Chapter 15: SolutionsChapter 15: Solutions

SolutionSolution• Solution – a homogeneous mixture of Solution – a homogeneous mixture of

two or more substances in a single two or more substances in a single physical state. physical state.

• Properties of solutions – Properties of solutions – • The particles are very small (atoms, The particles are very small (atoms,

molecules or ions)molecules or ions)• the particles in a solution are evenly the particles in a solution are evenly

distributed or uniformly mixed (a distributed or uniformly mixed (a spoonful of lemonade tastes the spoonful of lemonade tastes the same as the whole glass)same as the whole glass)

Parts of solutionsParts of solutions•Solute – the thing that gets Solute – the thing that gets

dissolveddissolved

•Solvent – the thing that does the Solvent – the thing that does the dissolving dissolving

•Soluble – something that can be Soluble – something that can be dissolved (salt and sugar)dissolved (salt and sugar)

• Insoluble – a substance that Insoluble – a substance that cannot be dissolved (Hg and oil)cannot be dissolved (Hg and oil)

Types of solutionsTypes of solutions

• Solid Solutions – alloys – made by melting Solid Solutions – alloys – made by melting different metals and then cooling themdifferent metals and then cooling them

• Makes material stronger, higher melting Makes material stronger, higher melting points and greater resistance to corrosionpoints and greater resistance to corrosion

• Ex. Dental filling – mercury in silverEx. Dental filling – mercury in silver• Sterling silver – copper in silverSterling silver – copper in silver• 16 karat gold – gold, copper and 16 karat gold – gold, copper and

silversilver• Coinage bronze – copper, tin and zinc Coinage bronze – copper, tin and zinc

Types of solutionsTypes of solutions•Gaseous solutions – air we Gaseous solutions – air we

breathe. Nitrogen with oxygen breathe. Nitrogen with oxygen in it in it

• If gas molecules mix they If gas molecules mix they become a solution. Anytime become a solution. Anytime gases are near each other they gases are near each other they will mix, therefore any gas will mix, therefore any gas mixture is a solution.mixture is a solution.

Types of solutionsTypes of solutions

• Liquid Solutions – MOST COMMON! Liquid Solutions – MOST COMMON! the solvent and the solution are both the solvent and the solution are both liquids. The solute can be a gas, liquids. The solute can be a gas, liquid or solid. liquid or solid.

• Miscible – liquids that can be mixed Miscible – liquids that can be mixed in any amount (water and ethanol)in any amount (water and ethanol)

• Immiscible – liquids that cannot mix Immiscible – liquids that cannot mix in any proportion (oil and water)in any proportion (oil and water)

Types of solutionsTypes of solutions

• Aqueous solutions – solutions with water Aqueous solutions – solutions with water as the solvent as the solvent

• 2 types 2 types

• 1. When ionic compounds dissolve, 1. When ionic compounds dissolve, ions are present and make an Electrolyte. ions are present and make an Electrolyte. (NaCl)(NaCl)

• 2. When molecular compounds 2. When molecular compounds dissolve, no ions are present and make a dissolve, no ions are present and make a nonelectrolyte. (Sugar) nonelectrolyte. (Sugar)

15-2 Concentration of 15-2 Concentration of SolutionsSolutions

• Concentration – the amount of solute Concentration – the amount of solute in a given amount of solvent. in a given amount of solvent.

• Molarity = Molarity = moles of solutemoles of solute

• Liters of solutionLiters of solution

• What is the Molarity of a NaOH What is the Molarity of a NaOH solution if 10.0g of NaOH is dissolved solution if 10.0g of NaOH is dissolved in enough solvent to make 0.100L of in enough solvent to make 0.100L of solution?solution?

You try it!!!You try it!!!

• Find the Molarity of a solution formed Find the Molarity of a solution formed by mixing 10.0g of Hby mixing 10.0g of H22SOSO44 with with enough water to make 100.0mL of enough water to make 100.0mL of solutionsolution

• M = M = 10.0g H10.0g H22SOSO44 x 1 x 1 mol Hmol H22SOSO44 = 1.02 mol/L = 1.02 mol/L

• 0.100L sol0.100L sol 98.1 g H 98.1 g H22SOSO44

• Molality = Molality = moles of solutemoles of solute

• Kg of Solvent Kg of Solvent

• If 18.0g of CIf 18.0g of C66HH1212OO66 is dissolved in 1Kg is dissolved in 1Kg of water. of water.

• 18.0g C6H12O618.0g C6H12O6 x x 1mol C6 H12O6 1mol C6 H12O6 = 0.100 mol C6 H12O6 / 1Kg H2O= 0.100 mol C6 H12O6 / 1Kg H2O

• 1Kg H2O1Kg H2O 180g C6 H12O6 180g C6 H12O6

• Mole Fraction = Mole Fraction = moles of componentmoles of component

• Total moles of solutionTotal moles of solution

• What is the mole fraction of SOWhat is the mole fraction of SO22 in a in a gas containing 128.0g of SOgas containing 128.0g of SO22 dissolved in every 1500.g of COdissolved in every 1500.g of CO2 2

• 1st change grams to moles 1st change grams to moles • 128g SO128g SO22 x 1 x 1 mole SOmole SO22 = 1.999 moles SO = 1.999 moles SO22

64.04 g SO64.04 g SO22

• 1500 g CO1500 g CO22 x x 1 mole CO1 mole CO22 = 34.09 moles = 34.09 moles COCO22

• 43.99 g CO43.99 g CO22

• XSO2 = XSO2 = moles of SOmoles of SO22 • Total moles of solutionTotal moles of solution• 1.999moles SO1.999moles SO22

• (1.999 moles SO(1.999 moles SO22 + 34.09 moles CO + 34.09 moles CO22))• = 0.05539= 0.05539

Example. Example.

• A gas mixture contains 50.4 g of NA gas mixture contains 50.4 g of N22O O and 65.2g of Oand 65.2g of O22 what is the mole what is the mole fraction of Nfraction of N22O? (0.360)O? (0.360)

15-3 The formation of 15-3 The formation of SolutionsSolutions• How they formHow they form

• Why does salt dissolve? The particles Why does salt dissolve? The particles are attracted to the water. The water are attracted to the water. The water attaches itself on the face of NaCl attaches itself on the face of NaCl and pulls it apart. Then the water and pulls it apart. Then the water surrounds the ions. surrounds the ions.

• solvation - the interaction between solvation - the interaction between solute and solventsolute and solvent

• hydration – when the solvent is water hydration – when the solvent is water

• energy is needed when bonds are energy is needed when bonds are broken between the solute and broken between the solute and solventsolvent

• when the solvent and solute are when the solvent and solute are attracted energy is released.attracted energy is released.

• Cold packs = NHCold packs = NH44OHOH

• Hot packs = NaHot packs = Na22SS22OO33

• Solubility – the amount of solute that Solubility – the amount of solute that will dissolve in a specific solvent will dissolve in a specific solvent under given conditions. under given conditions.

• Depends on…Depends on…• 1. Nature of solute and solvent – polar 1. Nature of solute and solvent – polar

compounds dissolve polar compounds dissolve polar compounds. IE “likes dissolve likes” compounds. IE “likes dissolve likes” cholesterol is nonpolar and fat is cholesterol is nonpolar and fat is nonpolar nonpolar

• 2. Temperature – As temperature 2. Temperature – As temperature increases, gas particles get more increases, gas particles get more energy and solubility of a gas energy and solubility of a gas decreases (better to keep soda cold!) decreases (better to keep soda cold!)

• Solubility of a solid – as temp increases, Solubility of a solid – as temp increases, solubility increases (sugar to iced tea or solubility increases (sugar to iced tea or hot tea)hot tea)

• Pressure – solubility of a gas is Pressure – solubility of a gas is increases as the pressure increases increases as the pressure increases because when the pressure is increases, because when the pressure is increases, the gas particles hit the solution faster. the gas particles hit the solution faster.

• Scuba divers – nitrogen is dissolved in Scuba divers – nitrogen is dissolved in your blood because of the extra your blood because of the extra pressure, if they come up too fast the pressure, if they come up too fast the nitrogen will bubble out of your blood. nitrogen will bubble out of your blood. (bends)(bends)

Factors affecting the Rate of Factors affecting the Rate of DissolvingDissolving

• surface area – the greater the surface area – the greater the surface area, the faster the solid can surface area, the faster the solid can be dissolved. (big ice cubes or little be dissolved. (big ice cubes or little ones)ones)

• Stirring – allows the solute to be Stirring – allows the solute to be exposed to all of the solvent not just exposed to all of the solvent not just the top the top

• Temperature – warmer is fasterTemperature – warmer is faster

•Saturation – if a solution contains Saturation – if a solution contains as much solute as can possibly as much solute as can possibly be dissolved under the existing be dissolved under the existing conditions of temperature and conditions of temperature and pressurepressure

•Saturated and concentrated are Saturated and concentrated are not the same things. A solution not the same things. A solution can be saturated with only a can be saturated with only a little amount of solute.little amount of solute.

•Unsaturated – a solution that Unsaturated – a solution that has less than the maximum has less than the maximum amount of solute amount of solute

•Supersaturated – contains a Supersaturated – contains a greater amount of solute than greater amount of solute than needed to form a saturated needed to form a saturated solution. The extra solute will solution. The extra solute will eventually go back into eventually go back into solution. (rock candy) solution. (rock candy)

• 15-4 Colligative Properties – depend 15-4 Colligative Properties – depend on the collective effort f the solute on the collective effort f the solute properties and not their identity. properties and not their identity.

• 1. Vapor pressure reduction – 1. Vapor pressure reduction – when a nonvolatile solute is added to when a nonvolatile solute is added to a solvent, the solute takes up space a solvent, the solute takes up space at the surface which prevents some at the surface which prevents some of the solvent from leaving. Gases of the solvent from leaving. Gases are still returning to the liquid at the are still returning to the liquid at the same rate. This reduces the vapor same rate. This reduces the vapor pressure of the solution.pressure of the solution.

• 2. Boiling point Elevation – antifreeze 2. Boiling point Elevation – antifreeze is added to a car to make the water is added to a car to make the water not boil. Antifreeze is a nonvolatile not boil. Antifreeze is a nonvolatile substance so it reduces the vapor substance so it reduces the vapor pressure and increases the boiling pressure and increases the boiling point because it takes longer to point because it takes longer to reach atmospheric pressurereach atmospheric pressure

• Eq. ∆Tb = Kbm Eq. ∆Tb = Kbm • Where ∆Tb = boiling point elevation Where ∆Tb = boiling point elevation • Kb = molal boilin point constant Kb = molal boilin point constant

(varies with solvent)(varies with solvent)• m= molality m= molality

• Problem. Water with sugar added to it Problem. Water with sugar added to it will boil at a higher temperature than will boil at a higher temperature than pure water. By how much will the pure water. By how much will the boiling point of water be elevated if boiling point of water be elevated if 100g of C100g of C1212HH2222OO1111 is added to 500 g of is added to 500 g of water. Kb = 0.52 C/mwater. Kb = 0.52 C/m

• Molality = Molality = mol solutemol solute

• Kg solution Kg solution • = = 100. g sucrose100. g sucrose x 1 x 1 mole sucrose mole sucrose = 0.584mol/kg= 0.584mol/kg

• 0.500kG H0.500kG H22O O 342.3g sucrose 342.3g sucrose

• ∆∆Tb = Kbm = 0.52C/m x 0.584 m = Tb = Kbm = 0.52C/m x 0.584 m = 0.30 C 0.30 C

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

• Eq. ∆Tf = Kfm Eq. ∆Tf = Kfm

• Problem. Calculate the freezing point Problem. Calculate the freezing point depression of a solution of 100.g of depression of a solution of 100.g of CC1212HH66OO22 in 0.500Kg of water. The Kf is in 0.500Kg of water. The Kf is 1.86C/m1.86C/m

• Molality = mol solute / Kg solutionMolality = mol solute / Kg solution

• 100.g C100.g C22HH66OO22 x x 1 mol 1 mol

• 0.500Kg 0.500Kg 62.0g 62.0g

• =3.23m=3.23m

• ∆∆Tf = Kfm = 1.86C/m x 3.23m = 6.01CTf = Kfm = 1.86C/m x 3.23m = 6.01C