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Unit 10: Solutions

Unit 10: Solutions. Chapter 15-16. This tutorial is designed to help students understand scientific measurements. Objectives for this unit appear on the next slide. Each objective is linked to its description. Select the number at the front of the slide to go directly to its description.

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Unit 10: Solutions

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  1. Unit 10: Solutions Chapter 15-16

  2. This tutorial is designed to help students understand scientific measurements. • Objectives for this unit appear on the next slide. • Each objective is linked to its description. • Select the number at the front of the slide to go directly to its description. • Throughout the tutorial, key words will be defined. • Select the word to see its definition.

  3. Objectives 18Define solubility including the terms soluble/insoluble and miscible/immiscible; unsaturated, saturated, and supersaturated; solute, solvent, and aqueous 19 Define conductivity including how electrolytes and dissociation affect conductivity 20Describe heterogeneous mixtures including suspensions, colloids, and emulsions 21 Explain how mixtures can be separated by six methods 22Explain saturation points regarding liquids and gases and read a solubility curve chart 23State Henry’s Law and give examples of the law 24Define concentration in terms of molarity, molality, and solve problems 25Solve dilution problems involving molarity 26Identify and define colligative properties including calculations for boiling point elevation and freezing point depression

  4. 18 Defining a Solution • Two chemicals can often be mixed together to form a solution. • In this case, the chemical that has the larger quantity is known as a solvent. • If water is the solvent, the solution is known as aqueous. • The chemical that has the smaller quantity is known as a solute. • It is stated that a solute will dissolve in the solvent • This is not limited to a solid dissolved into a liquid. • Gases can be dissolved in liquids (pop). • Liquids can be dissolved in liquids (gasoline). • Gases can be dissolved in gases (air in the room). • Etc.

  5. Defining a solution • When identifying solutes, they are often labeled as: • Soluble: will dissolve in a certain solvent • Insoluble: will not dissolve in a certain solvent • If it is a liquid in a liquid, the solutes are labeled as: • Miscible: will dissolve in a certain liquid • Immiscible: will not dissolve in a certain liquid

  6. Defining a solution • When a solution is made, the solute dissolves in the solvent. • The amount of solute can vary. • Solvents can only hold so much solute. • Once too much is added, the remaining solute sinks to the bottom. • The terms saturated, unsaturated, and supersaturated explain this situation.

  7. Unsaturated • A solution in which more solute can be added.

  8. Saturated • A solution in which the solution is holding as much solute as possible with any extra settling to the bottom.

  9. Supersaturated • A unique case where a solution has more solute dissolved than in should be allowed to.

  10. 19 Conductivity • The conductivity of a solution is dependent on the presence of electrolytes. • An electrolyte is an ion produced when a solid dissociates into a solvent. • If electrolytes are present, the solution will conduct electricity.

  11. Dissociation • Ions are found in solution when a compound dissociates during the dissolving process. • Each compound will break into a distinct number of parts. • Covalent molecules remain as one. • Ionic molecules break into their ions.

  12. Dissociation • When an ionic compound is placed in water, the partially positive hydrogens of the water molecule will attract the anion. • The partially negative oxygen of the water molecule will attract the cation. • These attractions will pull apart the compound thus separating it into ions.

  13. Dissociation

  14. Dissociation Factors • Each compound will break into certain numbers of parts when they dissociate. • This is the dissociation factor. • For covalent molecules, the molecule will not break apart so the dissociation factor is always 1. • For ionic compounds, the compound will break into its ions so it is determined by adding the ions together.

  15. Dissociation Factors • Consider NaCl • There is one Na+ and one Cl-. • Therefore, the dissociation factor is 2. • Here are some more examples:

  16. 20 Heterogeneous Mixtures • Not all mixtures will make solutions. • Some are not uniform and are called heterogeneous mixtures. • Two common varieties of heterogeneous mixtures are: • Suspensions: will separate if not agitated • Surfactants and micelles will make up suspensions • These describe soaps and detergents. • Colloids: will not separate

  17. 21 Separations • With the different homogenous (solutions) and heterogeneous mixtures that exist, it is important to be able to separate the parts. • There are six main ways to separate mixtures: • Decanting • Filtration • Evaporation • Centrifuge • Distillation • Chromatography

  18. Decanting • Decanting is a separation technique where the solid is allowed to settle to the bottom. • The liquid is then poured off leaving two parts. Image from: http://www.sciencequiz.net/jcscience/jcchemistry/practicals/decanting.htm

  19. Filtration • Filtration occurs by pouring a mixture through a porous paper. • The larger particles are caught in/above the paper. • The smaller particles pass through to the collection container. Image from: http://en.wikipedia.org/wiki/File:Vacuum-filtration-diagram.png

  20. Evaporation • Evaporation is the removal of the liquid from a solution. • The liquid is heated past its boiling point driving it away. • The solute then sinks to the bottom. Image from: http://www.school-for-champions.com/science/evaporation.htm

  21. Centrifuge • The centrifuge is an instrument in which the mixture is separated based on its densities. • The sample is spun rapidly forcing the more dense portion to the bottom of the tube. • After using a centrifuge, the sample is generally decanted. Image from: http://www.daviddarling.info/encyclopedia/C/centrifuge.html

  22. Distillation • Distillation is used when two or more liquids are present with distinct boiling points. • The mixture is heated to just above the boiling point of one liquid. • As it boils, the gas is funneled down a cooling tube causing the gas to condense. • It is then collected in an additional flask. • This allows both liquids to be kept. Image from: http://glossary.periodni.com/glossary.php?en=distillation

  23. Chromatography • Chromatography is a widely used technique to separate mixtures. • It is typically used for dyes and organic molecules. • The sample is added to a mobile phase which is passed over a stationary phase. • The stationary phase will interact with one part of the mixture thus slowing it down. • As this occurs, the mobile phase pulls the other part of the mixture further down the column. Image from: http://www.waters.com/waters/nav.htm?cid=10048919&locale=en_US

  24. Chromatography • There are several types of chromatography. • Here are a few: • Gas chromatography • High Pressure Liquid Chromatography • Ion Chromatography • Size-Exclusion Chromatography • Thin-Layer Chromatography

  25. Separation Recap

  26. 22 Solubility • Solubility describes how much solute a solution is allowed to hold during certain conditions. • The solubility of certain substances can be shown using a solubility curve. • The curve shows how a solute dissolves in a solvent given certain conditions.

  27. Solubility Curves • The line represents the saturation point at each temperature. • For example: at 20°C, sugar is saturated when 200 grams are added to 100 grams of water. • Any point below the line is unsaturated. • Any point above the line is saturated. • The curves often show more than one solute for each solvent. • In this case, water is the solvent. Image taken from: http://www.btinternet.com/~chemistry.diagrams/solubility_curves.htm

  28. 23 Henry’s Law • Henry’s Law explains a phenomena that occurs when a gas is dissolved in a liquid. • The gas in the liquid is directly proportional to the partial pressure of the gas above that liquid. • If the gas above the liquid is removed, gas that is dissolved will escape to fill this space.

  29. Henry’s Law Example - POP • One of the key components of pop is the carbonation. • The carbonation is caused by dissolving carbon dioxide in the liquid. • When the lid is removed from a bottle of pop, the gas above the liquid is removed. • This will cause some gas in solution to escape and fill the space above the liquid meaning less carbon dioxide is dissolved. • After time, this will cause the pop to go flat.

  30. Henry’s Law Example-POP The pop starts with some gas dissolved and some in the space above the liquid. Once the can is opened, gas escapes out the top of the can. To return to the correct relationship, some gas will leave the solution to fill the space again.

  31. 24 Concentration • Since solutions can have more or less solute, it is important to know how much solute is dissolved. • This is measured with concentration. • Concentration is typically given in either molarity or molality.

  32. Molarity • Molarity provides a relationship between the moles of solute and liters of solution. • Its mathematical relationship is: Molarity = For example: Assume each red dot represents one mole. Assume there is 500 ml of water. The molarity of this solution would be:

  33. Molality • Molarity provides a relationship between the moles of solute and kilograms of solvent. • Its mathematical relationship is: Molality = For example: Assume each red dot represents one mole. Assume there is 450 grams of water. The molality of this solution would be:

  34. Solving concentration problems • Assume you 200 ml have a 5 molar solution and you want to know how many moles this would be. • 5 molar can be written as • 200 ml would have to be converted to liters • So 200 ml x = 0.200 liters • Dimensional analysis will allow moles to be calculated. 0.200 liters x = 1 mole

  35. 25 Dilutions • Often, it is necessary to take a concentrated solution and make it less concentrated. • This is called a dilution. • To make a dilution, a portion of the concentrated amount is taken and more solvent is added.

  36. Dilutions • Suppose a 2 molar solution was desired but only a 4 molar solution was present. • Assume that water was the solvent. • To make the 2 molar solution, take a sample of the concentrated and double the water. 2 molar solution 4 molar solution

  37. Dilutions • The previous problem can be calculated mathematically as well. • Assume a 500 ml sample of a 2 molar solution was desired. • If this was the case, we would need 1 mole of solute. • 0.500 liters x 1.0 moles • To get 1 mole of solute from the 4 molar solution, we would need 250 ml. • 1.0 moles x • Once we take the 250 ml, we would add an additional 250 ml of water to bring the solution to 500 ml.

  38. Dilutions • The calculation used on the previous slide can be condensed to the following: M1V1 = M2V2 • It is important to realize that the volume of the concentrated is only the volume needed. Enough solvent would have to be added to reach the desired concentration. Volume of concentrated required Desired concentration Molarity of concentrated sample Desired volume

  39. 26 Colligative Properties • Colligative properties are physical properties that are affected by the amount of solute rather than the actual identity of that solute. • Two of the more common properties that are affected by the solute are: • Freezing points • Boiling points

  40. Freezing Point Depression • The freezing point of the solvent will decrease based on the amount of solvent present. • The change in the freezing point is calculated using the equation: ∆T = kfdm ∆T = change in temperature Kf = freezing constant (find on the PT) d = dissociation factor m = molality

  41. Boiling Point Elevation • The boiling point of the solvent will increase based on the amount of solvent present. • The change in the boiling point is calculated using the equation: ∆T = kbdm ∆T = change in temperature Kb = boiling constant (look on back of PT) d = dissociation factor m = molality

  42. Colligative Properties • Recall that water boils at 100°C and freezes at 0°C. • If we dissolve NaCl into water (assume 2 m) the freezing a boiling point will change. FreezingBoiling ∆T = kfdm∆T = kbdm ∆T = 1.86°C/m x 2 x 2 m ∆T = 0.52°C/m x 2 x 2 m ∆T = 7.44 °C ∆T = 2.08°C Therefore, the new freezing point is -7.44 °C and the new boiling point is 102.08 °C.

  43. This concludes the tutorial on solutions. • To try some practice problems, click here. • To return to the objective page, click here. • To exit the tutorial, hit escape.

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