CH 302 – Exam 1 Review PHYSICAL EQUILIBRIUM, Doing Well on Exam One • Learn through repeon: the more you pracce, the beer you will do on the exam • Know how to use your calculators. Be comfortable with large exponents and natural log. • Be able to make calculaons ; be able to explain what the calculaons are saying and why they are important • For example: colligave properes are easy quantave quesons. Do you understand why boiling point increases, how we make a hypertonic soluon, etc.? • Don’t forget the basics: thermodynamics is sll important for conceptual quesons, especially for solubility Exam One Learning Outcomes

Learning Outcomes for Physical Equilibrium Describe how T and P (Henry’s Law) each affect solubility.

Students will be able to... Define and perform calculaons for common concentraon units molarity, , and mole fracon. Demonstrate mastery of and compound and reacon stoichiometry (mole to mole conversion and grams to mole conversions). Perform calculaons and understand the concepts of the 4 colligave properes: vapor pressure lowering, boiling point elevaon, freezing point depression, and osmoc pressure. Predict the sign of ΔG, ΔH, and/or ΔS for physical change Describe the dissociaon of ionic compounds in soluon and the effects on colligave Understand the concept of spontaneous change and equilibrium in the context of properes (van’t Hoff factor, i ). changes, including calculang phase transion from standard thermodynamic data. Learning Outcomes for Solubility Equilibrium Interpret heang curves and calculate heat required for phase transions and changes. Students will be able to...

Describe phase transions (macroscopically and microscopically). Understand the concept of the solubility product, Ksp.

Understand phase in the context of Boltzmann distribuon. Write total and net ionic equaons to idenfy spectator ions.

Understand how intermolecular forces, temperature, and solute concentraon affect Quantavely determine solubility from Ksp and vice versa. vapor pressure. Esmate molar solubility from Ksp. Interpret phase diagrams and idenfy normal boiling and melng point, crical point, and triple point. Understand and apply the "common ion effect" on solubility. Describe the factors that favor the dissoluon process in terms of intermolecular forces and thermodynamics (eg.: enthalpies of soluon, hydraon, lace energy, entropies of Given concentraons of specific ions, predict if a precipitate will form (amount or soluon, free energy of soluon, and temperature). concentraon) using Qsp vs Ksp. Ksp vs. Qsp SOLUBILITY PRODUCT, , FREE ENERGY Ksp , Qsp , and Saturaon - Definions • Solubility Product (K ): K is a constant that represents the product of all ion concentraons at equilibrium sp sp specific to a given salt.

• Molar Solubility: Solved from Ksp, molar solubility represents the maximum amount of solute that can dissolve for a reacon in terms of Molarity (M). This term is represented by the “x” in Ksp calculaons. • Common Ion Effect: When an ion is already present in soluon, the molar solubility of a salt containing that ion significantly decreases.

• Reacon Quoent Solubility Product (Qsp): Qsp is a variable that is calculated by the product of all ion concentraons at a point away from equilibrium. Solved in the same way as Ksp, Qsp uses experimental values rather than “ideal” values.

• Saturaon: when the maximum amount of ions are present in soluon (Ksp = Qsp) • Saturaon is an equilibrium posion. • Dissoluon: when Qsp < Ksp and your reacon moves forward (solid becomes ions) NOT equilibrium • Precipitaon: when Qsp > Ksp and your reacon moves backward (ions become solid) Saturaon: Qsp vs. Ksp

• Ksp represents the ion product of a saturated soluon in terms of molar solubility (x). You can think of it as a measurement of the maximum saturaon capacity of a soluon.

• Qsp represents the ion product of the actual concentraons of ions at any given me. You can control these concentraons experimentally. You can think of Qsp like a starng point

• Ksp is a fixed value ; Qsp is dependent on your actual concentraons. Therefore, your value of Qsp in relaonship to Ksp will describe what happens:

1. Qsp < Ksp (unsaturated) ; more solid can dissolve if added to the soluon

2. Qsp = Ksp (perfectly saturated) ; your reacon is at equilibrium

3. Qsp > Ksp (over saturated) ; precipitaon occurs unl Qsp = Ksp Final Point: Free Energy and Ksp

• When Ksp is very small (less than 1), only a ny When the reactant is favored, fracon of your solid the free energy change for product dissociated into the reacon is posive. ions. This means that the solid reactant is favored.

• When Ksp is large (greater than 1), a greater amount of ions are formed than When the product is favored, the amount of solid that the free energy change for remains. This means that the reacon is negave. the dissociated ions are favored. The exact relaonship will be discussed next unit; but understand this conceptually

Ksp to Molar Solubility -9 What is the molar solubility of Li3PO4? The Ksp of Li3PO4 is 3.2 x 10 .

+ 3 3− Ksp = [Li ] [PO4 ] 3 4 Ksp is the “Solubility Product,” K = (3x) x = 27x which is a constant unique to sp a parcular compound that x represents the “Molar Solubility,” represents the product of ion K which is a direct measurement of concentraons that are 4 sp = x solubility. Molar solubility is the present at equilibrium 27 concentraon of a solute that dissolves in molarity (M) for a reacon. x = 3.3⋅10−3 M

Ksp and the common ion effect

What is the apparent molar solubility of Li3PO4 when added to a 0.5M soluon of LiCl? The Ksp of -9 Li3PO4 is 3.2 x 10 . + 3 3− Ksp = [Li ] [PO4 ] 3 Ksp = (0.5) x Ksp remains constant (because it is a constant). Therefore, you should predict that the K K presence of a common ion sp sp Noce how the molar solubility here is x = 3 = decreases the overall (0.5) .125 much less than that of the last problem “apparent” molar solubility of we solved. your compound. x = 2.56 ⋅10−8 M

Ksp vs. Qsp What happens when you mix 135mL 0.2M lithium nitrate and 250mL 0.1M potassium phosphate? The -9 Ksp of Li3PO4 is 3.2 x 10 .

Here you are given “starng + 3 3− point” concentraons of Qsp = [Li ] [PO4 ] lithium and phosphate ions. Therefore, your ion product 3 −5 will be Qsp. Qsp = (0.07M ) (0.065M ) = 2.2 ⋅10

Qsp > Ksp

You have oversaturated your soluon. The reacon will run backwards unl equilibirium is

reached, resulng in a solid Li3PO4 precipitate.

Ksp Solubility Comparison Which of the following salts is more soluble?

-9 -16 Li3PO4 (Ksp = 3.2 x 10 ) or Fe(OH)2 (Ksp = 8 x 10 ) Ksp Solubility Comparison Which of the following salts is more soluble?

-9 -16 Li3PO4 (Ksp = 3.2 x 10 ) or Fe(OH)2 (Ksp = 8 x 10 ) 3 4 2 3 Ksp = (3x) x = 27x Ksp = (x)(2x) = 4x K K 4 sp = x 3 sp = x 27 4

−4 x = 3.3⋅10−3 M x =1.2 ⋅10 M Phase Changes EQUILIBRIUM CALCULATIONS, HEATING CURVES, PHASE DIAGRAMS Free Energy: All Condions Summary

Freezing, condensaon, deposion Occur at relavely low temperatures ΔH < 0 ; ΔS <0

Fusion (melng), boiling, sublimaon Occur at relavely high temperatures ΔH > 0 ; ΔS > 0 Free Energy: Physical Equilibrium Equaons

ΔG = ΔHtrans −TΔStrans ΔG = 0 Be able to calculate the ΔH, ΔS, and T for a phase ΔHtrans = TΔStrans change

ΔS = ΔHtrans T = ΔHtrans trans Ttrans trans ΔStrans

The temperature change of a phase transion The entropy change of a phase transion given an (boiling point, freezing points, etc.) given the enthalpy and temperature value enthalpy and entropy Heang Curves Checklist • Heang Curves are a 301 concept that show the two equaons necessary to calculate the total heat of a single or mulple phase changes for a substance.

• Calculang Heat: • Use your two calculaons ( q = mcΔT and q = mΔH ) • Be able to do this same calculaon for cooling, even though we are used to doing it for heang (many mistakes on 301 exams) • Reading the graph: • Understanding heat capacity / slope of the heang curves • Determine the value of heat without making a calculaon Heang Curves Checklist – Heat Capacity Heat capacity can be thought of as a substance’s resistance to change in temperature. Consider the heang curve for a liquid being heated to the gas phase:

A low heat capacity results in a steeper slope on a heang curve (substance does not do a good job resisng change in A high heat capacity results temperature when heat is in a smaller slope on a applied) heang curve (substance resists change in temperature when heat is applied)

Therefore, the gas of this substance has a lower heat capacity than the liquid. Checklist • Phase diagrams show the lowest free energy phase of a substance at a given temperature and pressure. • Idenfy the key features of the diagram: • What is the stable phase at a certain temperature and pressure? • Idenfy the triple point • Idenfy the crical point • What phase transion does a specific line represent? • Moving along the diagram: • What phase transions do you go through if you go from point A to point B on the graph? Phase Diagram Checklist • Phase diagrams show the lowest free energy phase of a substance at a given temperature and pressure. • Idenfy the key features of the diagram: • What is the stable phase at a certain temperature and pressure? • Idenfy the normal boiling point, melng point, etc. • Idenfy the triple point • Idenfy the crical point • What phase transion does a specific line represent? • How does this change for a soluon versus a pure ? • Moving along the diagram: • What phase transions do you go through if you go from point A to point B on the graph? Vapor Pressure • For any given liquid sample in a closed container, a certain amount of that sample exists is the gas phase. This is what is known as the vapor pressure. • These molecules have enough kinec energy to “escape” the aracons of other molecules on the surface. • Vapor pressure is a funcon of between the gas and the surface molecules of a liquid.

• The Vapor Pressure is a funcon of the IMF’s of the liquid (which results in a higher ΔHvap) and the temperature. Vapor Pressure and IMF’s

• If Vapor Pressure is the pressure of gas above a liquid in a closed container at a given temperature, we can easily determine that a sample with High IMF’s will have a Low VP (less molecules able to overcome the IMFs and be “lied into the gas phase).

Strong IMF - > low VP (and high boiling point, high ΔHvap)

Weak IMF -> high VP (and low boiling point, low ΔHvap)

IMF Review Dispersion Forces < Dipole Forces < Hydrogen-Bonding

You can see in this equaon also that as ΔHvap increases, Pvap decreases Vapor Pressure and Temperature

• Mathemacally, the vapor pressure for a given liquid is only dependent on a change in ! temperature. −ΔH vap R⋅T Higher Temperature = Exponenally Higher Vapor Pressure Pvap = K ⋅ e • Changing the surface area, size of the container, amount of water, and so on does not change the pressure. • Of all the derivaons used to model this relaonship, the most important equaon to you will be the Clausius-Clapeyron Equaon: ΔH ln( P2 ) = vap ( 1 − 1 ) P1 R T1 T2

Note: the order of P2 / P1 and (1 / T1 – 1/T2) is important. If you forget the order, remember that as you increase temperature, you are increasing vapor pressure. You are taking the difference of the inverse temperature, so you need to switch the order. Soluons QUALITATIVE: DOES IT DISSOLVE? Introducon to Soluons

• A soluon is a homogenous consisng of a solute (can be gas, liquid, or solid) dissolved in a liquid solvent. • Solubility is a measurement of how much solute dissolves in a parcular solvent • The most important term you will learn here is “like-dissolves-like” • Solubility is highest when the IMF’s of the solute match the IMF’s of the solvent • Some basic rules: • with dominant hydrogen bonding don’t dissolve nonpolar solutes (and vice-versa) • Large organic molecules that are dominated by dispersion forces (example: fats) dissolve best in nonpolar solvents + + • Salts formed with sodium (Na ), potassium (K ), and nitrate (NO3-) ions are soluble in water.

Enthalpy of Soluon – Ion Dissoluon

ΔHsolution = ΔHlattice + ΔHsolvation Usually + Always + Always - But can be -

+ - + + - + -

+ ΔHlace (posive value) + ΔHsolvaon (negave value) Step one: Lace Energy Step two: Solvaon Energy is released when solvent breaks apart ionic compound dissolves posive and negave charges Enthalpy of Soluon – Gas Dissoluon

ΔHsolution = ΔHlattice + ΔHsolvation Always 0 Always - Always -

+

ΔH = 0 ; boring lace + ΔHsolvaon (negave value) Step one: There is no lace energy, Step two: Solvaon Energy is released when gas so nothing really happens here! dissolves in water Free Energy of a Salt Dissolving in Water: +ΔH , +ΔS • The majority of salts have a posive ΔHsoluon

• In all cases, the ΔSsoluon is posive (solid to aqueous is a posive entropy change)

• Dissoluon is favored at high temperatures for all salts dissolving in water. Free Energy of a Gas Dissolving in Water: -ΔH , -ΔS • For a gas dissolving in liquid, the reacon is exothermic (ΔHsoluon < 0) because there is no lace energy and a negave solvaon energy.

• There is a negave change in entropy (ΔS < 0) because you are going from gas to aqueous

• Therefore, dissoluon is favored at low temperatures.

Henry’s Law Gas Dissolving in a Liquid • Gases dissolving in a liquid is favored at low temperatures. • For example: an open bole of soda will stay carbonated longer if you keep it chilled. • The solubility of a gas also depends on the paral pressure of the gas above a liquid in a closed container – This is Henry’s Law. • For example: a closed bole of soda will stay carbonated unl you open it (release pressure) The mole fracon of the • There is a mathemacal relaonship here: gas dissolved in a liquid (Xgas) is directly proporonal to the Pgas = KXgas paral pressure of the gas above the liquid

(Pgas). Units of Soluons

• Some of the concentraon units you mol should know for soluons are (gchem M = fundamentals chapter): L • Molarity (M): moles of solute per liter soluon (mol/L) mol • Molality: moles of solute per kilogram m = solvent (mol / kg) kg • Mole Fracon: moles of a parcular species per total moles of soluon / mixture (dimensionless) molA XA = moltotal Yes, there will be a queson that is a pure unit conversion. Colligave Properes CREATING A INCREASES THE STABILITY OF A SOLVENT

Colligave Property Summary Vapor Pressure Lowering (Raoult’s Law):

◦ Solves for the new vapor pressure (PA) of a soluon, based on the mole fracon of the solvent (XA)

◦ PA = XAP°A Freezing Point Depression:

◦ Solves for the negave change in the freezing point (ΔTf), based on the molality of the solute (m)

◦ ΔTf = i kf m Boiling Point Elevaon:

◦ Solves for the posive change in the boiling point (ΔTb), based on the molality of the solute (m)

◦ ΔTb = i kb m Osmoc Pressure: ◦ Solves for the pressure exerted by a fluid to restore osmoc equilibrium, based on the molarity of the solute (M) ◦ Π = iMRT ΔTf = i kf m van’t Hoff Factor • Colligave properes depend on the Examples: number of moles of solute in the final • NaCl: I = 2 soluon. • : I = 1 • Therefore, in the case of electrolytes • PbI2: I = 3 (salts), colligave properes depend on • Na SO : I = 3* the concentraon of ions rather than the 2 4 inial amount of the solid salt. *Remember that polyatomic • We make this “correcon” by using the ions stay together and don’t Van’t Hoff Factor (i) in our colligave dissociate in soluon properes calculaons • The Van’t Hoff Factor is the total number of solute species in soluon Boiling Point Elevaon, Freezing Point Depression • Creang a soluon from a pure solvent increases the stability of your substance in the liquid phase. • This has two effects: 1. The freezing point decreases 2. The boiling point increases • You can see in the diagram to the right that the liquid phase is favored in a larger range of temperatures Boiling Point Elevaon, Freezing Point Depression • Using the following equaons, you are solving for the change in the freezing point or boiling point. Remember: you are solving for a change (ΔTf is always negave, ΔTb is always posive) and you are NOT solving for the final temperatures.

Freezing Point Depression: Boiling Point Elevaon: Solves for the negave change in Solves for the posive change in

the freezing point (ΔTf), based on the boiling point (ΔTb), based on the molality of the solute (m) the molality of the solute (m)

ΔTf = i kf m ΔTb = i kb m Tf = T°f - ΔTf Tb = T°b + ΔTb Vapor Pressure Lowering • Vapor Pressure is lower when a liquid is more stable. Therefore, it should make sense that when you create a lower free energy soluon from a pure solvent, the vapor pressure goes down. • Mathemacally, vapor pressure looks like this: You WILL have to account for electrolytes in the “total moles” of Vapor Pressure Lowering (Raoult’s Law): this mole fracon term, Solves for the new vapor pressure (PA) of a even though you don’t soluon, based on the mole fracon of the see a Van’t Hoff Factor here. solvent (XA) and the vapor pressure of the pure solvent (P°A) PA = XAP°A Vapor Pressure Lowering – Mixing Volale Liquids • The vapor pressure of a binary liquid (A + B) must take into account the vapor pressure of both liquids. • Mathemacally, this follows Dalton’s Law of Paral Pressures:

Vapor Pressure Lowering (Raoult’s Law) + Dalton’s Law:

PA = XAP°A PB = XBP°B

Ptotal = PA + PB Vapor Pressure Lowering: Mixing Liquids • When you mix two miscible liquids, you will get a new vapor pressure value for your soluon. There are two ways you might be asked to determine this value. • Consider two liquids, A and B. Liquid A has a vapor pressure of 200 torr. Liquid B has a vapor pressure of 400 torr. Calculate the vapor pressure when the concentraons of A and B are equal. Then use a vapor pressure diagram to prove your answer. Vapor Pressure Lowering (Raoult’s Law): Solves for the total vapor pressure

PA = XAP°A Dalton’s Law (CH301) PB = XBP°B Ptotal = PA + PB Osmoc Pressure • Osmoc pressure is the pressure exerted by a fluid to restore osmoc equilibrium. • The basic idea is that fluid will flow from the less concentrated to the high concentrated side of a semipermeable membrane (only allows the flow of the solvent; blocks solute). • Again, this property is due to the fact that the soluon is lower in energy. This is why water flows toward the soluon (going from high to low energy). Osmoc Pressure – Biological Membranes • In biology, osmosis is the process of water flowing from low to high solute concentraons. • The driving force of this process is the osmoc pressure and the chemical ion gradient established between the inside and outside of a cell. • The direcon and outcome of osmosis depends on the ion gradient of the soluon: 1. Hypertonic soluons: high solute concentraon on the outside of a cell ; water flows OUT ; results in cells shrinking 2. Hypotonic soluons: high solute on the inside of the cell ; water flows IN ; results in cells swelling 3. Isotonic soluons: equal solute concentraons on the inside and outside of a cell ; no net flow of water ; no change