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Chapter 15: Solutions

15.1 Solubility 15.2 Solution Composition 15.3 Mass Percent 15.4 Molarity 15.7 Neutralization Reactions. Chapter 15: Solutions. Solutions. Solutions are homogeneous mixtures. Mixtures in which the components are uniformly intermingled

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Chapter 15: Solutions

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  1. 15.1 Solubility15.2 Solution Composition15.3 Mass Percent15.4 Molarity 15.7 Neutralization Reactions Chapter 15: Solutions

  2. Solutions • Solutions are homogeneous mixtures. • Mixtures in which the components are uniformly intermingled • Solvent: the substance present in the highest percentage • Solute: the dissolved substance, which is present in lesser amount • Aqueous solutions: solutions with water as the solvent

  3. Solutions (cont.)

  4. The Solution Process: Ionic Compounds • When ionic compounds dissolve in water they dissociate into ions and become hydrated. • When solute particles are surrounded by solvent molecules we say they are solvated.

  5. Solution: Solid in Liquid Figure 15.2: Polar water molecules interact with the positive and negative ions of a salt. These interactions replace the strong ionic forces holding the ions together in the undissolved solid, thus assisting in the dissolving process. • Salt in water - separate into ions

  6. H O H H C O H H H H O H The Solution Process: Covalent Molecules • Covalent molecules that are small and have “polar” groups tend to be soluble in water. • The ability to H-bond with water enhances solubility.

  7. Solubility • When one substance (solute) dissolves in another (solvent), it is said to be soluble. • Salt is soluble in water • Bromine is soluble in methylene chloride • When one substance does not dissolve in another, it is said to be insoluble. • Oil is insoluble in water

  8. Solubility (cont.) • There is usually a limit to the solubility of one substance in another. • Gases are always soluble in each other • Some liquids are always mutually soluble

  9. Solutions & Solubility • Molecules that are similar in structure tend to form solutions: “like dissolves like”

  10. Solutions & Solubility (cont.) • The solubility of the solute in the solvent depends on the temperature. • Higher temp = greater solubility of solid in liquid • Lower temp = greater solubility of gas in liquid • The solubility of gases depends on the pressure. • Higher pressure = greater solubility

  11. Figure 15.6: An oil layer floating on water.

  12. Describing Solutions Qualitatively • A concentrated solution has a high proportion of solute to solution. • A dilute solution has a low proportion of solute to solution.

  13. Describing Solutions Qualitatively (cont.) • A saturated solution has the maximum amount of solute that will dissolve in the solvent. • Depends on temp • An unsaturated solution has less than the saturation limit. • A supersaturated solution has more than the saturation limit. • Unstable

  14. Describing Solutions Quantitatively (cont.) • Solutions have variable composition. • To describe a solution accurately, you need to describe the components and their relative amounts. • Concentration: the amount of solute in a given amount of solution • Occasionally amount of solvent

  15. Solution Concentration mass percent = mass of solute X 100% mass of solution Molarity = moles of solute liters of solution (mass of solute + mass of solvent)

  16. Solution Concentration Percentage • Mass percent = grams of solute per 100 g of solution • 5.0% NaCl has 5.0 g of NaCl in every 100 g of solution • Mass of solution = mass of Solute + mass of solvent • Divide the mass of solute by the mass of solution and multiply by 100% to get mass percent.

  17. Mass Percent #15.1 A solution is prepared by mixing 2.50 g if calcium chloride with 50.0g of water. Calculate the mass percent of calcium chloride in this solution mass percent = mass of solute X 100% mass of solution 4.76% CaCl2 mass percent = 2.50g X 100 = 2.50 + 50.0g

  18. Mass Percent #15.2 Concentrated hydrochloric acid solution contains 37.2% by mass HCl. What mass of HCl is contained in 35.5g of concentrated HCl? mass percent = mass of solute X 100% mass of solution 13.2 g HCl 37.2 = mass of solute X 100 = 35.5g

  19. moles of solute liters of solution molarity = Solution Concentration Molarity • Moles of solute per 1 liter of solution • Used because it describes how many moles of solute in each liter of solution • If a sugar solution concentration is 2.0 M, 1 liter of solution contains 2.0 moles of sugar, 2 liters = 4.0 moles sugar, 0.5 liters = 1.0 mole sugar, etc.

  20. Molarity #15.3 Calculate the molarity of a solution prepared by dissolving 15.6 g of solid KBr in enough water to make 1.25 L of solution Molarity = moles of solute liters of solution 15.6 g KBr (1 mol /119.0g) = 0.131 mol KBr 0.105 M Molarity = 0.131 mol = 1.25 L

  21. Molarity #15.4 Calculate the molarity of a solution prepared by dissolving 2.80 g of solid NaCl in enough water to make 135 mL of solution Molarity = moles of solute liter of solution 2.80 g NaCl (1 mol /58.44g) = 0.0479 mol KBr 135 mL (1 L /1000 mL) = 0.135 L 0.355 M Molarity = 0.0479 mol = 0.135 L

  22. Molarity & Dissociation • The molarity of the ionic compound allows you to determine the molarity of the dissolved ions.

  23. Molarity & Dissociation (cont.) • CaCl2(aq) = Ca+2(aq) + 2 Cl-1(aq) • A 1.0 M CaCl2(aq) solution contains 1.0 moles of CaCl2 in each liter of solution • Because each CaCl2 dissociates to one Ca+2, 1.0 M CaCl2 = 1.0 M Ca+2 • Because each CaCl2 dissociates to 2 Cl-1, 1.0 M CaCl2 = 2.0 M Cl-1

  24. Calculating Ion Concentration from Molarity #15.5 • Give the concentrations of all ions in each of the following solutions: • 1.20 M Na2SO4 • 0.750 M K2CrO4 2.40M Na+, 1.20M SO42- 1.50M K+, 0.750M CrO42- • Na2SO4 Na+(aq) + SO42-(aq) 1 2 1 • 1.20 Na2SO4 2.40 Na+(aq) + 1.20 SO42-(aq) • K2CrO4 K+(aq) + CrO42-(aq) 1 2 1 • 0.750 K2CrO4 1.50 K+(aq) + 0.750 CrO42-(aq)

  25. Dilution Dilution: adding solvent to decrease the concentration of a solution The amount of solute stays the same, but the concentration decreases.

  26. Dilution (cont.) Dilution Formula M1 x V1 = M2 x V2 # Moles/L · # L = # moles In dilution we take a certain number of moles of solute and dilute to a bigger volume. Concentrations and volumes can be most units as long as they are consistent.

  27. Dilution (cont.) M = moles of solute volume (L) M1 x V1 = moles of solute = M2 x V2 remains constant decreases Add water, therefore increases Initial conditions Final conditions Molarity before Dilution Volume before Dilution Molarity after Dilution Volume after Dilution

  28. Dilution #15.6 A What volume of 19M NaOH must be used to prepare 1.0 L of a 0.15M NaOH solution? M1 x V1 = M2 x V2 19 M x V1 = 0.15M x 1.0 L V1 = 0.15M x 1.0 L = 0.0079 L = 7.9 mL 19 M

  29. Dilution #15.6 B What volume of 19M NaOH must be used to prepare 1.0 L of a 0.15M NaOH solution? M1 x V1 = M2 x V2 19 M x V1 = 0.15M x 1.0 L V1 = 0.15M x 1.0 L = 0.0079 L = 7.9 mL 19 M

  30. 15.7 Neutralization Reactions

  31. Neutralization Reactions Acid-Base reactions are also called neutralization reactions. Often we use neutralization reactions to determine the concentration of an unknown acid or base. The procedure is called a titration. With this procedure we can add just enough acid solution to neutralize a known volume of a base solution. Or vice-versa

  32. Neutralization Reactions When a strong acid and a strong base react, the net ionic reaction is H+(aq) + OH-(aq)  H2O(l)

  33. Neutralization Reactions # 15.7a What volume of 0.150 M HNO3 solution is needed to neutralize 45.0 mL of a 0.550 M KOH solution? Answer = 165 mL

  34. Neutralization Reactions # 15.7b What volume of 1.00 X 10-2 M HCL solution is needed to neutralize 35.0 mL of a 5.00 X 10-3 M Ba(OH)2 solution? Answer = 35.0 mL

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