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Chapter 16 Solutions

Chapter 16 Solutions. Solution. Solution is a homogenous mixture . A solution can exist in any of the three states: Solid solution: steel, brass, bronze Liquid solution: alcohol in water, sugar water. Gaseous solution: air, any mixture of gases . Solution Terminology. Solute

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Chapter 16 Solutions

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  1. Chapter 16Solutions

  2. Solution Solution is a homogenous mixture. A solution can exist in any of the three states: Solid solution: steel, brass, bronze Liquid solution: alcohol in water, sugar water. Gaseous solution: air, any mixture of gases .

  3. 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.

  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 Terminology Saturated solution A solution that can exist in equilibrium with undissolved solute is a saturated solution. Solubility The concentration of the saturated solution is called the solubility of the solute. Unsaturated A solution is unsaturated if its concentration is smaller than the solubility.

  6. Solution Terminology Supersaturated A solution is said to be supersaturated if its concentration is greater than the solubility. Supersaturated solution is not stable, a slight physical disturbance can start crystallization

  7. The Formation of a Solution The water molecule is polar. A polar molecule is one with an asymmetrical distribution of charge, resulting in positive and negative poles.

  8. The Formation of a NaCl Solution

  9. The Formation of a NaCl Solution The formation of a sodium chloride solution from sodium chloride crystals is made possible by the interaction of water molecules with sodium and chloride ions. NaCl (s) ↔ Na+ (aq) + Cl- (aq) The sodium and chloride ions in solution are surrounded by polar water molecules, and are said to be hydrated.

  10. The Formation of a Solution The rate of dissolving depends on 1 The surface area. A finely divided solid dissolves more rapidly. 2 The diffusion of the solute from the surface. Stirring increases the rate. 3 Temperature. The higher the temperature, the higher the rate.

  11. The Formation of a Solution Crystallization rate The rate per unit of surface area increases as the solution concentration at the surface increases. Equilibrium in a Saturated Solution When the rates of dissolving and crystallization are the same the solution is saturated.

  12. The Formation of a Saturated Solution When crystallization rate is equal to the dissolving rate the solution is saturated.

  13. Factors That Determine Solubility Solubility depends on three factors: Intermolecular forces Gas pressure (for a gas) Temperature

  14. Factors That Determine Solubility Intermolecular Forces Generally, if the forces between molecules A are about the same as the forces between molecules B, A and B will probably dissolve in each other. The rule of solubility is like dissolves like. A nonpolar solvent can dissolve a nonpolar solute. A polar solvent can dissolve a polar or ionic solute

  15. Factors That Determine Solubility Ethylene glycol HO-CH2–CH2–OH, which is polar and can form hydrogen bond, is soluble in water. Motor oil is not soluble in water because hydrocarbon molecules are nonpolar.

  16. Factors That Determine Solubility Partial Pressure of Solute Gas Over Liquid Solution • Pressure has little effect on the solubility of solids or liquids but has a pronounced effect on the solubility of gases. • For an ideal solution, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas over the surface of the liquid.

  17. Factors That Determine Solubility Solubility of carbon dioxide decreases when partial pressure of carbon dioxide drops

  18. Factors That Determine Solubility Temperature The solubility of most solids increases with rising temperature (but there are notable exceptions). The solubilities of gases in liquids are generally lower at higher temperatures.

  19. Factors That Determine Solubility

  20. Factors That Determine Solubility

  21. Percentage by Mass Percentage by mass can be used as conversion factor between mass of solution and mass of solute.

  22. Percentage by Mass • When 125 grams of a solution is evaporated to dryness, 42.3 grams of solute was recovered. What was the percentage of the solute?

  23. Molarity Molarity, M Moles of solute per liter of solution: Molarity can be used as conversion factor between liters of solution and moles of solute. Volume of solution x molarity = number of moles

  24. Molarity Example: Calculate the molarity of a solution made by dissolving 15 g of NaOH in water and diluting to 1.00 x 102mL Solution: First calculate the number of moles NaOH

  25. Molarity Next calculate the volume of solution in liters 1.00 x 102 Calculate the answer using the definition

  26. Molarity Example: • How many grams of silver nitrate must be dissolved to prepare 5.00 x102mL of 0.150 M AgNO3 • 5.00 x102mL x ( 1 L/ 1000 mL) x(0.150 mol AgNO3/L) x(169.9 g AgNO3/mol AgNO3) = 12.7 g AgNO3

  27. Molarity Example Find the volume of a 1.40 M solution that contains 0.287 mole of ammonia. Solution 0.287 mole NH3 x ( 1L/ 1.40 mole NH3) = 0.205 L

  28. 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.

  29. Preparation of 250 mL Solution KMnO4 0.0100M 250 mL x (1L/1000 mL) x (0.00100 mol KMnO4 /1L) x (158.04 g KMnO4/ 1 mol KMnO4) = 0.395 g KMnO4

  30. 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.

  31. Normality: Equivalent 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.

  32. Normality: Equivalent eq /mol H3PO4 + NaOH→ NaH2PO4 + H2O 1 H3PO4 + 2 NaOH→ Na2HPO4 + 2 H2O 2 H3PO4 + 3 NaOH→ Na3PO4 + 3 H2O 3 Number of equivalents of acid = Number of equivalents of base

  33. Normality: Equivalent Mass Equivalent mass: The number of grams per equivalent. H3PO4 + NaOH→ NaH2PO4 + H2O 97.99 g/eq H3PO4 + 2 NaOH→ Na2HPO4 + 2 H2O 97.99 g/2 eq H3PO4 + 3 NaOH→ Na3PO4 + 3 H2O 97.99 g/3 eq Equivalent mass of H3PO4 for the three reactions are respectively: 97.99 g/eq, 49.00 g/eq, and 32.66 g/eq .

  34. Normality Normality, N The number of equivalents, eq, per liter of solution:

  35. Normality Example: What is the normality of a solution made by dissolving 2.50 grams of sodium hydroxide in water to make 5.00 × 102mL of solution? Solution: By definition, Therefore, we need to find equivalents of solute and liters of solution.

  36. Normality What is the normality of a solution made by dissolving 2.50 grams of sodium hydroxide in water to make 5.00 × 102mL of solution? Equivalents of solute: GIVEN: 2.5 g NaOHWANTED:eqNaOH

  37. Normality What is the normality of a solution made by dissolving 2.50 grams of sodium hydroxide in water to make 5.00 × 102mL of solution? GIVEN: 5.00 × 102mLWANTED: L

  38. Normality What is the normality of a solution made by dissolving 2.50 grams of sodium hydroxide in water to make 5.00 × 102mL of solution? We now have both parts of the normality fraction, 0.0625 eqNaOHand 0.500 L solution:

  39. Solution Concentration: Summary

  40. Dilution of Concentrated Solutions Concentrated solutions are diluted by adding more solvent. The amount of solute remains the same before and after a dilution. Mc x Vc = number of moles = Md x Vd Mc x Vc= Md x Vd

  41. Dilution of Concentrated Solutions Example: How many milliliters of concentrated hydrochloric acid, which is 11.6 M, should be used to prepare 5.50 liters of 0.500 M HCl ? Vc x Mc = Vd x Md Vc x 11.6 M = 5.50 L x 0.500 M Vc = 5.50 L x 0.500 M /11.6 M = 0.237 L = 237 mL

  42. Dilution of Concentrated Solutions If the volume of solution is increased 10 times, the concentration of diluted solution is 10 times smaller.

  43. Solution Stoichiometry How many grams of lead(II) iodide will precipitate when excess potassium iodide solution is added to 50.0 mL of 0.811 M lead(II) nitrate. Pb(NO3)2 (aq) + 2 KI(aq)  PbI2(s) + 2 KNO3(aq) Given: 50.0 mL of 0.811 M lead(II) nitrate Wanted: gram of lead(II) iodide. Need to convert volume of solution to moles of solute

  44. Solution Stoichiometry How many grams of lead(II) iodide will precipitate when excess potassium iodide solution is added to 50.0 mL of 0.811 M lead(II) nitrate. Pb(NO3)2 (aq) + 2 KI(aq)  PbI2(s) + 2 KNO3(aq) mL of Pb(NO3)2liters of Pb(NO3)2moles of Pb(NO3)2 moles of PbI2gram of PbI2 50.0 mLof Pb(NO3)2 x (L of Pb(NO3)2/ 1000 mLofPb(NO3)2) x (0.811 mol of Pb(NO3)2/L Pb(NO3)2) x (1 mol PbI2 /1 mol Pb(NO3)2) x (461.0 g PbI2 /mol PbI2) = 18.7 g PbI2

  45. Titration Using Molarity Titration: the very careful addition of one solution to another by a burette.

  46. 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.

  47. Titration Using Molarity 1.18 g oxalic acid, H2C2O4. 2H2O (126.07 g/mol), are dissolved in water and the solution is titrated with a solution of NaOH of unknown concentration. 28.3 ml NaOH are required to neutralize the acid. Calculate the molarily of the NaOH solution. H2C2O4+ 2 NaOH→ Na2C2O4+ 2 H2O Given 1.18 g H2C2O4. 2H2O, 28.3 mL (0.0283L) NaOH Wanted mol/L NaOH g H2C2O4. 2H2O mol H2C2O4. 2H2Omol H2C2O4  mol NaOH molarityNaOH

  48. Titration Using Molarity 1.18 g oxalic acid, H2C2O4. 2H2O (126.07 g/mol), are dissolved in water and the solution is titrated with a solution of NaOH of unknown concentration. 28.3 ml NaOH are required to neutralize the acid. Calculate the molarily of the NaOH solution.  H2C2O4 + 2 NaOH→ Na2C2O4 + 2 H2O 1.18 g H2C2O4. 2H2O x (1 mol H2C2O4. 2H2O/ 126.07g H2C2O4. 2H2O ) x (1 mol H2C2O4. / 1 mol H2C2O4. 2H2O) x ( 2 mol NaOH/1 mol H2C2O4.) x ( 1/ 0.0283L NaOH) = 0.661 mol/L

  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 A 25.0 mL sample of an electroplating solution is analyzed for its sulfuric acid concentration. It takes 46.8 mL of the 0.661 N NaOH to neutralize the sample, find the normality of the acid. Na x Va= Nb x Vb Na x 25.0 mL= 46.8 mL x 0.661 N Na = (46.8 mL / 25.0 mL ) x 0.661 N = 1.24 N H2SO4

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