AP Chemistry Unit 7 Notes PDF

Unit 7: Solutions and Descriptive Chemistry I. Solutions A. Overview 1\. Solution = homogeneous mixture of two or more substances in a single phase. 2\. Solute = substance dissolved in a solvent to form a solution 3\. Solvent = medium in which a solute is dissolved to form a solution Ex.) soli...

Unit 7: Solutions and Descriptive Chemistry I. Solutions A. Overview 1\. Solution = homogeneous mixture of two or more substances in a single phase. 2\. Solute = substance dissolved in a solvent to form a solution 3\. Solvent = medium in which a solute is dissolved to form a solution Ex.) solid in liquid = salt water Liquid in liquid = antifreeze Solid in solid = brass, bronze Gas in liquid = soda Gas in gas = air Adding a solute changes the properties of the solvent 4\. Colligative properties = properties of a s olution that depend only on the number of solute particles per solvent particle and not on the nature of the solute or solvent. 5\. Outline Units of solution concentration How and why solutions form on the molecular/ionic level Colligative properties Suspensions and Colloids B. Units of Concentration 1\. Molarity (M) = moles of solute/liters of solution Problems with molarity: Temperature-dependent Volume proportional to temperature Does not tell us exact amount of solvent Ex.) How do you prepare 500 mL of 43 mM succinic acid (C4H6O4)? 43 mM (1 M/1000 mM)(0.500 L) = 0.0215 mol 0.0215 mol (118 g/mol) = 2.54 g 2\. Molality (m) = moles of solute/kg of solvent Temperature-independent Ex.) You dissolve 45.0 g camphor (C10H16O) in 425 mL ethanol (C2H5OH). What is the molality of the solution? (Density of ethanol = 0.785 g/mL) (425 mL)(0.785 g/mL)(1 kg/1000 g) = 0.334 kg C2H5OH (45.0 g)(1 mol C10H16O/152 g C10 H16O) = 0.296 mol C10H16O m = 0.296 mol C10H16O/0.334 kg C2H5OH = 0.886 m solution 3\. Mole Fraction (X) = moles of part/total moles Sum of mole fractions of all components in a solution = 1 Ex.) 45.0 g camphor in 425 mL ethanol Mole fraction of each component? (45.0 g camphor)(1 mol camphor/152 g camphor) = 0.296 mol camphor Ethanol: (425mL)().785 g/mL)(1 mol/ 46 g) = 7.25 mol ethanol Xcamphor = 0.296/(7.25 + 0.296) = 0.0392 Xethanol = 7.25/(7.25 + 0.296) = 0.961 4\. Mass% = mass of part/mass of whole X 100 Ex.) An alcohol-water mixture contains 1 mol of ethanol and 9 mol of water. What is the mass % of each? ethanol: (1 mol ethanol)(46 g ethanol/1 mol ethanol) = 46 g ethanol water: (9 mol water)(18 g water/1 mol water) = 162 g water \% ethanol = 46 g/(46 + 162) x 100 = 22.1% \% water = 162 g/(46 + 162) x 100 = 77.9% If we know mass %, we can calculate mole fraction and molality (or vice versa) Ex.) Assume you add 1.2 kg ethylene glycol, HOCH2CH2OH, as an antifreeze to 4.0 kg water in the radiator of your car. Calculate the mole fraction, molality, and mass % of the ethylene glycol. ethylene glycol: (1.2 kg)(1000g/1 kg)(1 mol/62 g) = 19 mol water: (4.0 kg)(1000 g/1 kg)(1 mol/18 g) = 220 mol Xethylene glycol = 19/(19 + 220) = 0.079 m = 19 mol/4.0 kg = 4.8 m \% ethylene glycol = 1.2/(4.0 + 1.2) x 100 = 23% C. Solution Process If solid CuCl2 is added to a beaker of water, the salt will begin to dissolve Amount of solid decreases. \[Cu2+\] and \[Cl-\] increases. If we continue to add CuCl2, a point will be reached when no additional CuCl2 seems to dissolve \[Cu2+\] and \[Cl-\] will remain the same Any more CuCl2 added will precipitate. Solution is saturated. 1\. When a solution is saturated a\. no change is observed on the macroscopic level b. on the particulate level two things are taking place 1\. dissolving (CuCl2(s) Cu2+(aq) + 2 Cl-(aq)) 2\. precipitation (Cu2+(aq) + 2 Cl-(aq) CuCl2(s)) 3\. rates of each are equal 4\. equilibrium 5\. no net change 2\. Solubility = concentration of solute in equilibrium with undissolved solute in a saturated solution at a given temperature 3\. Liquids Dissolving in Liquids a\. miscible = when two liquids can mix to an appreciable extent to form a solution b\. immiscible = when liquids cannot mix to form a solution c\. "like dissolves like" Molecular basis for "like dissolves like" Why? Ex.) In pure water and pure ethanol, major intermolecular force is H bonding. When they are mixed, H bonding can occur between ethanol and water molecules, aiding the solution process. Ex.) Pure octane and pure CCl4 nonpolar major intermolecular force = London dispersion forces. When they are mixed, the energy associated with these forces of attraction is similar in value to the energy due to the forces between octane and CCl4 Energy change ≈ 0 But, ΔS \> 0 4\. Solids Dissovling in Liquids a\. "like dissolves like" b\. ion solids are soluble in water water dipoles attract + and -- ions c\. ions become completely surrounded by water molecules (hydrated) (diagram) D. Heat of Solution Both NH4NO3 and NaOH dissolve readily in water, but the solution becomes colder when NH4NO3 dissolves, and it becomes when NaOH dissolves. Why do energy changes occur when substances dissolve? 1\. At the particulate level Ex.) KF dissolving in water Two forces at play a\. attractive forces between ions in the crystal lattice b\. attractive forces between ions and water molecules We can represent the first force in the following way KF(s) K+(aq) + F-(aq) ΔE = +821 kJ This is just the negative of the lattice energy. Next, we represent the second force, which is just the water molecules surrounding the ions K+(g) + F-(g) K+(aq) + F-(aq) ΔHhydration = -819 kJ enthalpy of hydration Overall process: KF(s) K+(aq) + F-(aq) Estimated ΔHsolution = lattice energy + enthalpy of hydration +2 kJ = +821 kJ + -819 kJ This means dissolving KF is slightly endothermic c\. Relationship between ΔHsolution and water solubility As ΔH decreases (becomes more negative), solubility increases Ex.) AgCl is quite insoluble and has a very positive ΔHsolution (endothermic) d\. ΔHhydration depends on 3 factors: 1\. distance between ion and dipole (H2O) closer they are stronger the attraction ΔHhydration more negative 2\. charge of ion higher charge stronger attraction ΔHhydration more negative 3\. polarity of solvent stronger dipole stronger attraction ΔHhydration more negative e\. ΔHsolution and Thermodynamics Standard conditions for a solution refer to 1 m. NaCl(s) NaCl(aq, 1m) ΔHsolution = ∑ΔH°~f~(products) - ∑ΔH°~f~(reactants) = ∑ΔH°~f~ NaCl(aq, 1m) - ∑ΔH°~f~ NaCl(s) = -407.3 kJ/mol -- (-411.2 kJ/mol) = +3.9 kJ/mol E. Factors Affecting Solubility What affects the solubility of gases in liquids? Pressure and Temperature Only temperature affects solubility of solids in liquids 1\. Dissolving Gases in Liquids a\. Solubility of a gas is directly proportional to the gas pressure Henry's Law: Sg = KH · Pg where Sg = gas solubility Pg = partial pressure of gaseous solute KH = Henry's Law constant (different for each gas) (Worksheet) Aside: Henry's Law has important consequences in SCUBA diving (Self-Contained Underwater Breathing Apparatus) Pressure of air you breathe must balance the external pressure of water At great depths, gas pressure must be several atmospheres. Increase pressure more gas dissolves in the blood If you ascend too rapidly, you may experience a paiful and potentially lethal condition known as "the bends" Pressure increases Solubility (N2) decreases N2 gas bubbles form in the blood Divers may use a Helium-Oxygen mixture (rather than N2---O2) because He is not as soluble as N2. 2\. temperature Effects on Solubility a\. LeChatelier's Principle usually dissolving gases is exothermic Gas + Liquid solvent Saturated solution + heat add heat energy, equilibrium shifts Left Less gas dissolved lower solubility at higher temperature F. Colligative Properties properties of a solution that depend only on the number of solute particles per solvent particle 1\. Vapor Pressure a\. Vapor pressure of a solution is always lower than the vapor pressure of pure solvent. b\. Raoult's Law Psolution = Xsolvent · P°solvent where Psolution = vapor pressure of a non-volatile solution Xsolvent = mole fraction of solvent P°solvent = vapor pressure of pure solvent Ex.) 651 g of ethylene glycol, HOCH2CH2OH, is dissolved in 1.50 kg water. What is the vapor pressure of the solution at 90°C? (Pwater at 90°C = 525.8 mmHg) (1.50 kg H2O)(1000 g/1 kg)(1 mol/18 g) = 83.3 mol H2O (651 g C2H6O2)(1 mol/62 g) = 10.5 mol C2H6O2 XH2O = 83.3/(83.3 + 10.5) = 0.888 Psolution = XH2O · P°H2O = (0.888)(525.8 mmHg) Psolution = 467 mmHg So, the addition of solute decreases the vapor pressure from 525.8 mmHg to 467 mmHg. c\. Ideal solution Raoult's Law is an approximation for real solutions ideal solutions obey Raoult's law ΔHsolution = 0 forces of attraction between all molecules are equal Real solutions deviate from Raoult's Law to hold, the forces between solute and solvent molecules must be the same as those between the solvent molecules in the pure solvent. So, solutions involving similar structures (such as C6H14 dissolved in C8H18) will follow Raoult's law closely. If solvent-solute interactions are stronger than solvent-solvent interactions, the actual vapor pressure will be lower than calculated by Raoult's Law (negative deviations) If solvent-solute interactions are weaker than solvent-solvent interactions, the vapor pressure will be higher (positive deviations) A B (A dissolving in B) If A---B \> B---B negative deviation If A---B \< B---B positive deviation 2\. Boiling Point Elevation Raoult's Law tells us something interesting Psolution = Xsolvent · P°solvent If we add more solute, Xsolvent decreases. This means, vapor pressure of solution decreases It is less than vapor pressure of pure solvent (diagram) A decrease in vapor pressure leads to an increase in boiling point. Graphically: red line between L and G Slope of S-L equilibrium is unchanged, but shifts left to meet new triple point a\. ΔTbp = Kbp · msolute where ΔTbp = boiling point elevation Kbp = molal boiling point elevation constant (solvent) m = molality of solute 3\. Freezing Point Depression a\. ΔTfp = Kfp · msolute (Colligative Properties Worksheet) 4\. Colligative Properties of Solutions Containing Ions a\. Colligative properties depend only on the number of solute particles per solvent particle. When 1 mol of NaCl dissolves, 2 mol of ions are formed, which means that the effect on freezing point of water should be twice as large as that expected for a mol of sugar. NaCl(s) Na+(aq) + Cl-(aq) So, 0.1 m NaCl produces two solutes, 0.1 m Na+ and 0.1 m Cl- ΔTfp = Kfp · m x 2 b\. ΔTfp = Kfp · m · i i = van't Hoff factor = correction for observed values = ΔTmeasured/ΔTcalculated\* - If solute does not dissociate For NaCl, hypothetical van't Hoff factor, i = 2, if it dissociates 100%. This means ΔTfp should be twice what is expected. Na2SO4 i should be 3 LaCl3 i should be 4 Al2(SO4)3 i should be 5 \* Problem: Ionic compounds do not dissociate 100% completely. Reason: Strong attractions between ions Positive and negative ions associate in pairs, so the total molality of particles decreases. Increase charge Increase attraction Na2SO4 \> NaCl Na2SO4 will form more pairs Ex.) Calculate I for 0.106 m MgCl2(aq) if the freezing point is -0.517°C. (Kfp for water = -1.86°C/m) ΔTfp = Kfp · m · i i = ΔTfp/( Kfp · m) = -0.517°C/( -1.86°C/m · 0.106 m) i = 2.62 What should it be if MgCl2 100% dissociated? i = 3 Van't Hoff factor depends on 3 factors 1. Solute number of solute particles 2. \% dissociation number of solute particles 3. Concentration Dilute closer to expected value of i Concentrated more association of ions, so i will deviate more from expected value. 5\. Osmosis a\. Definition = movement of solvent molecules through a semipermeable membrane from a region of lower to a region of higher solute concentration b\. Osmotic pressure (Π) = the pressure exerted by osmosis in a solution system at equilibrium (osmosis diagram) c\. equilibrium when the pressure exerted by the column counterbalances the pressure of the water moving through the membrane d\. Π is measured by height of solution column e\. PV = nRT P = (n/V)RT P = MRT Π = MRT G. Colloids 1\. Solution = homogeneous mixture of two or more substances in a single phase Solute does not settle out Solute particles are very small 2\. suspension = homogeneous mixture of suspended particles that gradually settle Solute particles are very big 4. Colloid = state of matter intermediate between a solution and a suspension a\. properties of colloids colloids have high molar masses particles of a colloid are relatively large (\> 1000 nm) Tyndall effect = colloids scatter visible light, making the mixture appear cloudy Colloid particles do not settle out b\. Types of colloids (handout) 1\. emulsions = colloidal dispersions of one liquid in another Ex.) salad dressing, mayonnaise, milk Mayonnaise is an oil-in-water emulsion. (diagram) The charges of oil droplets prevent them from aggregating and settling out a. Emulsions are very stable due to emulsifying agents = substances that aid in the formation of emulsions b. Hydrophilic = attracted to water (polar) c. Hydrophobic = nonpolar Aside: the stabilization of colloids has an interesting application in our own digestive system. When fats in our diet reach the small intestine, a hormone causes the gallbladder to excrete a fluid called bile. A component of bile is a compound with a hydrophilic end and a hydrophobic end. The compound emulsifies the fats present in the intestine, which permits digestion and absorption of fat-soluble vitamins through the intestinal wall. 2\. Surfactants = surface-active agents = substances that affect the properties of surfaces = affect interaction between two phases a\. detergent = a surfactant used for cleaning b\. surfactants lower the surface tension of water c\. soap How it works: Soap has hydrophilic and hydrophobic ends. Hydrophilic end attached to water molecules. Hydrophobic ends attach to oil, dirt, grease, etc. As such, the oil can now be washed away. Ex.) sodium stearate (a soap) (diagram) II\. Reactions A. Question \#4 on Part II of AP Exam Complete and balance chemical equations, given reactants. Answer a short question about each reaction Choose (3) three reactions In all cases, a reaction occurs Assume that solutions are aqueous unless otherwise indicated. Represent substances in solution as ions if the substances are extensively ionized. Omit formulas for any ions or molecules that are unchanged by the reaction (spectator ions) You MUST balance the equation. Phases need not be indicated Point values for reactants products balanced equation short question B. Types of Reactions reactions can be categorized in the following 4 main groupings. 1\. Synthesis and Decomposition 2\. Precipitation know you solubility rules COLD 3\. Redox look for elements single replacement, combustion, or special cases 4\. Acid/Base traditional: Bronsted-Lowry anhydrides: oxides in water Lewis: complex ions C. Synthesis and Decomposition 1\. Synthesis look for: a\. gas blown over metal oxide to produce corresponding ionic compound Ex.) Solid calcium oxide is heated in the presence of sulfur trioxide gas CaO + SO3 CaSO4 Ex.) Carbon dioxide is bubbled through molten magnesium oxide. MgO + CO2 MgCO3 Ex.) Bromine gas is bubbled through propene. Br2 + C3H6 C3H4Br2 Ex.) Powdered strontium oxide is added to distilled water. SrO + H2O Sr2+ + OH- 2\. Decomposition look for: a\. 1 compound is "heated" b\. metal hydroxides/carbonates when heated form oxides and corresponding gases Ex.) Solid potassium chlorate is heated in the presence of manganese dioxide as a catalyst. KClO3 KCl + O2 Ex.) Solid ammonium carbonate is heated. (NH4)2CO3 NH3 + CO2 + H2O Ex.) Solid calcium carbonate is heated. CaCO3 CaO + CO2 D. Precipitation 1\. Know solubility rules Review Solubility Rules 1\. All compounds of the alkali metals (Group 1) are soluble. 2\. All salts containing NH~4~^+^, NO~3~^-^, ClO~4~^-^, ClO~3~^-^, or C~2~H~3~O~2~^-^ are soluble. 3\. All chlorides, bromides, and iodides (salts containing Cl^-^, Br^-^, or I^-^) are soluble, except when combined with Ag^+^, Pb^2+^, or Hg~2~^2+^ 4\. All sulfates (salts containing SO~4~^2-^) are soluble, except those of Pb^2+^, Ca^2+^, Sr^2+^, Hg~2~^2+^, and Ba^2+^. [Insoluble Compounds] 5\. All hydroxides (OH^-^ compounds) and all metal oxides (O^2-^ compounds) are insoluble, except those of Group 1 and of Ca^2+^, Sr^2+^, and Ba^2+^. 6\. All compounds that contain PO~4~^3-^, CO~3~^2-^, SO~3~^2-^, and S^2-^ are insoluble, except those of Group 1 and NH~4~^+^. Ex.) Solutions of lead(II) nitrate and sodium bromide are mixed. Pb2+ + Br- PbBr2 Ex.) Solutions of tin(IV) sulfate and sodium carbonate are mixed. Sn4+ + CO32- Sn(CO3)2 Ex.) An excess of potassium hydroxide solution is added to a solution of iron(III) nitrate. OH- + Fe3+ Fe(OH)3 E. Redox Reactions 1\. Single replacement an element must be present as a reactant and a product Ex.) A strip of zinc is added to a solution of 6.0 molar hydrochloric acid. Zn + H+ Zn2+ + H2 Ex.) A mixture of powdered iron(III) oxide and powdered aluminum metal is heated strongly. Fe2O3 + Al Al2O3 + Fe Ex.) Solid sodium is placed in distilled water. Na + H2O Na+ + OH- + H2 Ex.) Bromine gas is bubbled through a solution of nickel(II) chloride. Br2 + Cl- Cl2 + Br- 2\. Combustion look for "is burned in air" all products are oxides/sulfides ex.) metal oxides, nonmetal oxides, CO2, SO2, H2O, nonmetal sulfides Ex.) Sulfur powder is burned in air. S + O2 SO2 Hydrocarbons and organics produce CO2 and H2O Ex.) Butanol is burned in air. C4H9OH + O2 CO2 + H2O 3\. Special Cases permanganate, dichromate "in acidic/basic solution" H2O as a product Ex.) A solution of lead(II) chloride is added to an [acidified] solution of potassium [permanganate]. think redox MnO4- + H+ + Pb2+ Mn2+ + Pb4+ + H2O Ex.) A concentrated solution of hydrochloric acid is added to powdered manganese dioxide and gently heated. H+ + Cl- + MnO2 Mn2+ + Cl2 + H2O Ex.) A solution of potassium dichromate is added to an acidified solution of iron(II) chloride. Cr2O72- + H+ + Fe2+ Fe3+ + Cr3+ + H2O Ex.) Copper coil is placed in dilute nitric acid. Cu + H+ + NO3- Cu2+ + NO + H2O F. Acid/Base 1\. Traditional: hydroxides and acids look for stoichiometry "excess" or "equal volumes of equimolar" H2O product when OH- reactant Ex.) Equal volumes of equimolar solutions of disodium hydrogen phosphate and hydrochloric acid are mixed. HPO42- + H+ H2PO4- Ex.) Equal volumes of equimolar solutions of phosphoric acis and potassium hydroxide are mixed. H3PO4 + OH- H2PO4- + H2O 2\. Anhydrides mixed in water look for metal oxide in H2O form hydroxide base Remember: Slightly soluble hydroxides: Ca, Mg nonmetal oxide in H2O form acid Remember: Strong vs. Weak acids Ex.) Solid sodium hydride is mixed with water. NaH + H2O Na+ + OH- + H2 Ex.) A chunk of barium oxide is dropped in water. BaO + H2O Ba2+ + OH- Ex.) Powdered magnesium nitride is placed in water. Mg3N2 + H2O Mg(OH)2 + NH3 slightly soluble Ex.) Solid dinitrogen pentoxide is added to water. N2O5 + H2O H+ + NO3- 3\. Lewis Acid/Base: Complex Ions Lewis acid = electron pair acceptor Lewis base = electron pair donor Complex ion = combination of one or more anions or neutral molecules (ligands) with a metal ion look for: transition metal cation and an electron-rich ligand common ligands: NH3, CN-, OH-, SCN- ligand often found in "excess" Ex.) Cu(NH3)42+ coordination number of ligand 2 times charge on metal to get coordination number of ligand Ex.) Excess sodium cyanide solution is added to a solution of silver nitrate. CN- + Ag+ Ag(CN)~2~^-^ G. Final Thoughts balance, net-ionic reaction no NR categorize by reaction type start with precipitate remember slightly soluble hydroxides (Ca/Mg) and unstable acids: H2CO3, H2SO3 sulfuric acid appears as either H+ + HSO4- or H+ + SO42- remember common spectator ions Ex.) Na+, K+, Cl-, etc.