pressure
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VI. Vapor Pressure
• Vapor pressure is the pressure of a gas in dynamic equilibrium w/ its liquid.
• It’s a measure of how many molecules go into the gas phase.
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VI. Vapor Pressure of Solutions
• All liquid solutions of nonvolatile solutes have lower vapor pressures than the pure solvents.
• Why?*
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VI. Striking Example
• Why is the volume on the right increasing?*
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VI. Raoult’s Law
• The vapor pressure of a solution can be calculated using Raoult’s Law.
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VI. Sample Problem 13.3
• What’s the vapor pressure at 20 °C of a solution of 20.0 g dibutyl phthalate (MW = 278 g/mole) in 50.0 g of octane (MW = 114 g/mole) if pure octane has a vapor pressure of 10.5 torr at 20 °C?
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VI. Ionic Solutes
• An aqueous solution of MgCl2 will have a lower vapor pressure than an aqueous solution of NaCl – why?*
• Recall the definition of colligative properties to understand why!
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VI. Sample Problem 13.4
• A solution contains 0.115 mole H2O and an unknown number of moles of sodium sulfate. The vapor pressure of the solution at 30 °C is 25.7 torr. If the vapor pressure of pure water at 30 °C is 31.8 torr, how many moles of sodium sulfate are in solution?
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VI. Ideal and Nonideal Liquid/Liquid Solutions
• Ideal solutions obey Raoult’s law at all concentrations for both solute and solvent.
• In an ideal solution, ΔHsoln = 0, which means forces of attraction between all molecules are identical.
• We can calculate vapor pressures of each component…
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VI. Rauolt’s Law for 2 Liquids
• Stronger or weaker solute-solvent interactions will cause deviations as shown in the curves in the next slide…
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VI. Raoult’s Law Curves
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VI. Effect of Vapor Pressure Lowering
• Vapor pressure lowering occurs at all temperatures.
• This lowering alters the appearance of phase diagrams of a solution when compared to a pure solvent.
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VI. Altered Phase Diagram
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VI. Changes in FP and BP
• We see that there’s a freezing point depression and a boiling point elevation.
• Since they are caused by vapor pressure lowering, these are also colligative properties.
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VI. Changes in FP and BP
• Equations for calculating freezing point depression and boiling point elevation are very simple.
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VI. Sample Problem 13.5
• At what temperature will a 10% aqueous solution of sugar (C12H22O11) boil if the Kb for water is 0.51 °C/m?
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VI. Sample Problem 13.6
• A solution made by dissolving 3.46 g of an unknown compound in 85.0 g of benzene froze at 4.13 °C. What’s the molar mass of the compound if Kf for benzene is 5.07 °C/m and the normal freezing point of benzene is 5.45 °C?
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VI. Osmosis • osmosis: the flow of solvent from a solution of
lower solute concentration to one of higher solute concentration
• Why does osmosis occur?*
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VI. Osmotic Pressure
• The flow of solvent can be stopped by applying external pressure. This pressure is called the osmotic pressure.
• The “gas” value of R is used.
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VI. Isotonic Solutions • The concept of osmotic pressure is important
in medicine. • IV solutions need to have the same osmotic
pressure as body fluids; if it’s off, there’s a “burning” sensation.
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VI. Ionic Solutions
• Since colligative properties only depend on number, we need to consider how many particles form when ionics form a solution.
• We have to adjust the molality/molarity values used in the equations.
• Problem – ionics do not dissociate completely!
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VI. Ion Pairs
• In an aqueous NaCl solution, some Na+ and Cl- pair up and behave as one entity.
• These are known as ion pairs.
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VI. Accounting for Ion Pairs
• We use the van’t Hoff factor (i) to account for ion pairs.
![Page 54: Ch. 13: Solutions - San Diego Miramar Collegefaculty.sdmiramar.edu/nsinkaset/powerpoints/Chap13.pdfVI. Sample Problem 13.6 • A solution made by dissolving 3.46 g of an unknown compound](https://reader036.vdocuments.net/reader036/viewer/2022071103/5fdc4a38e086324e23413741/html5/thumbnails/54.jpg)
VI. van’t Hoff Factors
![Page 55: Ch. 13: Solutions - San Diego Miramar Collegefaculty.sdmiramar.edu/nsinkaset/powerpoints/Chap13.pdfVI. Sample Problem 13.6 • A solution made by dissolving 3.46 g of an unknown compound](https://reader036.vdocuments.net/reader036/viewer/2022071103/5fdc4a38e086324e23413741/html5/thumbnails/55.jpg)
VI. Equations for Ionic Solutions
• Incorporating van’t Hoff factors is easy…