Chapter 18 Solutions http://online.ualr.edu/rebelford/chem1402/c1402lec/c1402lec.htm for reference!

Section 18.1 Properties of Solutions Objectives: Identify the factors that determine the rate at which a solute dissolves Calculate the solubility of a gas in a liquid under various pressure conditions

Definitions Soluble - capable of being dissolved. Solution - homogeneous mixture of 2 or more substances in a single phase. Solvent - the dissolving medium in a solution. Solute - the substance dissolved in a solution. Phase – a physically distinct & mechanically separable portion of a dispersion or solution. Combine slides 2 & 3 or fix 3

Factors that Increase the Rate of Dissolving A substance dissolves faster if- It is stirred or shaken (agitation). This enables fresh solvent to come in contact with the solute. The particles are made smaller. This increases the surface area. The temperature is increased. This increases the kinetic energy, hence increases particle movement.

Solubility Solubility- the amount that dissolves in a given quantity of a solvent at a given temperature to produce a saturated solution.

Solubility Saturated solution - Contains the maximum amount of solid dissolved. Unsaturated solution - Can dissolve more solute. Supersaturated - A solution that is temporarily holding more than it can, a seed crystal will make it come out.

Solubility UNSATURATED SOLUTION more solute dissolves no more solute dissolves SUPERSATURATED SOLUTION becomes unstable, crystals form concentration

Solubility For liquids dissolved in liquids: Miscible – Two liquids are said to be miscible if they dissolve in each other. Ex: Water and ethanol Immiscible – Liquids that are insoluble in each other. Ex: Oil and water

Factors Affecting Solubility Solids in liquids - temperature goes up the solubility goes up. Gases in a liquid - the temperature goes up the solubility goes down. Gases in a liquid - as the pressure goes up the solubility goes up.

Solubility Solids are more soluble at... high temperatures. Gases are more soluble at... low temperatures. high pressure (Henry’s Law)

Henry’s Law (A friend of Dalton’s) At a given temperature the solubility of a gas in a liquid (S) is directly proportional to the pressure of the gas above the liquid (P)

S2 = 0.2 g/L S1 = P1 S2 P2 0.77 g/L= 350 kPa S2 100 kPa Ex #1: If the solubility of a gas in water is 0.77 g/L at 350 kPa of pressure, what is its solubility, in g/L at 100 kPa of pressure. (Temperature is constant) S1 = P1 S2 P2 0.77 g/L= 350 kPa S2 100 kPa (100 kPa)(0.77 g/L) = (350 kPa)S2 S2 = 0.22 g/L S2 = 0.2 g/L

P2 = 260 kPa S1 = P1 S2 P2 3.6 g/L = 100. kPa 9.5 g/L P2 Ex #2: A gas has a solubility in water at 0°C of 3.6 g/L at a pressure of 100. kPa. What pressure is needed to produce an aqueous solution containing 9.5 g/L of the same gas at 0 °C? S1 = P1 S2 P2 3.6 g/L = 100. kPa 9.5 g/L P2 (100. kPa)(9.5 g/L) = (3.6 g/L)P2 P2 = 263.89 kPa P2 = 260 kPa

Section 18.1 Properties of Solutions Did We Meet Our Objectives? Identify the factors that determine the rate at which a solute dissolves Calculate the solubility of a gas in a liquid under various pressure conditions

Section 18.2 Concentrations of Solutions Objectives: Solve problems involving the molarity of a solution Describe how to prepare dilute solutions from more concentrated solutions of known molarity Explain what is meant by percent by volume and percent by mass solutions

Molarity The concentration of a solution is a measure of the amount of solute that is dissolved in a given quantity of solvent. A dilute solution is one that contains only a low concentration of solute. A concentrated solution contains a high concentration of solute.

Molarity Molarity (M) - the number of moles of solute dissolved per liter of solution. Molarity is also known as molar concentration. It is read as the numerical value followed by the word “molar.” 6 M HCl

0.95 M NaF M = moles Volume (L) 10.0 g NaF x 1 mol NaF = 0.238 moles Ex #3: Find the molarity of a 250 mL solution containing 10.0 g of NaF. M = moles Volume (L) 10.0 g NaF x 1 mol NaF = 0.238 moles 41.98 g NaF M = 0.238 mol 0.25 L M = 0.952 0.95 M NaF

n = 4.5 mol NaCl M = moles Volume (L) 0.75 M NaCl = n . 6.0 L Ex #4: How many moles of NaCl are needed to make 6.0 L of a 0.75 M NaCl solution? M = moles Volume (L) 0.75 M NaCl = n . 6.0 L n = (0.75 mol)(6.0 L) L n = 4.5 mol NaCl

Ex #5: How many grams of CaCl2 are needed to make 625 mL of a 2 Ex #5: How many grams of CaCl2 are needed to make 625 mL of a 2.0 M solution? M = moles Volume (L) 2.0 M CaCl2 = n . 0.625 L n = (2.0 mol)(0.625 L) L . n = 1.25 mol CaCl2 1.25 mol CaCl2 x 110.98 g = 138.725 g CaCl2 1 mol . m = 140 g CaCl2

Moles before = the moles after Making Dilutions The number of moles of solute doesn’t change if you add more solvent. M = n/V n = M V Moles before = the moles after n1 = n2 M1 x V1 = M2 x V2

V1 = 95 mL M1V1 = M2V2 (15.8 M)V1 = (6.0 M)(250 mL) Ex #6: What volume of 15.8 M HNO3 is required to make 250 mL of a 6.0 M solution? M1V1 = M2V2 (15.8 M)V1 = (6.0 M)(250 mL) V1 = 94.93670886 mL V1 = 95 mL

0.46 M solution M1V1 = M2V2 (0.88 M)(2.0 L) = (M2)(3.8 L) Ex #7: 2.0 L of a 0.88 M solution are diluted to 3.8 L. What is the new molarity? M1V1 = M2V2 (0.88 M)(2.0 L) = (M2)(3.8 L) M2 = 0.463157895 0.46 M solution

Percent solutions Percent by volume %(v/v) – A convenient way to measure if both solute & solvent are liquids % by volume = Volume of solute x 100% Volume of solution

% by volume = Volume of solute x 100% Ex #8: What is the percent solution if 25 mL of CH3OH is diluted to a volume of 150 mL with water? % by volume = Volume of solute x 100% Volume of solution % by volume = 25 mL x 100% = 16.6667 % 150 mL . 17% CH3OH

Percent solutions Percent by mass % (m/v) - is used for solids dissolved in liquids (more common). % by mass = Mass of solute(g) x 100% Volume of solution(mL)

5.9% NaCl % by mass = Mass of solute(g) x 100% Volume of solution(mL) Ex #9: 4.8 g of NaCl are dissolved in 82 mL of solution. What is the percent of the solution? % by mass = Mass of solute(g) x 100% Volume of solution(mL) % by mass = 4.8 g NaCl x 100% 82 mL % by mass = 5.853658537 5.9% NaCl

Ex #10: How many grams of salt are there in 52 mL of a 6.3% solution? % by mass = Mass of solute(g) x 100% Volume of solution(mL) 6.3% = m x 100% 52 mL m = 3.276 g m = 3.3 g salt

Concentrations Describing Concentration % by mass - medicated creams % by volume - rubbing alcohol ppm, ppb - water contaminants molarity - used by chemists molality - used by chemists

Section 18.2 Concentrations of Solutions Did We Meet Our Objectives? Solve problems involving the molarity of a solution Describe how to prepare dilute solutions from more concentrated solutions of known molarity Explain what is meant by percent by volume and percent by mass solutions

Section 18.3 Colligative Properties of Solutions Objectives: Explain on a particle basis why a solution has a lower vapor pressure than the pure solvent of that solution Explain on a particle basis why a solution has an elevated boiling point and a depressed freezing point compared with the pure solvent.

Decrease in Vapor Pressure Colligative properties depend only on the number of particles dissolved in a given mass of solvent. Not based on the kind of particle Three important colligative properties of solutions are: Vapor pressure lowering Boiling point elevation Freezing point depression

Number of Dissolved Particles Add 3 formula units of KMnO4 Add 3 formula units of Na2CO3 Produces 3 K+ ions and 3 MnO4- ions, total of 6 ions Produces 6 Na+ ions and 3 CO32- ions, total of 9 ions

Number of Dissolved Particles KCl BeF2 KCl → K+ + Cl- BeF2 → Be2+ + 2F- KCl → 2 mols ions BeF2 → 3 mols ions MgO C6H12O MgO → Mg2+ + O2- C6H12O → C6H12O MgO→ 2 mols ions C6H12O → 1 mol molecule

Vapor Pressure Vapor pressure – the pressure above a substance in a sealed container

Vapor Pressure Vapor pressure is always lowered if a nonvolatile (not easily vaporized) solute is added The decrease is proportional to the # of particles in solution NaCl breaks into 2 ions, but sugar stays as 1 molecule, so NaCl will lower the vapor pressure more The higher the amount of moles of product, the lower the vapor pressure. When a ions become solvated, they become surrounded by shells of water Attractions keep molecules from escaping.

Boiling Point Elevation Boiling point – the temperature at which the vapor pressure of a liquid equals the atmospheric pressure. Adding a solute lowers the vapor pressure Lower vapor pressure - higher boiling point. Boiling Point Elevation (tb) – the b.p. of a soln is higher than the b.p. of the pure solvent.

Freezing Point Depression Adding a solute makes the freezing point lower. Solids form when molecules make an orderly pattern. The solute molecules break up the orderly pattern. Freezing Point Depression (tf) – the f.p. of a solution is lower than the f.p. of pure solvent

Concept Practice KF → K+ + F- = 2 mols MgF2 → Mg2+ + 2F- = 3 mols If equal numbers of moles of KF and MgF2 are dissolved in equal volumes of water, state which solution has the highest: Boiling point Freezing point Vapor pressure KF → K+ + F- = 2 mols MgF2 → Mg2+ + 2F- = 3 mols

Section 18.3 Colligative Properties of Solutions Did We Meet Our Objectives? Explain on a particle basis why a solution has a lower vapor pressure than the pure solvent of that solution Explain on a particle basis why a solution has an elevated boiling point and a depressed freezing point compared with the pure solvent.

Section 18.4 Calculations Involving Colligative Properties Objectives: Calculate the molality and mole fraction of a solution Calculate the molar mass of a molecular compound from the freezing-point depression or boiling-point elevation of a solution of the compound

Molality Another unit for concentration Molality (m) is the number of moles of solute dissolved in 1 kg of solvent

kilograms of solvent vs. liters of solution Molality vs. Molarity kilograms of solvent vs. liters of solution

3.2 m MgCl2 m = mol or m = n . mass (kg) m Ex #11: Find the molality of a solution containing 75 g of MgCl2 in 250 mL of water. m = mol or m = n . mass (kg) m The density of water is 1 g/mL, so 250 mL of water = 250 g 75 g MgCl2 x 1 mol MgCl2 = 0.787732381 mol MgCl2 95.21 g MgCl2 m = 0.787732381 mol .250 kg m = 3.15093 mol/kg 3.2 m MgCl2

m = 45.0 g NaCl m = mol or m = n . mass (kg) m 1.54 m = n . 0.500 kg Ex #12: How many grams of NaCl are required to make a 1.54 m solution using 0.500 kg of water? m = mol or m = n . mass (kg) m 1.54 m = n . 0.500 kg n = 0.77 mol 0.77 mol NaCl x 58.44 g NaCl = 44.9988 g NaCl 1 mol NaCl m = 45.0 g NaCl

Mole Fraction The ratio of the moles of a solute in solution to the total number of moles of solvent and solute is the mole fraction of that solute. XA = nA XB = nB . nA + nB nA + nB

Ex #13: Compute the mole fraction of each component in a solution of 1 Ex #13: Compute the mole fraction of each component in a solution of 1.25 mol of ethylene glycol (EG) and 4.00 mole water. 1.25 (1.25 + 4.00) 1.25 = 0.238 5.25 . 4.00 4.00 = 0.762 5.25 .

Ex #14: What is the mole fraction of each component in a solution made by mixing 300.0 g of ethanol (C2H5OH) and 500.0 g of water? 300.0 g C2H5OH x 1 mol C2H5OH = 6.510 mol C2H5OH 46.08 g C2H5OH 500.0 g H2O x 1 mol H2O = 27.75 mol H2O 18.02 g H2O XA = nA XB = nB . nA + nB nA + nB Xethanol = 6.510 Xwater = 27.75 . 6.510 + 27.75 6.510 + 27.75

Nonelectrolytes (covalent) Determining the Number of Particles in a Covalent and an Ionic Compound Electrolytes (ionic) dissociate into ions when dissolved 2 or more particles Nonelectrolytes (covalent) remain intact when dissolved 1 particle

Ex #15: Determine the total number of particles when the following dissolve. NaCl BaCl2 CCl4

Boiling-Point Elevation ΔTb = Kb x m Tb – Boiling-Point Elevation (°C) Kb – Molal Boiling-Point Elevation Constant (°C/m) m – Molality (m = moles/kg) Remember to multiply by the number of particles.

Ex #16: What is the boiling point of a 1.50 m NaCl solution? ΔTb = Kb x m ΔTb = (0.512 °C/m) x (1.50 m x 2) ΔTb =1.55 °C Tsolution = 100.00 °C + 1.54 °C Tsolution = 101.54 °C

Ex #17: At what temperature will a solution that is composed of 0 Ex #17: At what temperature will a solution that is composed of 0.73 moles of glucose in 225 g of phenol boil (b.p. of phenol 181.7°C)? mphenol = mol/kg = (0.73 mol/0.225 kg) = 3.2 m ΔTb = Kb x m ΔTb = (3.56 °C/m) x (3.2 m) ΔTb = 11.392 °C = 11 °C Tsolution = 181.7 °C + 11 °C = 192.7 °C Tsolution = 193 °C

Freezing-Point Depression ΔTf = Kf x m Tf (Freezing-Point Depression - °C) lowers Kf (Molal Freezing-Point Depression Constant - °C/m) m (Molality - m = moles/kg) Remember to multiply by the number of particles.

Ex #18: Find the freezing point of a saturated solution of NaCl containing 28 g NaCl in 100. mL water. mNaCl = mol/kg = (0.48 mol/0.100 kg) = 4.8 m ΔTf = Kf x m ΔTf = (1.86 °C/m) x (4.8 m x 2) ΔTf = 17.856 °C = 18 °C Tsolution = 0.0 °C - 18 °C = -18 °C Tsolution = -18 °C

Ex #19: What is the boiling point of a solution made by dissolving 1 Ex #19: What is the boiling point of a solution made by dissolving 1.20 moles of NaCl in 750 g of water? What is the freezing point? ΔTb = Kb x m ΔTb = (0.512 °C/m) x (1.6 m x 2) ΔTb = 1.6384 °C = 1.6 °C Tbp = 100.0 °C + 1.6 °C = 101.6 °C Tbp = 101.6 °C ΔTf = Kf x m ΔTf = (1.86 °C/m) x (1.6 m x 2) ΔTf = 5.952 °C = 6.0 °C Tfp = 0.0 °C – 6.0 °C = -6.0 °C Tfp = -6.0 °C

Molar Mass One can use the changes in boiling and freezing points to determine the molar mass of a substance. Molar Mass = Mass of solute Moles of solute

Ex #20: A solution containing 16 Ex #20: A solution containing 16.9 g of a nonvolatile molecular compound in 250 g of water has a freezing point of -0.744 ˚C. What is the molar mass of the substance? ΔTf = Kf x m 0.744 °C = (1.86 °C/m) x (m) m = 0.400 m m = mol mass (kg) 0.400 m = mol/(0.250 kg) n = 0.100 mol Molar Mass = Mass of solute Moles of solute M = 16.9 g = 169 g/mol 0.100 mol .

Section 18.4 Calculations Involving Colligative Properties Did We Meet Our Objectives? Calculate the molality and mole fraction of a solution Calculate the molar mass of a molecular compound from the freezing-point depression or boiling-point elevation of a solution of the compound