Solutions: Formation and Properties

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1 New Jersey Center for Teaching and Learning Slide 1 / 48 Progressive Science Initiative This material is made freely available at 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. Click to go to website: Slide 2 / 48 Solutions: Formation and Properties Solution Formation Slide 3 / 48 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. C 6H 12O 6(s) --> C 6H 12O 6(aq) Step What happens? Intermolecular forces between glucose (solute) molecules must weaken. Intermolecular forces between solvent (water) molecules must weaken. New coulombic attractions will form between the solute and solvent Enthalpy Change 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.

2 Heats of Solution Slide 4 / 48 The heat of solution will vary depending on the affinity of the solute for the solvent. Ethanol (CH 3CH 2OH) has very different heats of solution when dissolved in water (H 2O) and hexane (C 6H 14). Ethanol dissolved in water ( H = 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. Heats of Solution Slide 5 / 48 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 CH 3OH and CH 3CH 2OH C 6H 14 and C 7H 14 C 6H 6 and C 7H 8 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. Heats of Solution Slide 6 / 48 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 NH 4Cl are dissolved in 100 grams of 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? Energy lost by solution = g x 3.3 C x 4.2 J = 1460 J g C Expressed in kj/mol = kj /0.1 mol NH 4Cl = 14.6 kj/mol This process is endothermic and therefore will likely become more favorable as the temperature increases.

3 Solution Formation Slide 7 / 48 Ionic solutes dissociate into ions in aqueous solvent while covalent molecular solutes do not. Dissolution of NaCl in H 2O NaCl(s) --> Na + (aq) + Cl - (aq) Each ion becomes solvated by water molecules Dissolution of glucose (C 6H 12O 6) in H 2O C 6H 12O 6(s) --> C 6H 12O 6(aq) The entire glucose molecule becomes solvated by water molecules Ionic solutes are called electrolytes. Why? Since ionic solutes produce ions in solution resulting in increased electrical conductivity, there are referred to as electrolytes Electrolytes Slide 8 / 48 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. Comparing equimolar NaCl(aq) and MgCl 2(aq) NaCl(s) --> Na + (aq) + Cl - (aq) MgCl 2(s) --> Mg 2+ (aq) + 2Cl - (aq) Which compound is the stronger electrolyte? MgCl 2 produces Move 1 ion to of see Mg 2+ answer and 2 ions of Cl - when it dissociates, so it is the stronger electrolyte. Electrolytes Slide 9 / 48 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. Comparing equimolar HF(aq) and HBr(aq) HF(aq) --> H + (aq) + F - (aq) HBr(aq) --> H + (aq) + Br - (aq) Which compound is a stronger electrolyte? Since HBr is a strong acid, it produces many more ions compared to HF, a Move weak acid to see in which answer very few of the HF molecules have ionized.

4 1 Which of the following is NOT true regarding the formation of an aqueous glucose solution? Slide 10 / 48 Covalent bonds within the glucose molecule must A be broken Intermolecular coulombic forces will form between B glucose molecules and water molecules The hydrogen bonding network between water C molecules must be disrupted D The glucose molecule remains un-ionized E All of these are true 1 Which of the following is NOT true regarding the formation of an aqueous glucose solution? Slide 10 () / 48 Covalent bonds within the glucose molecule must A be broken Intermolecular coulombic forces will form between B glucose molecules and water molecules The hydrogen bonding network between water C molecules must be disrupted A D The glucose molecule remains un-ionized E All of these are true 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. Slide 11 / 48

5 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. Slide 11 () / kj/mole 3 How much would the temperature of a solution prepared by dissolving 10.6 grams of LiNO 2 in 300 grams of water increase? Assume a specific heat of solution of 4.2 J/gC and a heat of solution of LiNO 2 of kj/mol. Slide 12 / 48 3 How much would the temperature of a solution prepared by dissolving 10.6 grams of LiNO 2 in 300 grams of water increase? Assume a specific heat of solution of 4.2 J/gC and a heat of solution of LiNO 2 of kj/mol. Slide 12 () / C

6 4 Which of the following would be the strongest electrolyte when dissolved in water? Slide 13 / 48 A HCN B CH 3OH C C 6H 12O 6 D H 2SO 4 E HC 2H 3O 2 4 Which of the following would be the strongest electrolyte when dissolved in water? Slide 13 () / 48 A HCN B CH 3OH C C 6H 12O 6 D H 2SO 4 E HC 2H 3O 2 D 5 Which of the following correctly ranks the solutions from highest to lowest conductivity? Slide 14 / 48 A 0.1 M NaF > 0.1 M CH 3OH > pure water B 0.2 M AlCl 3 > 0.2 M NaF > pure water C pure water > 0.1 M NaF > 0.1 M CH 3OH D 0.1 M CH 3OH > 0.1 M AlCl 3 > 0.1 M NaF E None of these

7 5 Which of the following correctly ranks the solutions from highest to lowest conductivity? Slide 14 () / 48 A 0.1 M NaF > 0.1 M CH 3OH > pure water B 0.2 M AlCl 3 > 0.2 M NaF > pure water C pure water > 0.1 M NaF > 0.1 M CH 3OH B D 0.1 M CH 3OH > 0.1 M AlCl 3 > 0.1 M NaF E None of these 6 Which of the following pairs of liquids would form the most IDEAL solution? Slide 15 / 48 A C 6H 14(l) and H 2O(l) B CH 3OH(l) and C 6H 14(l) C CH 3OH(l) and CH 3COCH 3(l) D C 5H 12(l) and C 6H 14(l) E None of these 6 Which of the following pairs of liquids would form the most IDEAL solution? Slide 15 () / 48 A C 6H 14(l) and H 2O(l) B CH 3OH(l) and C 6H 14(l) C CH 3OH(l) and CH 3COCH 3(l) D C 5H 12(l) and C 6H 14(l) E None of these D

8 Colligative Properties Slide 16 / 48 Colligative properties of solutions depend exclusively on the number of solute particles in the solution, not on their kind. Examples of colligative properties Vapor pressure lowering Boiling point elevation 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. Vapor Pressure Slide 17 / 48 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. Vapor Liquid Vapor Pressure Slide 18 / 48 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. H2O CH3COCH3 VP = 55.3 mm 40 C H-Bonds less evaporation VP = 400 mm 40 C no H-bonds more evaporation

Vapor Pressure Slide 19 / 48 The vapor pressure is directly proportional to the temperature. As heat is added, more evaporation results, leading to a higher vapor pressure.

When a liquid's vapor pressure equals the atmospheric/ external pressure it will boil.

9 Vapor Pressure Slide 19 / 48 The vapor pressure is directly proportional to the temperature. As heat is added, more evaporation results, leading to a higher vapor pressure. VP H2O vs Temp Note the relationship is not linear. Vapor Pressure Slide 20 / 48 VP of Various Substances vs. Temp. Note the larger hydrocarbons have a lower vapor pressure at a given temperature due to higher LDF's. 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. Vapor Pressure Slide 21 / 48 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 Vapor Pressure 1 atm external pressure C mm Hg not boiling! C 760 mm Hg 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.

10 7 Which of the following would NOT occur when a solute is added to a solvent? Slide 22 / 48 The freezing point of the solution will be lower than A the pure solvent The boiling point of the solution will be higher than B the pure solvent The osmotic pressure of the solution will be higher C than the pure solvent D The vapor pressure of the solution will be lower than the pure solvent E All of these would occur 7 Which of the following would NOT occur when a solute is added to a solvent? Slide 22 () / 48 The freezing point of the solution will be lower than A the pure solvent The boiling point of the solution will be higher than B the pure solvent E The osmotic pressure of the solution will be higher C than the pure solvent D The vapor pressure of the solution will be lower than the pure solvent E All of these would occur 8 Which of the following would have the lowest vapor pressure at a given temperature? Slide 23 / 48 A C 6H 14(l) B C 4H 10(l) C H 2O(l) D CH 3COCH 3(l)

11 8 Which of the following would have the lowest vapor pressure at a given temperature? Slide 23 () / 48 A C 6H 14(l) B C 4H 10(l) C H 2O(l) D CH 3COCH 3(l) B 9 Which of the following would be TRUE? Slide 24 / 48 A Water will boil at a lower temperature at high altitude because its Coulombic attractions are stronger at these pressures B Water will boil at a lower temperature at high altitude because its vapor pressure is higher at these altitudes. Water will boil at a lower temperature at high C altitude because atmospheric pressure is lower D Water will boil at a higher temperature at high altitude because atmospheric pressure is higher 9 Which of the following would be TRUE? Slide 24 () / 48 A Water will boil at a lower temperature at high altitude because its Coulombic attractions are stronger at these pressures 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 C altitude because atmospheric pressure is lower D Water will boil at a higher temperature at high altitude because atmospheric pressure is higher

12 Vapor Pressure of Solutions Adding solute to create a solution lowers the solutions vapor pressure and raises the boiling point. Slide 25 / 48 When solute is mixed with the solvent, attractions form between the two, resulting in less solvent molecules evaporating and thereby lowering the vapor pressure. Higher VP Cl- Na+ Lower VP Pure H 2O 0.1 M NaCl(aq) Attractions between the ions and water reduce evaporation and the vapor pressure. Vapor Pressure of Solutions Slide 26 / 48 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 CH 3OH, C 6H 12O 6, CH 3COCH 3 Ionic solutes have Van't Hoff factors that correlate to the number of ions present NaCl(aq) --> Na + (aq) + Cl - (aq) VHF = 2 Al(NO 3) 3(aq) --> Al 3+ (aq) + 3NO 3- (aq) VHF = 4 A given quantity of Al(NO 3) 3 will have 4x the impact of a molecule solute and 2x the impact of NaCl(aq) on the vapor pressure of a solution. Vapor Pressure of Solutions The vapor pressure of a solution can be calculated using Raoult's Law. Slide 27 / 48 VP solution = (X solvent)(p solvent) where... X solvent = mole fraction of solvent P solvent = pressure of pure solvent 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.

13 Vapor Pressure of Solutions The vapor pressure of a solution can be calculated using Raoult's Law. Slide 28 / 48 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? Find mole fraction of solvent 2.3 g NaCl = 0.05 moles NaCl x VHF(2) = 0.10 mol solute 108 g H 2O = 6 moles H 2O mole fraction = 6/6.1 = 0.98 Find VP solution VPsolution = (0.98)(18.9 mm Hg) = 18.6 mm Hg Vapor Pressure of Solutions The vapor pressure of solution made from two liquids can be determined using an expanded version of Raoult's Law Slide 29 / 48 VP solution = (X liquida)(p liquid A) + (X liquidb)(p liquid B) Both liquids evaporate and therefore contribute to the vapor pressure of the solution Example: What is the vapor pressure of a solution made by mixing 1 mole of acetone (C 3H 6O) with 4 moles of ethanol (CH 3CH 2OH); assuming vapor pressures of each being 28 mm Hg and 17.4 mm Hg at a given temperature. VP solution = (1/5)28 mm Hg + (4/5)17.4 mm Hg = 19.5 mm Hg 10 When a solute is mixed with a solvent, the resulting solution will have... Slide 30 / 48 A A higher vapor pressure and boiling point B A higher vapor pressure and lower boiling point C A lower vapor pressure and a higher boiling point D A lower vapor pressure and a lower boiling point

14 10 When a solute is mixed with a solvent, the resulting solution will have... Slide 30 () / 48 A A higher vapor pressure and boiling point B A higher vapor pressure and lower boiling point C A lower vapor pressure and a higher boiling point D A lower vapor pressure and a lower C boiling point 11 Which of the following substances/mixtures would have the highest vapor pressure? Slide 31 / 48 A 0.1 M NaCl B 0.2 M CH 3OH C 0.1 M CH 3CH 2OH D 0.1 M MgCl 2 E 0.05 M Al(NO 3) 3 11 Which of the following substances/mixtures would have the highest vapor pressure? Slide 31 () / 48 A 0.1 M NaCl B 0.2 M CH 3OH C 0.1 M CH 3CH 2OH C D 0.1 M MgCl 2 E 0.05 M Al(NO 3) 3

15 12 Which of the following substances/mixtures would have the lowest vapor pressure? Slide 32 / 48 A 0.2 M LiC 2H 3O 2 B 0.3 M HNO 3 C 0.4 M HC 2H 3O 2 D 0.1 M Al(NO 3) 3 E Pure H 2O 12 Which of the following substances/mixtures would have the lowest vapor pressure? Slide 32 () / 48 A 0.2 M LiC 2H 3O 2 B 0.3 M HNO 3 C 0.4 M HC 2H 3O 2 D 0.1 M Al(NO 3) 3 E Pure H 2O B 13 What is the vapor pressure of a solution prepared by adding 1.8 grams of glucose (C 6H 12O 6) to 200 ml of water (D =1 g/ml). Assume water vapor has a vapor pressure of 12.3 mm Hg at this temperature. Slide 33 / 48

16 13 What is the vapor pressure of a solution prepared by adding 1.8 grams of glucose (C 6H 12O 6) to 200 ml of water (D =1 g/ml). Assume water vapor has a vapor pressure of 12.3 mm Hg at this temperature. Slide 33 () / mmhg 14 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? Slide 34 / 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? Slide 34 () / mmhg

17 15 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) Slide 35 / 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) Slide 35 () / Boiling Point Elevation Slide 36 / 48 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). T b = K b*m*i Lower BP Higher BP Pure H 2O 0.1 m NaCl(aq) The solute impedes evaporation thereby requiring more heat to be added to get the vapor pressure = atmospheric pressure.

Boiling Point Elevation Slide 37 / 48 Measuring the boiling point elevation of a solution can be used to determine the molar mass of a non-volatile molecular solute.

5 C x 1 m = 0.198 m 2.53 C Use molality and kg of solvent to get moles solute 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.

18 Boiling Point Elevation Slide 37 / 48 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 Tb and Kb to find molality 0.5 C x 1 m = m 2.53 C Use molality and kg of solvent to get moles solute kg solvent x mol solute = mol 1 kg solvent Divide the grams by moles 3.4 g/0.059 mol = 57.6 g/mol 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. Slide 38 / 48 The degree to which the freezing point has been depressed can be calculated by... Tf = K f*m*i where K f = freezing point constant of solvent The molar mass of solute can be calculated as done with boiling point elevation. Freezing Point Depression Slide 39 / 48 Depressing the freezing point of a solvent has many applications. 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!

19 Osmotic Pressure Slide 40 / 48 The osmotic pressure of a solution is the pressure required to prevent the osmosis (flow) of water into it. Pure H 2O Pure H 2O No net movement of water Osmotic Pressure = 0 Pure H 2O 0.1 M NaCl Pure H 2O 0.1 M NaCl Net movement of water into solution Osmotic Pressure of 2.44 mmhg needed to prevent net-flow of water Osmotic Pressure The osmotic pressure is directly proportional to the M of the solution, the Van't Hoff factor of the solute, and the temperature. Slide 41 / 48 Osmotic Pressure = im*r*t M = Molarity, R = L*atm/mol K, T = Kelvin temperature, i = VHF Plants use high concentrations of solute within their cells to draw water in to create the turgor pressure necessary to keep the cells rigid. 16 Compared to the pure solvent, a solution will have a... Slide 42 / 48 A B C D E higher boiling point and lower freezing point higher boiling and freezing point lower boiling point and higher freezing point lower boiling point and lower freezing point the same freezing and boiling points

20 16 Compared to the pure solvent, a solution will have a... Slide 42 () / 48 A B C D E higher boiling point and lower freezing point higher boiling and freezing point lower boiling point and higher freezing point lower boiling point and lower freezing point the same freezing and boiling points A 17 Which of the following solutions would have the lowest freezing point? Slide 43 / 48 A 0.02 M CH 3 OH B 0.02 M KNO 3 C 0.02 M HC 2 H 3 O 2 D 0.02 M CaCl 2 E Pure water 17 Which of the following solutions would have the lowest freezing point? Slide 43 () / 48 A 0.02 M CH 3 OH B 0.02 M KNO 3 C 0.02 M HC 2 H 3 O 2 D 0.02 M CaCl 2 E Pure water D

21 18 Which of the following solutions would have the lowest boiling point? Slide 44 / 48 A 0.45 M HC 2 H 3 O 2 B 0.45 M HI C 0.15 M Al(NO 3 ) 3 D 0.15 M HC 2 H 3 O 2 E 0.15 M HI 18 Which of the following solutions would have the lowest boiling point? Slide 44 () / 48 A 0.45 M HC 2 H 3 O 2 B 0.45 M HI C 0.15 M Al(NO 3 ) 3 D 0.15 M HC 2 H 3 O 2 E 0.15 M HI D 19 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. Slide 45 / 48

22 19 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. Slide 45 () / C 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 C. Assuming a Kf of C/m for water, what is the molar mass of the solute? Slide 46 / 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 C. Assuming a Kf of C/m for water, what is the molar mass of the solute? Slide 46 () / 48 5 g/mol

23 21 What is the osmotic pressure of a solution created by adding 5.8 grams of NaCl to water to produce a 340 ml C? Slide 47 / What is the osmotic pressure of a solution created by adding 5.8 grams of NaCl to water to produce a 340 ml C? Slide 47 () / mmhg Slide 48 / 48 Next up...equilibrium!!!

PSI AP Chemistry: Solutions Practice Problems

PSI AP Chemistry: Solutions Practice Problems Name Solutions: Mixtures, Solubility and Concentration Classwork 1. A student determined that there were 0.032 grams of oxygen gas dissolved in a 200.0 ml