Big Idea Three Topics 1. Molecular, Ionic, Net Ionic Equations 2. Stoichiometry 3. Synthesis, Decomposition Reactions 6. Chemical Change Evidence 7. Endothermic & Exothermic Reactions 8. Electrochemistry 4. Neutralization Reactions 5. Redox Reactions
Properties of Solutions November 28, 2016
Why Solutions? Importance of studying solutions Because many reactions take place in solutions Because mixing reactants in solid form often do not result in reactions. Reactions require collisions at the atomic/molecular level, and in the solid state, this does not occur at a significant rate.
Solutions Review: H-Chemistry Solutions are homogeneous mixtures *Do NOT always involve liquids* Solvent vs. Solute Solvent: Present in greater amount, Does the dissolving, Water is the universal Solvent Solute: Present in LESSER amount, Is the one dissolved
Solutions Review: H-Chemistry If a solution is a mixture, do the solutes and the solvents chemically react or physically mix?
Solution Composition moles of solute Molarity ( M) = liters of solution mass of solute Mass (weight) percent = 100% mass of solution Mole fraction ( χ moles A A ) = total moles of solution moles of solute Molality ( m) = kilogram of s olvent
Molarity Calculations Review: You have 1.00 mol of sugar in 125.0 ml of solution. Calculate the concentration in units of molarity. Consider separate solutions of NaOH and KCl made by dissolving 100.0 g of each solute in 250.0 ml of solution. Calculate the concentration of each solution in units of molarity.
Mass Percent Calculation What is the percent-by-mass concentration of glucose in a solution made my dissolving 5.5 g of glucose in 78.2 g of water?
Mole Fraction Calculation A solution of phosphoric acid was made by dissolving 8.00 g of H 3 PO 4 in 100.0 ml of water. Calculate the mole fraction of H 3 PO 4. (Assume water has a density of 1.00 g/ml.)
Molality Calculation A solution of phosphoric acid was made by dissolving 8.00 g of H 3 PO 4 in 100.0 ml of water. Calculate the molality of the solution. (Assume water has a density of 1.00 g/ml.)
Solution Stoichiometry 0.125 L of 0.100 M HCl is added to Zn. What mass of ZnCl 2 is formed? 25mL of 0.5M Ba(NO 3 ) 2 solution is combined with excess Na 2 SO 4. How many grams of precipitate formed?
Solution Stoichiometry FRQ 1. How many grams of precipitate would form if 30mL of a 0.25M Pb(NO 3 ) 2 solution was added to 20mL of a 0.50M NaCl solution? 2. How many moles of the excess reactant are left over after the reaction? 3. What is the molarity of the excess reactant after the reaction?
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Formation of a Liquid Solution 1. Separating the solute into its individual components (expanding the solute). 2. Overcoming intermolecular forces in the solvent to make room for the solute (expanding the solvent). 3. Allowing the solute and solvent to interact to form the solution.
Formation of a Liquid Solution Steps 1 and 2 require energy, since forces must be overcome to expand the solute and solvent. Step 3 usually releases energy. Steps 1 and 2 are endothermic, and step 3 is often exothermic.
Enthalpy (Heat) of Solution Enthalpy change associated with the formation of the solution is the sum of the ΔH values for the steps: ΔH soln = ΔH 1 + ΔH 2 + ΔH 3 ΔH soln may have a positive sign (energy absorbed) or a negative sign (energy released).
Enthalpy (Heat) of Solution Calculation Hess Law: The enthalpy of a given chemical reaction is constant, regardless of the reaction happening in one step or many steps. Use Hess Law ΔH rxn = Σ ΔH f (products) -Σ ΔH f (reactants) ΔH rxn = Σ ΔH f (lattice energy) -Σ ΔH f (hydration)
Enthalpy (Heat) of Solution Calculation The lattice energy of CaCl 2 (s) is -2247 kj/mol, and the enthalpy of hydration is -2293 kj/mol. Calculate the enthalpy of solution per mole of solid CaCl 2. The lattice energy of CaI 2 (s) is -2059 kj/mol, and the enthalpy of solution is -104 kj/mol. Calculate the enthalpy of hydration of solid CaI 2.
Formation of a Liquid Solid Determined by the ability to dissolve in solution What is meant by dissolve? Dissolution Like dissolves Like
Formation of a Liquid Solid Miscible vs. Immiscible Miscible = dissolves Immiscible = does NOT dissolve
Solubility Formation of a Liquid Solid The maximum amount of a solute that can dissolve at a specified temperature and pressure Solubility can be affected by altering the temperature, pressure, or structure
Factors Affecting Solubility - Structure Hydrophobic (water fearing) Non-polar substances Hydrophilic (water loving) Polar substances Likes Dissolve Likes à Polar solute will dissolve in a polar solvent
Factors Affecting Solubility - Temperature Although the solubility of most solids in water increases with temperature, the solubilities of some substances decrease with increasing temperature. Predicting temperature dependence of solubility is very difficult. Solubility of a gas in solvent typically decreases with increasing temperature
Solubility Aqueous Solution >> General Rule: Increase in temperature increase in the solubility
Solubility Gaseous Solution >> General Rule: Increase in temperature decrease in the solubility
Factors Affecting Solubility - Pressure Li=le effect on solubility of solids or liquids Henry s law: C = kp C = concentracon of dissolved gas k = constant P = parcal pressure of gas solute above the solucon Amount of gas dissolved in a solucon is directly proporconal to the pressure of the gas above the solucon.
Henry s Law Calculations If the solubility of a gas in water is 0.77g/L at 350kPa of pressure, what is its solubility, in units of grams/liter, at 100kPa? A gas has a solubility of 3.6g/L at a pressure of 100kPa. What pressure is needed to produce an aqueous solution containing 9.5g/L of the same gas?
Vapor Pressures of Solutions Nonvolatile solute lowers the vapor pressure of a solvent. Raoult s Law: χ P soln = observed vapor pressure of solution solv = mole fraction of solvent o Psolv = vapor pressure of pure solvent
A Solution Obeying Raoult s Law
Vapor Pressure Nonideal Solutions Liquid-liquid solucons where both components are volacle. Modified Raoult s Law: P = χ P + χ P o o Total A A B B Nonideal solucons behave ideally as the mole fraccons approach 0 and 1.
Sample AP MC Question For each of the following solutions, would you expect it to be relatively ideal (with respect to Raoult s Law), show a positive deviation, or show a negative deviation? a) Hexane (C 6 H 14 ) and chloroform (CHCl 3 ) Positive deviation; Hexane is non-polar, chloroform is polar. b) Ethyl alcohol (C 2 H 5 OH) and water Negative deviation; Both are polar, and will form strong hydrogen bonds between the two molecules c) Hexane (C 6 H 14 ) and octane (C 8 H 18 ) Ideal; Both are non-polar with similar molar masses.
Vapor Pressure Calculations Glycerin, C 3 H 8 O, is a nonvolatile liquid. What is the vapor pressure of a solution made by adding 164 g of glycerin to 338 ml of H 2 O at 40 0 C? The vapor pressure of pure water at 40 0 C is 0.992 g/cm 3.
Colligative Properties Depend only on the number, not on the idencty, of the solute parccles in an ideal solucon: Boiling-point elevacon Freezing-point depression OsmoCc pressure
BP & FP Temperatures The magnitude of BP elevation and FP depressions is proportional to the number of solute particles dissolved in the solvent.
1. Boiling Point Boiling Point Elevation the difference in temperature between the boiling points of a solution and of the pure solvent. ΔT = K b m solute ΔT = boiling-point elevation K b = molal boiling-point elevation constant m solute = molality of solute
BP Calculation What is the boiling point of a 1.5 mol solution that in 800g of water?
2. Freezing Point Freezing Point Depression is the difference in temperature between the freezing points of a solution and of the pure solvent. ΔT = K f m solute ΔT = freezing-point depression K f = molal freezing-point depression constant m solute = molality of solute
FP & BP Calculation What is the BP and FP of a 1.40 mol solution of Na 2 SO 4 in 1750g of water?
Determining Molar Mass from BP/FP 7.5 g of solute is added to 22.60 g of water. The water boils at 100.78 degrees C. What is the molar mass of the solute?
3. Osmotic Pressure Osmosis flow of solvent into the solucon through a semipermeable membrane. Π = MRT Π = osmocc pressure (atm) M = molarity of the solucon R = gas law constant T = temperature (Kelvin)
Osmotic Pressure Calculations When 33.4 mg of a compound is dissolved in 10.0 ml of water at 25 C, the solution has an osmotic pressure of 558 torr. Calculate the molar mass of this compound.
Colligative Properties of Electrolytes Van t Hoff Factor The relationship between the moles of solute dissolved and the moles of particles in solution is usually expressed as: i = moles of particles in solution moles of solute dissolved
Colligative Properties of Electrolytes Ion-Pairing At a given instant a small percentage of the sodium and chloride ions are paired and thus count as a single particle.
Colligative Properties of Electrolytes The expected value for i can be determined for a salt by noting the number of ions per formula unit (assuming complete dissociation and that ion pairing does not occur). NaCl KNO 3 Na 3 PO 4
Ion Pairing Ion pairing is most important in concentrated solutions. As the solution becomes more dilute, the ions are farther apart and less ion pairing occurs. Ion pairing occurs to some extent in all electrolyte solutions. Ion pairing is most important for highly charged ions.
Colligative Properties of Electrolytes Modified Equations for Electrolyte Solutions ΔT = imk Π = imrt
Colloids A suspension of tiny particles in some medium. Tyndall effect scattering of light by particles. Suspended particles are single large molecules or aggregates of molecules or ions ranging in size from 1 to 1000 nm.
Coagulation Destruction of a colloid. Usually accomplished either by heating or by adding an electrolyte.