Chapter 12 solutions and molarity

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Chapter 12 solutions and molarity

  1. 1. Solutions and MolarityDescribe the types of solutions.Define the vocabulary words.List and explain the factors that influencesolubility and the rate at which a solutedissolves in a solvent.Explain and calculate molarity.
  2. 2. Types of Solutions Gas in a gas – Gases mix freely and will always form a solution unless they react (Example: air) Solid in a solid – Alloy: a homogeneous mixture of metals (Example: brass is a mixture of copper and zinc) Liquid in a liquid Miscible – when liquids can be mixed together to form a solution (Example: ethylene glycol and water form antifreeze) Immiscible – when liquids cannot be mixed (Example: oil and water) “Like dissolves like” – polar dissolves polar Polar and nonpolar (vinegar and oil) are immiscible
  3. 3.  Solid in a liquid Energy is required to separate particles of a solid (Endothermic) Energy is released when solute particles and solvent particles are attached (Exothermic) Goes back and forth: Dynamic equilibrium Differences are responsible for different solubilities
  4. 4.  Gases in liquids – when gas is attracted to solvent particles---release energy Free---move toward entropy Examples: CO2 in soda and O2 in seawater Entropy (S) – disorder or randomness Systems tend to go from a state of order (low entropy) to a state of maximum disorder (high entropy)
  5. 5. Solubility CurveSolubility - quantity of solute thatwill dissolve in a specific amount ofsolvent at a certain temperature.(pressure must also be specified forgases). -soluble vs. insoluble, -saturated (on or above) vs. unsaturated (below) - solubility should NOT be confused with the rate at which a substance dissolves Reading the curve: At 30°C approximately 10g of KClO3 will dissolve in 100g of water
  6. 6. Refer to Solubility Curves At 10 °C, 135 grams of KI will dissolve At 50 °C, 85 grams of KNO3 will dissolve At 30 °C, 42 grams of NH4Cl will dissolve At 75 °C, 50 grams of KCl will dissolve
  7. 7. Factors Influencing the Rate at which a Solute Dissolves in a Solvent 1. Agitation – stirring; a surface phenomenon 2. Particle size 3. Temperature – the only factor that affects both the rate of solution and the solubility
  8. 8.  Solid in a liquid Energy is required to separate particles of a solid (Endothermic) Energy is released when solute particles and solvent particles are attached (Exothermic) Goes back and forth: Dynamic equilibrium Differences are responsible for different solubilities
  9. 9.  Gases in liquids – when gas is attracted to solvent particles---release energy Free---move toward entropy Examples: CO2 in soda and O2 in seawater Entropy (S) – disorder or randomness Systems tend to go from a state of order (low entropy) to a state of maximum disorder (high entropy)
  10. 10. Temperature Effects onSolubility In general, an increase in temperature increases the solubility of solids in liquids In general, an increase in temperature decreases the solubility of gases in liquids (gases escape)
  11. 11. Pressure Effects  Pressure increases the solubility of gases in liquids (nail being hammered into wood)  Henry’s Law – At a given temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid (Page 506)S = solubility S1 S2 =P = pressure P1 P2
  12. 12. Concentration of Solutions Dilute – contains relatively little solute Concentrated – contains relatively large amount of solute
  13. 13. Vocabulary Words Solubility: the amount of a substance that dissolves in a given quantity of solvent at specified conditions of temperature and pressure to produce a saturated solution Solute: dissolving particles Solvent: the dissolving medium in a solution (Example- water)
  14. 14.  Saturated – a solution that holds as much solid as it normally can at a given temperature (If more solid is added, it will not dissolve) Unsaturated – a solution which has not yet reached the limit of solubility at a given temperature Supersaturated – rare solution that contains MORE dissolved solute than it can normally hold at a given temperature (Crystallization)
  15. 15.  Suspension: if the particles are so large that they settle out unless the mixture is constantly stirred or agitated.suspension  Colloid: particles that are intermediate in size between those in solutions and suspensions. AKA- colloidal suspensions colloid
  16. 16. Molarity The number of moles of solute dissolved per liter (1000mL) of solution Water: 1 mL = 1 g; 1000 mL = 1 Kg mol of solute molarity (M) = liters/Kg of solution Moles of solute = liters of solution x molarity Problem: What is the molarity of the solution obtained by dissolving 90g of glucose (C6H12O6) in 1000 grams of water?
  17. 17.  Answer 1 mole of glucose = 180 g 90g ÷ 180 g = .5 mole 1000 g = 1000 mL .5M per 1000 mL water Problems: How many grams are needed to make a molar solution of a. 1M glucose b. 2M glucose c. .5M glucose
  18. 18. Answer 1 mole of glucose = 180 grams a. 1M = 180 grams b. 2M = 2 x 180 g = 360 grams c. .5M = .5 x 180 g = 90 grams
  19. 19. Preparing Molar Solutions The solute will take up some of the available space in the volumetric flask. Steps 1. The solute should be added to some of the solvent and dissolved. 2. Then solvent is added to the 1L mark on the volumetric flask. If these steps are not followed, the total volume of the mixture is likely to exceed the desired volume. A volumetric pipet measures volumes even more accurately.
  20. 20. Making Dilutions Dilution reduces the moles of solute per unit volume, however, the total moles of solute in solution does not change. Moles of solute = molarity (M) x liters of solution (V) Moles of solute = M1 x V1 = M2 x V2 Problem: How many milliliters of a stock solution of 2.00M MgSO4 would you need to prepare 100.0 mL of 0.400 M MgSO4?
  21. 21. Answer 0.400M x 100.0 mL ÷ 2.00M = 20.0 mL Thus, 20.0 mL of the initial solution must be diluted by adding enough water to raise the volume to 100.0 mL OR 0.400M is 1/5th of 2.00M 1/5th of 100mL is 20 mL
  22. 22. Percent Solutions If both the solute and solvent are liquids, a convenient way to make a solution is to measure volumes. Example: 20 mL of pure alcohol is diluted with water to a total volume of 100 mL – The concentration of alcohol is 20% (v/v) A commonly used relationship for solutions of solids dissolved in liquids is percent (mass/volume).
  23. 23. Problems 1. What is the percent by volume of ethanol (C2H6O), or ethyl alcohol, in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water? 2. How many grams of glucose (C6H12O6) would you need to prepare 2.0 L of 2.8% glucose (m/v) solution?
  24. 24. Answers 1. 85/250 x 100% = 34% ethanol (v/v) 2. In a 2.8% solution, each 100 mL of solution contains 2.8 grams of glucose 100mL is 1/10th of a liter, so you need 28 grams per liter 2 L x 28 g = 56 grams of glucose

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