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Unit Eleven: Solutions

Unit Eleven: Solutions. Solutions. A solution is a homogeneous mixture of two or more substances Homogeneous – uniform characteristics throughout Heterogeneous – different compositions; various throughout Solutions have at least two components Solutes – the minority component

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Unit Eleven: Solutions

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  1. Unit Eleven: Solutions

  2. Solutions • A solution is a homogeneous mixture of two or more substances • Homogeneous – uniform characteristics throughout • Heterogeneous – different compositions; various throughout • Solutions have at least two components • Solutes – the minority component • Solvents – the majority component • 75% isopropyl solution – solute is water, solvent is isopropyl • 3% H2O2 – solute is H2O2, solvent is water

  3. Nine types of mixtures • - ** means common type of solution • Amalgam: an alloy of mercury and another metal • Likes Dissolves Likes

  4. Non-Solutions - Suspensions • Suspensions • A mixture from which particles settle out upon standing • Particles are larger than in a solution • Suspended – temporarily out of school • Examples: Italian Dressing, muddy water, orange juice with pulp

  5. Non-Solutions - Colloids • Colloids • A permanent mixture whose particles are smaller than in a suspension and larger than in a solution • Particles will reflect light, cloudy appearance • Do not settle out and cannot be filtered • Milk, starch, dusty air, fog

  6. Tyndall Effect Laser pointed through a solution, colloid, and suspension – what will happen?

  7. Non-Electrolyte • Aqueous solutions containing a solute that dissolves as molecules • Do not conduct electricity • CH3OH  CH3OH • C12H22O11  C12H22O11

  8. Electrolyte • Aqueous solutions containing a solute that dissociates into ions • Conducts electricity • NaCl Na1+ + Cl1- or Al2(SO4)3  2Al3+ + 3SO42- • Strong electrolytes have large portions of solute existing as ions • Weak electrolytes have a fraction of the solute existing as ions

  9. Solubility • Solubility is the amount of compound (usually in grams) that will dissolve in a certain amount of liquid • There are three types of solutions: • Unsaturated • Saturated • Supersaturated

  10. Solubility • Saturated Solutions • Hold the maximum amount of solute under the solution conditions • If additional solute is added to a saturated solution, it will not dissolve • Unsaturated Solutions • Hold less than the maximum amount of solute under the solution conditions • If additional solute is added to an unsaturated solution, it will dissolve • Supersaturated Solutions • Hold more than the normal maximum amount of solute • Any disturbance will precipitate the solute or make it come out of solution

  11. Supersaturated Solution Example: Rock Candy

  12. Solubility Curves • 35 grams of NaCl per 100 grams at 25°C is a __ soln. • 25 g of KNO3 per 35 g of water at 50°C is a ___ soln. • 45 g of KNO3 per 100 g of water is cooled from 40°C to 0°C

  13. Solubility • Solubility depends on: • Identity of solute and solvent • Like dissolves like -> polar dissolves polar and nonpolar dissolves nonpolar • Temperature • For solids in liquids, solubility increases with increasing temperature • For gases in liquids, solubility decreases with increasing temperature

  14. Solubility • Pressure • For solids in liquids, a change in pressure has very little effect on the solubility • For gases in liquids, higher pressure increases the solubility of the gas in the liquid • When a can of soda is opened, there is less pressure so the gas is less soluble

  15. Worksheet One is due Friday (Tomorrow) • Worksheet Two is due Monday • Skip Question 10 • Questions 3, 4, and 5 you cannot answer until Friday’s notes

  16. Solubility • Rate of solution – • How fast a substance dissolves and how quickly the substance goes into solution • Factors that increase the rate of solution for solids • Decrease particle size • Stirring • Increase temperature

  17. Hydration versus Solvation • Hydration • A solute is dissolved by water molecules attaching to ions and moving them into solution • Solvation • Process of molecules of a solvent moving molecules or ions into solution

  18. Concentrations • Concentrations – the amount of solute in a solution • A dilute solution is one containing small amounts of solute relative to solvent • A concentrated solution is one containing large amounts of solute relative to solvent • Mass percent, molarity, and molality

  19. Mass Percents • Mass Percents – the number of grams of solute per 100 grams of solution Mass percent=(mass of solute)/(mass of solution)x100

  20. Practice Problems • Calculate the mass percent of NaCl in a solution containing 15.3 grams of NaCl and 155.0 grams H2O. • Calculate the mass percent of a solution containing 27.5 grams C2H6O and 175 mL of H2O • A soft drink contains 11.5% sucrose by mass. What volume of soft drink solution, in mL, contains 85.2 grams of sucrose?

  21. Molarity • Molarity – the number of moles of solute per liter of solution • A concentration of 6 MHCl contains 6 moles of HCl per 1 L of solution • Molar solutions are prepared in a volumetricflask • Molarity = M = moles of solute/liters of solution

  22. Practice Problems • Calculate the molarity of a sucrose solution made with 1.58 moles of solute diluted to a total volume of 5.0 L of solution. • Calculate the molarity of a solution made by putting 15.5 grams of NaCl into a beaker and adding water to make 1.50 L of NaCl solution.

  23. Practice problems • 6.7 grams of NH4Cl is dissolved in enough water to make 803 mL of solution. What is the molarity of the solution. • How many grams of NaOH are needed to make 500.0 mL of a 1.00 M solution? • What volume of a 1.0 M NaNO3 solution can be prepared from 170.0 grams of NaNO3?

  24. Solution Dilutions • Most solutions are bought as stock solutions, however most labs need diluted solutions. • We use M1V1=M2V2

  25. Practice Problems • A laboratory procedure calls for 5.00 L of a 1.50 M KCl solution. How should you prepare this solution from a 12.0 M stock solution? • To what volume should you dilute 0.100 L of a 15 M NaOH solution to obtain a 1.0 M NaOH solution?

  26. Practice Problems • How much 6.0 M NaNO3 solution should be used to make 0.585 L of a 1.2 M NaNO3 solution?

  27. Solution Stoichiometry • Tying it all together! How much 0.125 M NaOH solution is required to completely neutralize 0.225 L of 0.175 M H2SO4 solution? H2SO4(aq) + 2 NaOH(aq) → Na2SO4(aq) + 2 H2O(l)

  28. Solution Stoichiometry • How much 0.115 M KI solution in liters is required to completely precipitate the lead in 0.104 L of 0.225 M Pb(NO3)2? 2 KI(aq) + Pb(NO3)2(aq) → PbI2(s) + 2 KNO3(aq)

  29. Solution Stoichiometry • How many milliliters of 0.112 M Na2CO3 are necessary to completely react with 27.2 mL of 0.135 M HNO3 according to the following reaction? 2HNO3(aq)+ Na2CO3(aq) → H2O(l)+ CO2(g)+ 2NaNO3(aq)

  30. Molality • Molality – is the number of moles of solute per kilograms of solvent • A concentration of 6 m HCl contains 6 moles of HCl per 1 kg of solvent • To make 1 liter of 1 m solution, one mole is added to enough water to make 1 L • Which is more concentrated 1.0 m or 1.0 M? • 1.0 M • Molality = m = moles of solute/kg of solvent

  31. Practice Problems • What is the molality of the solution if 5.0 grams of KI are in 500.0 grams of water? • How many grams of calcium nitrate (Ca(NO3)2) must be added to 20.0 grams of water to make a 2.5 m solution?

  32. Colligative Properties • A property that depends on the number of solute particles and not the type of solute particles • How much you have not what you have • Two types • Freezing point depression • Boiling point elevation • These depend on quantity of solute not type of solute

  33. Colligative Properties • Freezing point depression – difference in temperature between the freezing point of a solution and the freezing point of the pure solvent • Solute disrupts the formation of the orderly pattern thus requiring more energy • Solution freezes at a lower temperature than the pure solvent • Magnitude is proportional to the number of solute particles dissolved

  34. ΔTf= Kf • m • (# of particles) • ΔTf= change in temperature • Kf = molal freezing point depression constant • m = molality • (# of particles) = how many particles of solute form in solution

  35. Colligative Properties • Boiling point elevation – difference in temperature between the boiling point of a solution and the boiling point of the pure solvent • Nonvolatile solute dissolved in the solvent disrupts the vapor pressure which increases the boiling point • Solution boil at a greater temperature than the pure solvent • Magnitude is proportional to the number of solute particles dissolved

  36. ΔTb = Kb • m • (# of particles) • ΔTb = change in temperature • Kb = molalboiling point depression constant • m = molality • (# of particles) = how many particles of solute form in solution

  37. Practice Problems • What is the freezing point depression and the boiling point elevation of a pinch of salt (0.25 grams NaCl) in a 2.00 L aqueous solution? Kf=1.86°C/m Kb=0.512°C/m

  38. Practice Problems • What is the freezing point depression and boiling point elevation of a 0.40 m solution of sucrose in ethanol? Kf=1.99°C/m Kb=1.22°C/m F.p.=-114.6°C B.p.=78.4°C

  39. Unit Eleven Test Tomorrow (Friday March 8th) • Worksheet Five due Tomorrow • Gizmo’s due Tomorrow • Lab Report Revisions due Tuesday • Will have optional computer time on Monday

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