Presentation on theme: "To go with Chapter 13: Silberberg Principles of General Chemistry"— Presentation transcript:
1 To go with Chapter 13: Silberberg Principles of General Chemistry AP Chemistry, Mrs PeckTopic 10: Solutions
2 Objectives/Study Guide Perform calculations with different solution concentrations such as molarity, mass percent, molality, and mole fraction.Discuss the effects of temperature, pressure, and structure on solubilityPerform calculations with Raoult’s lawUnderstand colligative properties such as boiling point elevation, freezing point depression, and osmotic pressureUse colligative properties to determine the molar mass of a soluteAP tip: Questions from this section may appear on both the F.R. and M.C.Sections of the AP exam. Question on the solubility or miscibility of substancesIn one another may appear in the essay section. Questions involving theDetermination of molar mass by freezing point depression could also appear inThe lab essay.
3 Solutions and their compositions Types of solutionsA solution is a homogeneous mixture..We will focus on liquid solutions in this topic.A solute, the substance being dissolved, is added to a solvent, which is present in the largest amount when referring to a liquid-liquid or gas-liquid solution.CompositionMolarity: ‘M’ of a solution is the number of moles of solute per liter of solutionMass percent: also called weight percent, is the percent by mass of solute in a solution.Mole fraction: ‘x’ is the ratio of moles of a given component to the total number of moles of solution. For a two component solution, where nA and nB are the moles of the two components:xA = nA/(nA + nB)Molality: ‘m’ is the number of moles of solute per kilogram of solventm = moles solute/kg of solvent
4 Example 1A solution is prepared by mixing 30.0mL of butane (C4H10, d= g/ml) with 65.0 mL of octane (C8H18, d= g/ml) Assuming that the volumes add in mixing, calculate the following for butane:A) molarityB) mass percentC) mole fractionD) molality30.0ml x 0.600g x 1 mol = mol C4H10 = 3.26 M1 ml g LMass solution = [30.0ml x 0.600g] + [65.0 ml x 0.700g] = 63.5g1ml mlMass solute = 30.0 ml x 0.600g = 18.0g1mlMass % = 18 g x 100 = 28.3% butane63.5gMoles octane = 65.0 ml x g x 1 mol = mol octane1ml gMole fraction butane = mol butane = 0.437(0.310 mol butane mol octane)0.310 moles butane/ kg = 6.81m
5 Factors affecting solubility The formation of a liquid solution begins by separating the solute into its individual components.Next, the solvent’s intermolecular forces must be overcome to make room for the solute.The solute and solvent then interact to form the solution.
6 Structural effectsThe phrase ‘like dissolves like’ means that solubility is favored if the solute and solvent have similar polarities, as determined by their structure.The table summarizes the solubilities of different types of solutes in different types of solvents.Type of soluteType of solventSolubilityExampleIonicPolarUsually solubleLiCl in H2OSoluble (miscible)CH3OH in H2ONonpolarImmiscibleC6H14 in H2OMiscibleC6H14 in CCl4
7 Example 2Discuss the solubility of each of the following solutes in carbon tetrachloride: ammonium nitrate, 1-pentanol, and pentane. Explain why each solute will or will not dissolve.Carbon tetrachloride is nonpolar and nonpolar solutesWill dissolve in it. Ammonium nitrate is ionic and will notDissolve in a nonpolar solvent. It will dissolve in a polar solventLike water. 1-propanol is polar and will not be soluble in carbonTetrachloride. Pentane is nonpolar and will be miscible (will formA homogeneous mixture) with carbon tetrachloride.
8 Pressure effectsPressure has little effect on the solubilities of liquids and solids.Gases become more soluble in liquids when the pressure of the gas above the liquid decreases.
9 Temperature effectsThe solubility of most solids increases with temperature.However, some solids decrease solubility with temperature.The energy needed to separate the solute-solute particles is usually greater than the energy released from the solute-solvent attractions.It is difficult to predict the temperature dependence of the solubilities of solids,.Gases, on the other hand, always decrease in solubility with increasing temperature.
10 The vapor pressures of solutions A nonvolatile solute lowers the vapor pressure of the solvent.The nonvolatile solute decreases the escaping tendency of the solvent molecules.Raoult’s law:State that the vapor pressure of a solution is directly proportional to the mole fraction of the solvent present.Psoln = xsolventP0solventPsoln is the observed vapor pressure of the solutionXsolvent is the mole fraction of the solventP0solvent is the vapor pressure of the pure solvent.For an ideal solution, a plot of Psoln vs xsolvent at constant temperature gives a straight line with a slope equal to P0solvent.
11 Example 3The vapor pressure of a solution containing 53.6g of glycerin, C3H8O3, in g of ethanol, C2H5OH, is 113 torr at 40*C. Calculate the vapor pressure of the pure ethanol at 40*C assuming that glycerin is a nonvolatile, nonelectrolyte solute in ethanol.Xethanol = mol ethanol/ total mol in solution Pethanol = xethanol P0ethanol133.7 g C2H5OH x 1 mol = 2.90 mol C2H5OH46.07 g53.6 g C3H8O3 x 1 mol = mol C3H8O392.09 gTotal mol = mol mol = 3.48 molP0ethanol = torr = 136 torr(2.90 mol/3.48 mol)
12 Real vs ideal solutions Ideal solutions obey Raoult’s law.Real solutions best approximate the behavior of ideal solutions when the solute concentration is low and the solute and solvent have similar types of intermolecular forces and molecular sizes.Deviations from Raoult’s law occur when the interactions between the solute and solvent are extremely strong, as in hydrogen bonding, and the vapor pressure of the solution is lower than predicted.Also, the vapor pressure of a real solution is greater than predicted when the intermolecular forces between the solute and solvent are weaker than the intermolecular forces between the solute-solute or solvent-solvent.
13 Colligative properties Colligative properties depend on the number, not the identity, of the solute particles in an ideal solution.Boiling point elevationA nonvolatile solute elevated the boiling point of a solvent.Because a nonvolatile solute lowers the vapor pressure of a solution, it must be heated to a temperature higher than the boiling point of the pure solvent.The normal boiling point is the point what the vapor pressure of liquid equals 1 atm.The change in boiling point due to the presence of the nonvolatile solute is represented byΔTb = KbmΔT is the difference between the boiling point of the solution and that of the pure solventKb is the molal boiling point elevation constant of the solventm is the molality of the solute in solution.
14 Example 42.00 g of a large biomolecule was dissolved in 15.0g of carbon tetrachloride. The boiling point of this solution was 77.85*C. Calculate the molar mass of the biomolecule. For carbon tetrachloride, the boiling point constant is 5.03*C kg/mol, and the boiling point of pure carbon tetrachloride is 76.5*CΔTb = 77.85*C – 76.5*C = 1.35*Cm = *C = mol/kg(5.03*C Kg/mol)Mol of biomolecule = kg solvent x mol hydrocarbon/kg solvent= 4.02 x 10-3 molMolar mass = 2.00 g = 498 g/mol4.02 x 10-3 mol
15 Freezing point depression A nonvolatile solute decreases the freezing point of a solvent.ΔTf = KfmΔT is the difference between the boiling point of the solution and that of the pure solvent.Kf is the molal freezing point depression constant of the solvent.
16 Example 5Which of the following solutions would have the largest freezing point depression? Explain your answer.A) m NaClB) m Na3PO4C) m Al2(SO4)3D) m C6H12O6The solution with the lowest freezing point depression is0.010 m Al2(SO4)3, the solution which yields the highest molality (or theMost) particles when dissolved m Al2(SO4)3 yields five particles(ions) when dissolved or the total concentration of solute particles is0.050 m. Two particles (or m) are present in a solution of mNaCl. Four particles (or m) are present in a solution of mNa3PO4. One particle (or m) is present in a solution of mC6H12O6
17 Osmotic pressureOsmosis is the flow of a pure solvent into a solution through a semipermeable membrane.Osmotic pressure is the pressure that must be applied to a solution to stop osmosisπ = MRTΠ is the osmotic pressure in atmospheresM is the molarity of the solutionR is the gas law constant, L*atm/mol*KT is the temperature in Kelvin
18 Example 6An aqueous solution of g of catalase, an enzyme found in the liver, has a volume of 1.00 L at 27*C. The solution’s osmotic pressure at 27*C is torr. Calculate the molar mass of the catalase.M = π = torr x (1 atm/760 torr) = x 10-5 M[ x atm/mol*K] 300K1.00 L x (3.98 x 10-5 mol/L) = 3.98 x 10-5 molMolar mass = g = 2.51 x 105 g/mol3.98 x 10-5 mol