Chapter 15: Solutions

Solution Solution – a homogeneous mixture of two or more substances in a single physical state. Solution – a homogeneous mixture of two or more substances in a single physical state. Properties of solutions – Properties of solutions – The particles are very small (atoms, molecules or ions) The particles are very small (atoms, molecules or ions) the particles in a solution are evenly distributed or uniformly mixed (a spoonful of lemonade tastes the same as the whole glass) the particles in a solution are evenly distributed or uniformly mixed (a spoonful of lemonade tastes the same as the whole glass)

Parts of solutions Solute – the thing that gets dissolved Solute – the thing that gets dissolved Solvent – the thing that does the dissolving Solvent – the thing that does the dissolving Soluble – something that can be dissolved (salt and sugar) Soluble – something that can be dissolved (salt and sugar) Insoluble – a substance that cannot be dissolved (Hg and oil) Insoluble – a substance that cannot be dissolved (Hg and oil)

Types of solutions Solid Solutions – alloys – made by melting different metals and then cooling them Solid Solutions – alloys – made by melting different metals and then cooling them Makes material stronger, higher melting points and greater resistance to corrosion Makes material stronger, higher melting points and greater resistance to corrosion Ex. Dental filling – mercury in silver Ex. Dental filling – mercury in silver Sterling silver – copper in silver Sterling silver – copper in silver 16 karat gold – gold, copper and silver 16 karat gold – gold, copper and silver Coinage bronze – copper, tin and zinc Coinage bronze – copper, tin and zinc

Types of solutions Gaseous solutions – air we breathe. Nitrogen with oxygen in it Gaseous solutions – air we breathe. Nitrogen with oxygen in it If gas molecules mix they become a solution. Anytime gases are near each other they will mix, therefore any gas mixture is a solution. If gas molecules mix they become a solution. Anytime gases are near each other they will mix, therefore any gas mixture is a solution.

Types of solutions Liquid Solutions – MOST COMMON! the solvent and the solution are both liquids. The solute can be a gas, liquid or solid. Liquid Solutions – MOST COMMON! the solvent and the solution are both liquids. The solute can be a gas, liquid or solid. Miscible – liquids that can be mixed in any amount (water and ethanol) Miscible – liquids that can be mixed in any amount (water and ethanol) Immiscible – liquids that cannot mix in any proportion (oil and water) Immiscible – liquids that cannot mix in any proportion (oil and water)

Types of solutions Aqueous solutions – solutions with water as the solvent Aqueous solutions – solutions with water as the solvent 2 types 2 types 1. When ionic compounds dissolve, ions are present and make an Electrolyte. (NaCl) 1. When ionic compounds dissolve, ions are present and make an Electrolyte. (NaCl) 2. When molecular compounds dissolve, no ions are present and make a nonelectrolyte. (Sugar) 2. When molecular compounds dissolve, no ions are present and make a nonelectrolyte. (Sugar)

15-2 Concentration of Solutions Concentration – the amount of solute in a given amount of solvent. Concentration – the amount of solute in a given amount of solvent. Molarity = moles of solute Molarity = moles of solute Liters of solution Liters of solution What is the Molarity of a NaOH solution if 10.0g of NaOH is dissolved in enough solvent to make 0.100L of solution? What is the Molarity of a NaOH solution if 10.0g of NaOH is dissolved in enough solvent to make 0.100L of solution?

You try it!!! Find the Molarity of a solution formed by mixing 10.0g of H 2 SO 4 with enough water to make 100.0mL of solution Find the Molarity of a solution formed by mixing 10.0g of H 2 SO 4 with enough water to make 100.0mL of solution M = 10.0g H 2 SO 4 x 1 mol H 2 SO 4 = 1.02 mol/L M = 10.0g H 2 SO 4 x 1 mol H 2 SO 4 = 1.02 mol/L 0.100L sol 98.1 g H 2 SO L sol 98.1 g H 2 SO 4

Molality = moles of solute Molality = moles of solute Kg of Solvent Kg of Solvent If 18.0g of C 6 H 12 O 6 is dissolved in 1Kg of water. If 18.0g of C 6 H 12 O 6 is dissolved in 1Kg of water. 18.0g C6H12O6 x 1mol C6­H12O6 = mol C6­H12O6 / 1Kg H2O 18.0g C6H12O6 x 1mol C6­H12O6 = mol C6­H12O6 / 1Kg H2O 1Kg H2O 180g C6­H12O6 1Kg H2O 180g C6­H12O6

Mole Fraction = moles of component Mole Fraction = moles of component Total moles of solution Total moles of solution What is the mole fraction of SO 2 in a gas containing 128.0g of SO 2 dissolved in every 1500.g of CO 2 What is the mole fraction of SO 2 in a gas containing 128.0g of SO 2 dissolved in every 1500.g of CO 2

1st change grams to moles 1st change grams to moles 128g SO 2 x 1 mole SO 2 = moles SO 2 128g SO 2 x 1 mole SO 2 = moles SO g SO g SO g CO 2 x 1 mole CO 2 = moles CO g CO 2 x 1 mole CO 2 = moles CO g CO g CO 2 XSO2 = moles of SO 2 XSO2 = moles of SO 2 Total moles of solution Total moles of solution 1.999moles SO moles SO 2 (1.999 moles SO moles CO 2 ) (1.999 moles SO moles CO 2 ) = =

Example. A gas mixture contains 50.4 g of N 2 O and 65.2g of O 2 what is the mole fraction of N 2 O? (0.360) A gas mixture contains 50.4 g of N 2 O and 65.2g of O 2 what is the mole fraction of N 2 O? (0.360)

15-3 The formation of Solutions How they form How they form Why does salt dissolve? The particles are attracted to the water. The water attaches itself on the face of NaCl and pulls it apart. Then the water surrounds the ions. Why does salt dissolve? The particles are attracted to the water. The water attaches itself on the face of NaCl and pulls it apart. Then the water surrounds the ions. solvation - the interaction between solute and solvent solvation - the interaction between solute and solvent hydration – when the solvent is water hydration – when the solvent is water

energy is needed when bonds are broken between the solute and solvent energy is needed when bonds are broken between the solute and solvent when the solvent and solute are attracted energy is released. when the solvent and solute are attracted energy is released. Cold packs = NH 4 OH Cold packs = NH 4 OH Hot packs = Na 2 S 2 O 3 Hot packs = Na 2 S 2 O 3

Solubility – the amount of solute that will dissolve in a specific solvent under given conditions. Solubility – the amount of solute that will dissolve in a specific solvent under given conditions. Depends on… Depends on… 1. Nature of solute and solvent – polar compounds dissolve polar compounds. IE “likes dissolve likes” cholesterol is nonpolar and fat is nonpolar 1. Nature of solute and solvent – polar compounds dissolve polar compounds. IE “likes dissolve likes” cholesterol is nonpolar and fat is nonpolar 2. Temperature – As temperature increases, gas particles get more energy and solubility of a gas decreases (better to keep soda cold!) 2. Temperature – As temperature increases, gas particles get more energy and solubility of a gas decreases (better to keep soda cold!)

Solubility of a solid – as temp increases, solubility increases (sugar to iced tea or hot tea) Solubility of a solid – as temp increases, solubility increases (sugar to iced tea or hot tea) Pressure – solubility of a gas is increases as the pressure increases because when the pressure is increases, the gas particles hit the solution faster. Pressure – solubility of a gas is increases as the pressure increases because when the pressure is increases, the gas particles hit the solution faster. Scuba divers – nitrogen is dissolved in your blood because of the extra pressure, if they come up too fast the nitrogen will bubble out of your blood. (bends) Scuba divers – nitrogen is dissolved in your blood because of the extra pressure, if they come up too fast the nitrogen will bubble out of your blood. (bends)

Factors affecting the Rate of Dissolving surface area – the greater the surface area, the faster the solid can be dissolved. (big ice cubes or little ones) surface area – the greater the surface area, the faster the solid can be dissolved. (big ice cubes or little ones) Stirring – allows the solute to be exposed to all of the solvent not just the top Stirring – allows the solute to be exposed to all of the solvent not just the top Temperature – warmer is faster Temperature – warmer is faster

Saturation – if a solution contains as much solute as can possibly be dissolved under the existing conditions of temperature and pressure Saturation – if a solution contains as much solute as can possibly be dissolved under the existing conditions of temperature and pressure Saturated and concentrated are not the same things. A solution can be saturated with only a little amount of solute. Saturated and concentrated are not the same things. A solution can be saturated with only a little amount of solute.

Unsaturated – a solution that has less than the maximum amount of solute Unsaturated – a solution that has less than the maximum amount of solute Supersaturated – contains a greater amount of solute than needed to form a saturated solution. The extra solute will eventually go back into solution. (rock candy) Supersaturated – contains a greater amount of solute than needed to form a saturated solution. The extra solute will eventually go back into solution. (rock candy)

15-4 Colligative Properties – depend on the collective effort f the solute properties and not their identity Colligative Properties – depend on the collective effort f the solute properties and not their identity. 1. Vapor pressure reduction – when a nonvolatile solute is added to a solvent, the solute takes up space at the surface which prevents some of the solvent from leaving. Gases are still returning to the liquid at the same rate. This reduces the vapor pressure of the solution. 1. Vapor pressure reduction – when a nonvolatile solute is added to a solvent, the solute takes up space at the surface which prevents some of the solvent from leaving. Gases are still returning to the liquid at the same rate. This reduces the vapor pressure of the solution.

2. Boiling point Elevation – antifreeze is added to a car to make the water not boil. Antifreeze is a nonvolatile substance so it reduces the vapor pressure and increases the boiling point because it takes longer to reach atmospheric pressure 2. Boiling point Elevation – antifreeze is added to a car to make the water not boil. Antifreeze is a nonvolatile substance so it reduces the vapor pressure and increases the boiling point because it takes longer to reach atmospheric pressure Eq. ∆Tb = Kbm Eq. ∆Tb = Kbm Where ∆Tb = boiling point elevation Where ∆Tb = boiling point elevation Kb = molal boilin point constant (varies with solvent) Kb = molal boilin point constant (varies with solvent) m= molality m= molality

Problem. Water with sugar added to it will boil at a higher temperature than pure water. By how much will the boiling point of water be elevated if 100g of C 12 H 22 O 11 is added to 500 g of water. Kb = 0.52 C/m Problem. Water with sugar added to it will boil at a higher temperature than pure water. By how much will the boiling point of water be elevated if 100g of C 12 H 22 O 11 is added to 500 g of water. Kb = 0.52 C/m Molality = mol solute Molality = mol solute Kg solution Kg solution = 100. g sucrose x 1 mole sucrose = 0.584mol/kg = 100. g sucrose x 1 mole sucrose = 0.584mol/kg 0.500kG H 2 O 342.3g sucrose 0.500kG H 2 O 342.3g sucrose ∆Tb = Kbm = 0.52C/m x m = 0.30 C ∆Tb = Kbm = 0.52C/m x m = 0.30 C

Freezing Point Depression – the temp at which the vapor pressure of the solid and liquid are the same. If the solute is nonvolatile (rock salt) Freezing Point Depression – the temp at which the vapor pressure of the solid and liquid are the same. If the solute is nonvolatile (rock salt)

Eq. ∆Tf = Kfm Eq. ∆Tf = Kfm Problem. Calculate the freezing point depression of a solution of 100.g of C 12 H 6 O 2 in 0.500Kg of water. The Kf is 1.86C/m Problem. Calculate the freezing point depression of a solution of 100.g of C 12 H 6 O 2 in 0.500Kg of water. The Kf is 1.86C/m Molality = mol solute / Kg solution Molality = mol solute / Kg solution 100.g C 2 H 6 O 2 x 1 mol 100.g C 2 H 6 O 2 x 1 mol 0.500Kg 62.0g 0.500Kg 62.0g =3.23m =3.23m ∆Tf = Kfm = 1.86C/m x 3.23m = 6.01C ∆Tf = Kfm = 1.86C/m x 3.23m = 6.01C