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Unit 9: Solutions

Unit 9: Solutions. The answer to all of life’s problems!. Castagno Chemistry Challenge #. Rules: 1) You are working as a CLASS ! 2) Each question is multiple choice. 3) There are 7 questions so the highest score gets the points. POINTS

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Unit 9: Solutions

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  1. Unit 9: Solutions The answer to all of life’s problems!

  2. Castagno Chemistry Challenge # • Rules: • 1) You are working as a CLASS! • 2) Each question is multiple choice. • 3) There are 7 questions so the highest score gets the points. • POINTS • Class:1st – 2pts, 2nd – 1pt, 3rd – 0pts, 4th – 0pts • Questions?

  3. Castagno Chemistry Challenge # • CHALLENGE • Ready?

  4. Mixing it up • We have already discussed mixtures and some are obvious to point out • Soil • Sand • Milk and saltwater are two mixtures • But which one is a solution?

  5. Mixture vs Solution • Milk is heterogeneous as the close-up view reveals different parts • Salt water on the other hand is homogeneous as the salt (Na+ and Cl- ions) are uniformly dispersed through the water. • This homogeneous mixture is known as a solution.

  6. Making a solution • Need something soluble: capable of being dissolved. • This substance, when in solution, will be known as the solute. • The medium in a solution (usually water) is called the solvent. • Water is known as the “universal solvent” as it can dissolve a wide range of substances.

  7. Types of Solutions • Solutions are not necessarily made of solids in liquids though certainly it seems most common.

  8. Types of Solutions II • There are different types of solutions involving the states of matter • 1) Gas – Gas: oxygen in nitrogen (air) • 2) Gas – liquid: carbon dioxide in water (soda) • 3) Liquid – liquid: alcohol in water • 4) Liquid – solid: mercury in silver (dental) • 5) Solid – liquid: sugar in water • 6) Solid – solid: copper in nickel (alloy)

  9. Solid – solid solutions • Alloys of metals usually have preferable properties when compared to individual metals. • Greater strength and resistance to corrosion • Gold • 24 K gold is too soft for use in jewelry • 18K, 14K gold have been alloyed with metals such as silver and copper. • The metal keeps the gold color, but has greater strength and hardness. • These are solutions because the distribution of atoms is even.

  10. Suspendable Offense • When left alone, mixtures can settle over time due to gravity. • Muddy water (flood water) • The soil particles are dense and many times bigger than water molecules, so gravity pulls them down when undisturbed. • This is known as a suspension. • Particles must be over 1000nm in diameter (or 1000x the diameter of an atom). • These can be filtered.

  11. Suspendable Offenses • The salad dressing on the left looks evenly distributed while the one on the right has separated into layers.

  12. The ‘Tween’ Years • Solutions have an even distribution. • Suspensions need particles 1000nm in diameter. • What about in between? • Colloids • Not a solution, not a suspension. • Particle size between 1nm and 1000nm. • Cannot be filtered. • But many are familiar • Paint, gelatin, mayonnaise, shaving/whipped cream, fog, clouds, aerosol spray, cheese, butter.

  13. The Inbetweeners • One of these is a glass of milk and the other is flour in water. • The milk (right) is typical white but the flour (white) in the water has turned blue due to the Tyndall effect – the colloid scatters incoming light.

  14. Comparison

  15. Electrifying • Just as there are different categories of solutions, there are for solutes as well. • The differentiating factor is conductivity.

  16. A Bright Idea • Solutes can cause a solution to conduct an electric current. • Electrolytes: solutes that give a solution the ability to conduct. • Ionic compounds • Acids/Bases • Nonelectrolytes: solutes that do not give a solution the ability to conduct. • Molecular compounds (sugars)

  17. The Solution Process • Solutions are pretty easy to create and it is simple to speed up the process as well. • Surface area • Increasing the surface area means more solute comes in contact with the solvent. • Agitation • Above the solute is a high concentration, stirring the solution disperse the solute. • Heat • Collisions between solute and solvent increase due to the addition of energy.

  18. Solubility • Not every substance dissolves in a solvent. These are said to have a low solubility. • Even substances with a high solubility will eventually reach a point where no more will dissolve in a given volume. • With gases, the solubility increases with pressure. This has no effect on the solubility of liquids and solids, which is usually increased with temperature.

  19. Saturation • Solutions can exist in 3 states • Saturated • A solution with the maximum amount of solute • Unsaturated • A solution that contains less than the maximum amount of solute • Supersaturated • A solution that contains MORE than the maximum amount of solute

  20. Concentration • Imagine two solutions, acetic acid (the acid in vinegar) and hydrochloric acid (in your stomach). • Now, imagine you have a complete lapse in judgment in pour both onto your arm. • Which would you say is guaranteed to burn a hole in your arm? • Did you say hydrochloric acid? • CONGRATULATIONS! You’re wrong. • In order to truly know if your arm will burn, you must know the concentration of the solution.

  21. Molarity (M) • The number of moles of solute in 1 liter of solution. • M = n/V • Example: what is the molarity of a 1.0 L solution made with 20.0 g of NaOH? • MM of NaOH = 40.0 g/mol • 20.0 g / 40.0 g/mol = 0.500 mol • M = 0.500 mol / 1.0 L • 0.50 M

  22. Molarity (M) II • Calculate the molarity of a 3.50 L solution that contains 90.0 g of NaCl. • You need to calculate moles of NaCl! • 90.0 g NaCl / 58. 44g/mol = 1.54 mol NaCl • M = n / V • M = 1.54 mol NaCl / 3.50 L • M = 0.440 M NaCl

  23. Molarity (M) III • How many moles of NaCl are in 1.25 L of 0.330 M NaCl solution? • M = n / V • 0.330 M = n / 1.25 L • 0.330 M x 1.25 L = n • n = 0.413 mol NaCl • So to solve for moles the equation can be rearranged to: n = M x V

  24. Molarity (M) IV • 1) What is the molarity of a 2.00 L solution that is made from 14.6 g of NaCl. • M = 0.125 M • 2) What is the molarity of a HCl solution that contains 10.0 g of HCl in 250 mL of solution? • M = 1.10 M • 3) How many moles of HCl exist in 500. mL of 0.50 M solution of HCl? • n = 0.25 mol • 4) What volume is necessary to create a 0.500 M solution prepared with 32.5 g of HBr? • 0.803 L

  25. Molarity (M) V • 1) How many grams of solute are needed to make 1.00 L of 3.5 M aqueous solution of H2SO4? • mass = 343 g • 2) How many grams of solute are needed to make 2.50 L of a 1.75 M solution of Ba(NO3)2? • mass = 1140 g • 3) What mass of (NH4)2SO4 is necessary to produce 50.0 mL of 4.0 M solution? • mass = 26.42 g • 4) What is the molarity of 6.0 L of Na2CO3 solution produced with 106 g of sodium carbonate? • M = 0.167 M

  26. Properties of Solutions • The presence of solute particles affects the properties of the solution. • Dissociation: the separation of ions when ionic compounds dissolve. • NaCl (s)  Na+ (aq) + Cl- (aq) • 1 mole of NaCl produces 2 moles of ions* • Colligative propertiesdepend on concentration, not identity.

  27. Colligative Properties Application • Every winter, municipalities take advantage of water using colligative properties. • The boiling point and freezing point of solvents can be altered. • Boiling Point Elevation • We will encounter this in our lab this week • Freezing Point Depression • “rock salt” dissociates in water, interferes with water’s ability to freeze at 0.0 *C. • This means salted roads should freeze well below water’s normal freezing point.

  28. Molality (m) • BPE and FPD can both be calculated, but a new concentration unit is necessary • Molality = moles solute/ kg solvent • m = n/kg • Luckily, 1.0 mL water = 1.0 g water • Example: What is the molality of a sucrose (MM = 342.34 g/mol) solution prepared by dissolving 17.1 g sucrose in 125 g of water? • n = 17.1 g / 342.34 g/mol = 0.05 mol • m = 0.05 mol / 0.125 kg • m = 0.400 m

  29. Molality (m) II • 1) What is the molality of acetone in solution prepared with 255 g of acetone, (CH3)2CO (MM = 58 g), dissolved in 200 mL water? • n= (255 g / 58 g/mol) = 4.4 mol • m = 4.4 mol / .200 kg = 22.0 m • 2) What is the molality of a solution composed of 13.0 g NaCl dissolved in 500 g of water? • m = (13.0/58.5)/.500kg = 0.444 m • 3) What mass of H2SO4 is necessary to produce a 4.50 m solution in 1.0 kg water? • mass = 441 g

  30. Boiling Point Elevation • Boiling point elevation and freezing point depression can both be calculated using molality. • ∆tb = Kbm • ∆tb = The change in boiling point • Kb = boiling point constant (0.51 *C/m for water) • m = molality of solution • Ionic compounds dissociate and have a greater effect on the solution • NaCl = m x 2 • CaCl2 = m x 3 • Sugars = m x 1

  31. Boiling Point Elevation II • ∆tb = Kbm • What is the boiling point elevation of a solution made from 20.1 g of a nonelectrolyte solute (MM = 62.0 g) and 400.0 g of water (Kb = 0.51 *C/m)? • 20.1 g / 62.0 g/mol = 0.324 mol • 0.324 mol / .4 kg = 0.810 m • ∆tb = (0.51 *C/m) x (0.810 m) • ∆tb = 0.41 *C • Note: you are not actually calculating the new boiling point, just the change.

  32. Boiling Point Elevation III • ∆tb = Kbm • 1) A solution contains 50.0g of sucrose (MM = 342.34 g) in 500.0 g of water. What is the boiling point elevation? • ∆tb = 0.15 *C • 2) What is the new boiling point of a solution with 450.o g sucrose dissolved in 250.0 g of water? • T = 102.7 *C • 3) What is the expected boiling point of water for a NaCl solution that contains 150.0 g dissolved in 600.0 g of water? • Don’t forget, NaCl = m x 2!!! • T = 104.36 *C

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