Slide 1 / 48 Slide 2 / 48 New Jersey Center for Teaching and Learning Progressive Science Initiative This material is made freely available at www.njctl.org and is intended for the non-commercial use of students and teachers. These materials may not be used for any commercial purpose without the written permission of the owners. NJCTL maintains its website for the convenience of teachers who wish to make their work available to other teachers, participate in a virtual professional learning community, and/or provide access to course materials to parents, students and others. Solutions: Formation and Properties Click to go to website: www.njctl.org www.njctl.org Slide 3 / 48 Slide 4 / 48 Heats of Solution Solution Formation Solute formation requires the weakening of the Coulombic attractions within the solvent and solute so new Coulombic attractions may form between the solvent and solute. Example: Formation of an aqueous glucose solution. C6H12O6(s) --> C6H12O6(aq) Step What happens? Enthalpy Change 1 Intermolecular forces between glucose (solute) molecules must weaken. + 2 3 Intermolecular forces between solvent (water) molecules must weaken. New coulombic attractions will form between the solute and solvent + - The net change in enthalpy for the solution formation process is called the heat of solution and is specific to a particular solute-solvent combination. Slide 5 / 48 Heats of Solution Ideal solutions are solutions in which when the solutes are mixed the heat of solution would be equal to zero. Solutions behave most ideally when the solute and solvent are extremely similar in molecular structure and polarity. Examples of nearly ideal solutions The heat of solution will vary depending on the affinity of the solute for the solvent. Ethanol (CH3CH2OH) has very different heats of solution when dissolved in water (H2O) and hexane (C6H14). Ethanol dissolved in water ( H = -10.7 kJ/mol) The hydrogen bonds between the solute and solvent release large amounts of energy when formed. Ethanol dissolved in hexane ( H = +23 kJ/mol) Since ethanol is polar and hexane is non-polar, very few coulombic attractions form to offset the energy required to weaken the solute-solute and solvent-solvent attractions. Slide 6 / 48 Heats of Solution The heat of solution can be calculated by monitoring the temperature change when the solute and solvent are mixed. Example: When 5.3 grams of NH4Cl are dissolved in 100 grams of water @22.0 C, the temperature of the solution drops to 18.7 C. Assuming the specific heat of the solution is 4.2 J/gC, what is the heat of solution? CH3OH and CH3CH2OH C6H14 and C7H14 C6H6 and C7H8 The heat of solution for these solutions is near zero because the the Coulombic attractions between the solute molecules and solvent molecules are almost identical to those that would form between solute and solvent. Energy lost by solution = 105.3 g x 3.3 C x 4.2 J = 1460 J gC Expressed in kJ/mol = 1.460 kJ /0.1 mol NH4Cl = 14.6 kJ/mol This process is endothermic and therefore will likely become more favorable as the temperature increases. Slide 7 / 48 Slide 8 / 48 Solution Formation Electrolytes Soluble ionic compounds and strong acids make excellent electrolytes. Covalent molecular materials make poor electrolytes as they do not dissociate into ions. Ionic solutes dissociate into ions in aqueous solvent while covalent molecular solutes do not. Dissolution of NaCl in H2O The more ions that are produced in solution, the stronger the electrolyte. NaCl(s) --> Na+(aq) + Cl-(aq) Each ion becomes solvated by water molecules Comparing equimolar NaCl(aq) and MgCl2(aq) NaCl(s) --> Na+(aq) + Cl-(aq) Dissolution of glucose (C6H12O6) in H2O C6H12O6(s) --> C6H12O6(aq) Which compound is the stronger electrolyte? The entire glucose molecule becomes solvated by water molecules Ionic solutes are called electrolytes. Why? MgCl2 producesMove 1 iontoofsee Mg2+ and 2 ions of Clanswer when it dissociates, so it is the stronger electrolyte. Since ionic solutes produce ions in solution resulting in increased electrical conductivity, there are referred to as electrolytes Slide 9 / 48 Slide 10 / 48 Which of the following is NOT true regarding the formation of an aqueous glucose solution? 1 Soluble ionic compounds and strong acids make excellent electrolytes. Covalent molecular materials make poor electrolytes as they do not dissociate into ions. The more ions that are produced in solution, the stronger the electrolyte. A Covalent bonds within the glucose molecule must be broken B Intermolecular coulombic forces will form between glucose molecules and water molecules C The hydrogen bonding network between water molecules must be disrupted Comparing equimolar HF(aq) and HBr(aq) HBr(aq) --> H+(aq) + Br-(aq) Which compound is a stronger electrolyte? Answer Electrolytes D The glucose molecule remains un-ionized Since HBr is a strong acid, it produces many more ions to see answer compared to HF, a Move weak acid in which very few of the HF molecules have ionized. E All of these are true Slide 11 / 48 Slide 12 / 48 Answer 2 What is the heat of solution in (kJ/mol) of KCl if when 14.8 grams of KCl was dissolved in 400 grams of water, the temperature dropped 4.3 C? Assume a specific heat of solution of 4.2 J/gC. 3 How much would the temperature of a solution prepared by dissolving 10.6 grams of LiNO2 in 300 grams of water increase? Assume a specific heat of solution of 4.2 J/gC and a heat of solution of LiNO2 of -11.0 kJ/mol. Answer HF(aq) --> H+(aq) + F-(aq) MgCl2(s) --> Mg2+(aq) + 2Cl-(aq) Slide 13 / 48 Slide 14 / 48 5 Which of the following correctly ranks the solutions from highest to lowest conductivity? A 0.1 M NaF > 0.1 M CH3OH > pure water B CH3OH B 0.2 M AlCl3 > 0.2 M NaF > pure water C C6H12O6 C pure water > 0.1 M NaF > 0.1 M CH3OH Answer A HCN D H2SO4 E HC2H3O2 D 0.1 M CH3OH > 0.1 M AlCl3 > 0.1 M NaF E None of these Slide 15 / 48 Slide 16 / 48 6 Which of the following pairs of liquids would form the most IDEAL solution? A C6H14(l) and H2O(l) Colligative properties of solutions depend exclusively on the number of solute particles in the solution, not on their kind. Vapor pressure lowering C CH3OH(l) and CH3COCH3(l) Boiling point elevation Answer E None of these Colligative Properties Examples of colligative properties B CH3OH(l) and C6H14(l) D C5H12(l) and C6H14(l) Answer 4 Which of the following would be the strongest electrolyte when dissolved in water? Freezing point depression Osmotic pressure elevation In essence, the addition of solute to any solvent will decrease the vapor pressure and hence raise the boiling point. The solution will freeze at a lower temperature and require more pressure to prevent osmosis into the solution. Slide 17 / 48 Vapor Pressure When a liquid or solid evaporates, the vapor above the liquid exerts pressure on the surface of the liquid. When condensation and evaporation occur at equal rates, the vapor is in equilibrium with its liquid. Slide 18 / 48 Vapor Pressure The vapor pressure is influenced by the strength of the solvent's particle interactions. The stronger the particle interactions, the lower the vapor pressure of a pure liquid at a given temperature. H 2O CH3COCH3 Vapor Liquid VP = 55.3 mm Hg @ 40 C H-Bonds less evaporation VP = 400 mm Hg @ 40 C no H-bonds more evaporation Slide 19 / 48 Slide 20 / 48 Vapor Pressure Vapor Pressure The vapor pressure is directly proportional to the temperature. VP of Various Substances vs. Temp. As heat is added, more evaporation results, leading to a higher vapor pressure. Note the larger hydrocarbons have a lower vapor pressure at a given temperature due to higher LDF's. VP H2O vs Temp When a liquid's vapor pressure equals the atmospheric/ external pressure it will boil. The stronger the particle interactions, the more energy must be added to raise the vapor pressure hence the higher boiling points. Note the relationship is not linear. Slide 21 / 48 Slide 22 / 48 Vapor Pressure To reach the boiling point, either the vapor pressure must be increased, the atmospheric pressure must be lowered, or both. The BP can be reached by raising the vapor pressure by heating H2O @25 C 23.88 mm Hg H2O @100 C 760 mm Hg Status @ 1 atm external pressure The freezing point of the solution will be lower than the pure solvent B The boiling point of the solution will be higher than the pure solvent C The osmotic pressure of the solution will be higher than the pure solvent D The vapor pressure of the solution will be lower than the pure solvent not boiling! boiling! The BP can also be reached by lowering the atmospheric pressure The atmospheric pressure at the peak of Mt. Everest is just 253 mm Hg so water must be heated only to roughly 72 C to obtain a vapor pressure of 253 mm Hg and thus boil. E All of these would occur Slide 23 / 48 Slide 24 / 48 8 Which of the following would have the lowest vapor pressure at a given temperature? 9 Which of the following would be TRUE? A Water will boil at a lower temperature at high altitude because its Coulombic attractions are stronger at these pressures A C6H14(l) C H2O(l) D CH3COCH3(l) Answer B C4H10(l) B Water will boil at a lower temperature at high altitude because its vapor pressure is higher at these altitudes. C Water will boil at a lower temperature at high altitude because atmospheric pressure is lower D Water will boil at a higher temperature at high altitude because atmospheric pressure is higher Answer Vapor Pressure A Answer 7 Which of the following would NOT occur when a solute is added to a solvent? Slide 25 / 48 Slide 26 / 48 Vapor Pressure of Solutions Adding solute to create a solution lowers the solutions vapor pressure and raises the boiling point. When solute is mixed with the solvent, attractions form between the two, resulting in less solvent molecules evaporating and thereby lowering the vapor pressure. Vapor Pressure of Solutions The amount of solute particles produced per solute molecule is called the Van't Hoff factor. The higher the Van't Hoff factor, the greater the impact on the colligative properties of the solution. Molecular solutes all have Van't Hoff factors of 1 CH3OH, C6H12O6, CH3COCH3 Na+ Lower VP NaCl(aq) --> Na+(aq) + Cl-(aq) VHF = 2 Cl- Pure H2O Al(NO3)3(aq) --> Al3+(aq) + 3NO3-(aq) VHF = 4 0.1 M NaCl(aq) Attractions between the ions and water reduce evaporation and the vapor pressure. A given quantity of Al(NO3)3 will have 4x the impact of a molecule solute and 2x the impact of NaCl(aq) on the vapor pressure of a solution. Slide 27 / 48 Slide 28 / 48 Vapor Pressure of Solutions Vapor Pressure of Solutions The vapor pressure of a solution can be calculated using Raoult's Law. VPsolution = (Xsolvent)(Psolvent) The vapor pressure of a solution can be calculated using Raoult's Law. Example: If 2.3 grams of NaCl is added to 108 grams of water, what is the vapor pressure of the resulting solution assuming the partial pressure of water vapor at this temperature is 18.9 mm Hg? where.... Find mole fraction of solvent Xsolvent = mole fraction of solvent 2.3 g NaCl = 0.05 moles NaCl x VHF(2) = 0.10 mol solute Psolvent = pressure of pure solvent 108 g H2O = 6 moles H2O As can be seen, as the mole fraction of solvent decreases - due to high moles of solute particles - the vapor pressure of the solution will diminish. mole fraction = 6/6.1 = 0.98 Find VP solution VPsolution = (0.98)(18.9 mm Hg) = 18.6 mm Hg Slide 29 / 48 Vapor Pressure of Solutions The vapor pressure of solution made from two liquids can be determined using an expanded version of Raoult's Law VPsolution = (XliquidA)(Pliquid A) + (XliquidB)(Pliquid B) Slide 30 / 48 10 When a solute is mixed with a solvent, the resulting solution will have... A A higher vapor pressure and boiling point B A higher vapor pressure and lower boiling point Both liquids evaporate and therefore contribute to the vapor pressure of the solution C A lower vapor pressure and a higher boiling point Example: D A lower vapor pressure and a lower boiling point What is the vapor pressure of a solution made by mixing 1 mole of acetone (C3H6O) with 4 moles of ethanol (CH3CH2OH); assuming vapor pressures of each being 28 mm Hg and 17.4 mm Hg at a given temperature. VPsolution = (1/5)28 mm Hg + (4/5)17.4 mm Hg = 19.5 mm Hg Answer Higher VP Ionic solutes have Van't Hoff factors that correlate to the number of ions present Slide 31 / 48 Which of the following substances/mixtures would have the highest vapor pressure? 12 Which of the following substances/mixtures would have the lowest vapor pressure? A 0.2 M LiC2H3O2 A 0.1 M NaCl B 0.3 M HNO3 B 0.2 M CH3OH Answer C 0.1 M CH3CH2OH C 0.4 M HC2H3O2 Answer 11 Slide 32 / 48 D 0.1 M Al(NO3)3 E Pure H2O D 0.1 M MgCl2 E 0.05 M Al(NO3)3 Slide 33 / 48 Equi-molar amounts of liquid methanol (VP = 19 mm Hg) and water (VP = 17 mm Hg) are mixed, what would be the vapor pressure of the solution? Answer 14 What is the vapor pressure of a solution prepared by adding 1.8 grams of glucose (C6H12O6) to 200 mL of water (D =1 g/mL). Assume water vapor has a vapor pressure of 12.3 mm Hg at this temperature. Answer 13 Slide 34 / 48 Slide 35 / 48 If equi-molar amounts of liquid methanol (VP = 19 mm Hg) and water (VP = 17 mm Hg) are mixed, what would be the mole fraction of methanol in the vapor phase? (Hint: For each liquid - its vapor pressure is calculated by Xsolvent*Psolvent. Remember also that the mole fraction can be determined as a pressure fraction (P/Ptot) Boiling Point Elevation The increase in boiling point of a solution compared to the pure solvent is directly proportional to the molal concetration (m) of solute, the boiling point constant (Kb) and the Van't Hoff factor of the solute (i). Tb = Kb*m*i Answer 15 Slide 36 / 48 Higher BP Lower BP Pure H2O 0.1 m NaCl(aq) The solute impedes evaporation thereby requiring more heat to be added to get the vapor pressure = atmospheric pressure. Slide 37 / 48 Slide 38 / 48 Boiling Point Elevation Measuring the boiling point elevation of a solution can be used to determine the molar mass of a non-volatile molecular solute. Example: What is the molar mass of a solute that when 3.4 grams of it are mixed with 300 grams of benzene (Tb = 80.1C, Kb = 2.53 C/m), the boiling point is 80.6 C. Use Freezing Point Depression The addition of solute disrupts the formation of the crystal lattice to freeze the solvent, thereby requiring a lower temperature to freeze the solution. The degree to which the freezing point has been depressed can be calculated by... Tb and Kb to find molality 0.5 C x 1 m Tf = Kf*m*i = 0.198 m where Kf = freezing point constant of solvent 2.53 C Use molality and kg of solvent to get moles solute The molar mass of solute can be calculated as done with boiling point elevation. 0.300 kg solvent x 0.198 mol solute = 0.059 mol 1 kg solvent Divide the grams by moles 3.4 g/0.059 mol = 57.6 g/mol Slide 39 / 48 Slide 40 / 48 Freezing Point Depression The osmotic pressure of a solution is the pressure required to prevent the osmosis (flow) of water into it. Antifreeze is a mixture of water and ethylene glycol and it allows the coolant in your engine to stay liquid even below the normal freezing point of either material. Some frogs release glucose into their bloodstream in cold temperatures so their blood doesn't freeze! Pure H 2O No net movement of water Osmotic Pressure = 0 Pure H 2O The osmotic pressure is directly proportional to the M of the solution, the Van't Hoff factor of the solute, and the temperature. Osmotic Pressure = iM*R*T M = Molarity, R = 0.0821 L*atm/mol K, T = Kelvin temperature, i = VHF 0.1 M NaCl Osmotic Pressure of 2.44 mmHg needed to prevent net-flow of water Slide 42 / 48 16 Compared to the pure solvent, a solution will have a... A higher boiling point and lower freezing point B higher boiling and freezing point C lower boiling point and higher freezing point D lower boiling point and lower freezing point E Plants use high concentrations of solute within their cells to draw water in to create the turgor pressure necessary to keep the cells rigid. Pure H 2O 0.1 M NaCl Net movement of water into solution Slide 41 / 48 Osmotic Pressure Pure H 2O the same freezing and boiling points Answer Depressing the freezing point of a solvent has many applications. Osmotic Pressure Slide 43 / 48 Slide 44 / 48 0.02 M CH3OH A 0.45 M HC2H3O2 B 0.02 M KNO3 B 0.45 M HI C 0.02 M HC2H3O2 C 0.15 M Al(NO3)3 D 0.02 M CaCl2 D 0.15 M HC2H3O2 E Pure water E 0.15 M HI Answer A Slide 45 / 48 Slide 46 / 48 What would be the boiling point of a 200 mL solution of 0.78 M MgCl 2? Assume a Kb of 0.51 C/m for water and a density of the solution of 1.01 g/mL. 20 When a 2.3 gram sample of a non-ionic solute is added to 250 grams of water, the freezing point is found to be -3.45 C. Assuming a Kf of -1.86 C/m for water, what is the molar mass of the solute? Slide 47 / 48 Answer Answer 19 Which of the following solutions would have the lowest boiling point? 18 Answer Which of the following solutions would have the lowest freezing point? 17 Slide 48 / 48 21 What is the osmotic pressure of a solution created by adding 5.8 grams of NaCl to water to produce a 340 mL solution @10 C? Answer Next up....Equilibrium!!!
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