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The Characteristics of a Soln

The Characteristics of a Soln. A homogenous mixture. A sample of matter containing two or more substances that has a uniform appearance and uniform properties throughout. Solid solution examples: Steel, brass, bronze Liquid solution examples: Alcohol in water, sugar water, coffee

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The Characteristics of a Soln

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  1. The Characteristics of a Soln A homogenous mixture. A sample of matter containing two or more substances that has a uniform appearance and uniform properties throughout. Solid solution examples: Steel, brass, bronze Liquid solution examples: Alcohol in water, sugar water, coffee Gaseous solution examples: An air sample, any mixture of gases

  2. Solution Terminology Solute The substance present in a relatively small amount in a solution; the solid or gas when a substance in that state is dissolved in a liquid to make a solution. Solvent The substance present in a relatively large amount in a solution; the liquid when a solid or gas is dissolved to make a solution.

  3. Microscopic View of a Sol’n

  4. Solution Terminology Concentrated Solution Has a relatively large quantity of a specific solute per unit amount of solution. Dilute Solution Has a relatively small amount of a specific solute per unit amount of solution.

  5. Solution Concentration AgNO3 (aq) + Cu0 Cu(NO3)2 (aq) + Ag0

  6. Solution Terminology Solubility A measure of how much solute will dissolve in a given amount of solvent at a given temperature. Saturated A solution whose concentration is at the solubility limit for a given temperature. Unsaturated A solution whose concentration is less than the solubility limit for a given temperature.

  7. Solution Terminology Supersaturated A solution whose concentration is greater than the normal solubility limit.

  8. Solution Terminology Miscible (Soluble) OR Immiscible (Insoluble) (usually in reference to solutions of liquids in liquids).

  9. The Formation of a Solution

  10. The Formation of a Solution Development of Equilibrium in Forming a Saturated Solution Dissolving rate: If temperature is held constant, the rate of dissolving per unit of solute surface is constant. Crystallization rate: The rate per unit of surface area increases as the solution concentration at the surface increases. Dynamic equilibrium: Dissolving rate is equal to crystallization rate.

  11. The Formation of a Solution Dynamic equilibrium

  12. Determination of Solubility The extent to which a particular solute dissolves in a given solvent depends on three factors: • The strength of intermolecular forces within the solute, within the solvent, and between the solute and the solvent • The partial pressure of a solute gas over a liquid solvent • The temperature Golden Rule of Solubility: LIKE DISSOLVES LIKE

  13. Determination of Solubility Miscible- dipole forces and H-bonding Immiscible- London forces only

  14. Pressure and Solubility The solubility of a gaseous solute in a liquid is directly proportional to the partial pressure of the gas over the surface of the liquid.

  15. Temperature and Solubility The solubility of most solids increases with rising temperature

  16. Concentration Units: Solution Concentration In general, concentration is

  17. Percentage by Mass Example: A solution is prepared by dissolving 1.23 g of sodium chloride in 500.0 mL of water. What is the percentage by mass? Solution: The density of water is 1 g/mL.

  18. Molarity Molarity, M Moles of solute per liter of solution: Molarity Cal’c Example # 1: Calculate the molarity of a solution prepared by dissolving 0.345 moles of NaCl in sufficient water to give 525 mL of solution. 0.345 mol / 0.525 L = 0.657 M (be sure to convert mL to Liters)

  19. Molarity Molarity Cal’c Example # 2: Calculate the molarity of a solution made by dissolving 13.0 grams of sugar, C12H22O11, in enough water to make 4.00  102 milliliters of solution. (convert grams to moles and mL to L) 0.0950 mol C12H22O11/L = 0.0950 M C12H22O11

  20. Molarity Example using molarity as a conversion factor: How many moles of methanol are in 45.3 mL of 0.550 M CH3OH? = 0.0249 mol CH3OH

  21. Molarity To prepare a solution of a specified molarity: • Weigh the appropriate amount of solute. • Add less than the total volume of solvent. • Mix to completely dissolve the solute. • Add additional solvent until the total solution volume is appropriate.

  22. Molarity- Using Volumetric Glassware

  23. Molality Molality, m The number of moles of solute dissolved in one kilogram of solvent: Molarity (mol/L) is temperature dependent; molality is temperature independent.

  24. Molality Calculate the molality of a solution that contains 7.72 g of NaBr in 500 mL of water. First convert grams to moles … Then calculate molality …

  25. Normality Normality, N The number of equivalents, eq, per liter of solution: Stop Here

  26. Normality Equivalent, eq One equivalent of acid is the quantity that yields one mole of hydrogen ions in a chemical reaction. Once equivalent of base is the quantity that reacts with one mole of hydrogen ions.

  27. Normality eq acid/moleq base/mol H3X + NaOH NaH2X + H2O 1 1 H3X + 2 NaOH Na2HX + 2 H2O 2 1 H3X + 3 NaOH Na3X + 3 H2O 3 1 2 HY + Ba(OH)2 BaY2 + 2 H2O 1 2 The number of equivalents of acid and base in each equation is the same.

  28. Solution Concentration General Form for Solution Concentration Ratios

  29. Solution Concentration

  30. Dilution of Solutions Concentrated solutions are diluted by adding more solvent particles. The number of solute particles remains the same before and after a dilution:

  31. Dilution of Solutions

  32. Dilution of Solutions Example: If 10.0 mL of a 16-M nitric acid solution is diluted to 1.00 L, what is the molar concentration of the dilute solution? Solution: Solve with algebra. M1 = 16 M M2 = ? V1 = 10.0 mL V2 = 1.00 L

  33. Dilution of Solutions Example: If 10.0 mL of a 16-M nitric acid solution is diluted to 1.00 L, what is the molar concentration of the dilute solution? M1 = 16 M M2 = ? V1 = 10.0 mL V2 = 1.00 L

  34. Dilution of Solutions

  35. Solution Stoichiometry For any reaction whose equation is known, the three steps for solving a stoichiometry problem are: • Convert the quantity of given species to moles. • Convert the moles of given species to moles of wanted species. • Convert the moles of wanted species to the quantity units required (usually volume, i.e., mL).

  36. Solution Stoichiometry

  37. Solution Stoichiometry Example: Aluminum shavings are dropped into 500.0 mL of 0.77 M hydrochloric acid until the reaction is complete. How many grams of hydrogen are produced? Solution: Solve with dimensional analysis. 2 Al + 6 HCl 2 AlCl3 + 3 H2 GIVEN: 500.0 mL of 0.77 M HCl WANTED: g H2

  38. Solution Stoichiometry Aluminum shavings are dropped into 500.0 mL of 0.77 M hydrochloric acid until the reaction is complete. How many grams of hydrogen are produced? 2 Al + 6 HCl 2 AlCl3 + 3 H2 GIVEN: 500.0 mL of 0.77 M HCl WANTED: g H2 PER: 0.77 mol HCl/1000 mL PATH: mL mol HCl 6 mol HCl/3 mol H2 2.016 g H2/mol H2 mol H2 g H2

  39. Solution Stoichiometry Aluminum shavings are dropped into 500.0 mL of 0.77 M hydrochloric acid until the reaction is complete. How many grams of hydrogen are produced? 2 Al + 6 HCl 2 AlCl3 + 3 H2 GIVEN: 500.0 mL of 0.77 M HCl WANTED: g H2

  40. Titration Using Molarity Titration The very careful addition of one solution to another by means of a device that can measure delivered volume precisely, such as a buret.

  41. Titration Using Molarity

  42. Titration Using Molarity Buret A glass tube of uniform width calibrated to accurately measure volume of liquid delivered through an adjustable-flow stopcock at the bottom of the tube. Indicator A substance that changes from one color to another, used to signal the end of a titration.

  43. Titration Using Molarity Standardize Determination of the concentration of a solution to be used in a titration by titrating it against a primary standard. Primary Standard A soluble solid of reasonable cost that is very stable and pure, preferably with a high molar mass, that can be weighed accurately for use in a titration.

  44. Titration Using Molarity Example: A student titrates 17.5 mL of 0.387 M nitric acid solution into a 25.0-mL sample of barium hydroxide solution. What is the molar concentration of the barium hydroxide solution? Solution: A titration is a solution stoichiometry problem. Use dimensional analysis to find mol Ba(OH)2 and algebra to find the molarity. GIVEN: 17.5 mL of 0.387 M HNO3WANTED: M Ba(OH)2 2 HNO3 + Ba(OH)2 2 H2O + Ba(NO3)2

  45. Titration Using Molarity A student titrates 17.5 mL of 0.387 M nitric acid solution into a 25.0-mL sample of barium hydroxide solution. What is the molar concentration of the barium hydroxide solution? GIVEN: 17.5 mL of 0.387 M HNO3WANTED: M Ba(OH)2 2 HNO3 + Ba(OH)2 2 H2O + Ba(NO3)2 PER: 0.387 mol HNO3/1000 mL PATH: mL mol HNO3 1 mol Ba(OH)2/2 mol HNO3 mol Ba(OH)2

  46. Titration Using Molarity A student titrates 17.5 mL of 0.387 M nitric acid solution into a 25.0-mL sample of barium hydroxide solution. What is the molar concentration of the barium hydroxide solution? PER: 0.387 mol HNO3/1000 mL PATH: mL mol HNO3 1 mol Ba(OH)2/2 mol HNO3 mol Ba(OH)2 = 0.00339 mol Ba(OH)2

  47. Titration Using Molarity A student titrates 17.5 mL of 0.387 M nitric acid solution into a 25.0-mL sample of barium hydroxide solution. What is the molar concentration of the barium hydroxide solution? = 0.00339 mol Ba(OH)2 = 0.136 mol Ba(OH)2/L = 0.136 M Ba(OH)2

  48. Titration Using Normality Goal 17 Given the volume of a solution that reacts with a known mass of a primary standard and the equation for the reaction, calculate the normality of the solution. Goal 18 Given the volumes of two solutions that react with each other in a titration and the normality of one solution, calculate the normality of the second solution.

  49. Titration Using Normality The number of equivalents of all species in a reaction is the same. For an acid–base reaction, equivalents of acid = equivalents of base

  50. Titration Using Normality Example: What is the normality of a sodium hydroxide solution if 33.16 mL of the solution reacts with 2.88 g of KHC8H4O4 in the reaction NaOH + KHC8H4O4 H2O + NaKC8H4O4? Solution: Find the number of equivalents with dimensional analysis, and then use algebra to determine the normality. GIVEN: 2.88 g KHC8H4O4WANTED: eq

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