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PRINCIPLES OF CHEMISTRY II CHEM 1212 CHAPTER 12

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PRINCIPLES OF CHEMISTRY II CHEM 1212 CHAPTER 12

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  1. PRINCIPLES OF CHEMISTRY II CHEM 1212CHAPTER 12 DR. AUGUSTINE OFORI AGYEMAN Assistant professor of chemistry Department of natural sciences Clayton state university

  2. CHAPTER 12 SOLUTIONS

  3. SOLUTION - A homogeneous mixture of two or more substances Solvent - The substance present in the greatest quantity Solute - The other substance(s) dissolved in the solvent

  4. SOLUTION - Solutions can exist in any of the physical states Solid Solution - dental fillings, metal alloys (steel), polymers Liquid Solution - sugar in water, salt in water, wine, vinegar Gas Solution - air (O2, Ar, etc. in N2), - NOx, SO2, CO2 in the atmosphere

  5. CONCENTRATION OF SOLUTIONS - The amount of solute dissolved in a given quantity of solvent or solution Molarity (M: molar) - The number of moles of solute per liter of solution - A solution of 1.00 M (read as 1.00 molar) contains 1.00 mole of solute per liter of solution

  6. CONCENTRATION OF SOLUTIONS • Calculate the molarity of a solution made by dissolving 2.56 g • of NaCl in enough water to make 2.00 L of solution • - Calculate moles of NaCl using grams and molar mass • Convert volume of solution to liters • - Calculate molarity using moles and liters

  7. CONCENTRATION OF SOLUTIONS After dissolving 1.56 g of NaOH in a certain volume of water, the resulting solution had a concentration of 1.60 M. Calculate the volume of the resulting NaOH solution - Convert grams NaOH to moles using molar mass - Calculate volume (L) using moles and molarity

  8. CONCENTRATION OF SOLUTIONS Mole Fraction (χ) - Fraction of moles of a component of solution The sum of mole fractions of all components = 1

  9. CONCENTRATION OF SOLUTIONS Given that the total moles of an aqueous solution of NaCl and other solutes is 1.75 mol. Calculate the mole fraction of NaCl if the solution contains 4.56 g NaCl.

  10. CONCENTRATION OF SOLUTIONS Percent Concentration - Percent by mass [mass-mass percent, %(m/m)] mass of solution = mass of solute + mass of solvent

  11. CONCENTRATION OF SOLUTIONS A sugar solution is made by dissolving 5.8 g of sugar in 82.5 g of water. Calculate the percent by mass concentration of sugar.

  12. CONCENTRATION OF SOLUTIONS Percent Concentration - Percent by volume [volume-volume percent, %(v/v)] volume of solution ≠ volume of solvent + volume of solute - Due differences in bond lengths and angles

  13. CONCENTRATION OF SOLUTIONS Calculate the volume percent of solute if 345 mL of ethyl alcohol is dissolved in enough water to produce 1257 mL of solution

  14. CONCENTRATION OF SOLUTIONS Percent Concentration - Mass-volume percent [%(m/v)] - Units are specified because they do not cancel

  15. CONCENTRATION OF SOLUTIONS The concentration of a solution of NaCl is 0.92 %(m/v) used to dissolve drugs for intravenous use. What is the amount, in grams, of NaCl needed to prepare 41.50 mL of the solution? g solute = [%(m/v)] x [volume of solution (mL)]/[100 %] = [(0.92 % g/mL) x (41.50 mL)]/(100 %) = 0.38 g

  16. PARTS PER MILLION (PPM) Percent can be defined as parts per hundred 1 ppm ≈ 1 µg/mL or 1 mg/L

  17. PARTS PER MILLION (PPM) If 0.250 L of aqueous solution with a density of 1.00 g/mL contains 13.7 μg of pesticide, express the concentration of pesticide in ppm ppm = µg/mL 0.250 L = 250 mL Density = 1.00 g/mL Implies mass solution = 250 g

  18. PARTS PER BILLION (PPB) 1 ppb ≈ 1 ng/mL or 1 µg/L

  19. PARTS PER MILLION (PPB) If 0.250 L of aqueous solution with a density of 1.00 g/mL contains 13.7 μg of pesticide, express the concentration of pesticide in ppb ppm = µg/L Volume of solution = 0.250 L Density = 1.00 g/mL Implies mass solution = 250 g

  20. MOLALITY (m) Moles of solute per kg of solvent Unit: m or molal

  21. MOLALITY (m) What is the molality of a solution that contains 2.50 g NaCl in 100.0 g water? - Calculate moles NaCl - Convert g water to kg water - Divide to get molality

  22. CONVERTING CONCENTRATION UNITS Calculate the molality of a 6.75 %(m/m) solution of ethanol (C2H5OH) in water Mass water = 100 g solution – 6.75 g ethanol = 93.25 g water

  23. CONVERTING CONCENTRATION UNITS Calculate the mole fraction of a 6.75 %(m/m) solution of ethanol (C2H5OH) in water Mass water = 100 g solution – 6.75 g ethanol = 93.25 g water

  24. CONVERTING CONCENTRATION UNITS Practice Question Given that the mole fraction of ammonia (NH3) in water is 0.088 Calculate the molality of the ammonia solution

  25. CONVERTING CONCENTRATION UNITS - Molarity is temperature dependent (changes with change in temperature) - Volume increases with increase in temperature hence molarity decreases On the other hand - Molality - Mass percent - Mole fraction are temperature independent

  26. SOLUBILITY - A measure of how much of a solute can be dissolved in a solvent - Grams of solute per 100 mL of solvent - Units: grams/100 mL Three factors that affect solubility - Temperature - Pressure - Polarity

  27. SOLUBILITY Unsaturated Solution - More solute can still be dissolved at a given temperature - Concentration of the solute is less than the solubility Saturated Solution - No more solute can be dissolved at a given temperature - Concentration of the solute is equal to the solubility - Dynamic equilibrium is reached Supersaturated Solution - Too much solute has temporarily been dissolved - Concentration of solute is temporarily greater than the solubility - Unstable condition

  28. DISSOLUTION The process of dissolving (known as dissolution) is contributed by factors such as - Enthalpy change due to solute-solvent interactions and - Change in disorder

  29. SOLUTE-SOLVENT INTERACTIONS - Change in enthalpy arises mainly from changes in intermolecular attractions Three types of intermolecular attractions are involved - Solute-solute - Solvent-solvent - Solute-solvent

  30. SOLUTE-SOLVENT INTERACTIONS - The relative strengths of these interactions determine the formation of a solution by two substances - Substances with similar properties (strong solute-solvent interactions) tend to form solutions - Like dissolves like

  31. SOLUTE-SOLVENT INTERACTIONS - Solvent molecules move apart to accommodate solute molecules - Energy is required to separate solvent molecules attracting each other (ΔH1) - Energy is also required to separate solute molecules (ΔH2) - Energy is released when solvent and solute molecules come together due to attractive forces between them (ΔH3)

  32. SOLUTE-SOLVENT INTERACTIONS Enthalpy of Solution - The overall enthalpy change that accompanies the dissolution of one mole of a solution ΔHsoln = ΔH1 + ΔH2 + ΔH3

  33. SOLUTE-SOLVENT INTERACTIONS Endothermic Heat of Solution - Energy released by solute-solvent interactions is less than the energy absorbed by separating the solvent and solute molecules ΔHsoln is positive ΔH3 < (ΔH1 + ΔH2) Example Ammonium nitrate in water

  34. ENTHALPY OF SOLUTION Endothermic Heat of Solution Separated solute Separated solvent + ∆H2 Enthalpy Solute + Separated solvent ∆H3 ∆H1 Solution ∆Hsoln Solute + Solvent

  35. SOLUTE-SOLVENT INTERACTIONS Exothermic Heat of Solution - Energy released by solute-solvent interactions is greater than the energy absorbed by separating the solvent and solute molecules ΔHsoln is negative ΔH3 > (ΔH1 + ΔH2) Example Sulfuric acid in water NaOH in water

  36. ENTHALPY OF SOLUTION Exothermic Heat of Solution Separated solute Separated solvent + ∆H2 Enthalpy Solute + Separated solvent ∆H3 ∆H1 Solute + Solvent ∆Hsoln Solution

  37. SOLUTE-SOLVENT INTERACTIONS Generally - Substances with similar intermolecular forces and hence similar properties have strong solute-solvent interactions - Such substances tend to form solutions - Like dissolves like Example CH3OH readily dissolves in H2O (hydrogen bonding in both) CCl4 readily dissolves in C7H16 (London forces in both)

  38. SOLUTE-SOLVENT INTERACTIONS - Increase in disorder on mixing is another contributing factor in the dissolution process - Increase in disorder tends to occur spontaneously in processes - The main driving force in the formation of solutions Consider NH4NO3 in H2O (used in cold packs) - Enthalpy change on mixing is positive (+28 kJ/mol) - NH4NO3 dissolves to form solution due to increase in disorder

  39. SOLUTE-SOLVENT INTERACTIONS Spontaneous Process - Takes place with no apparent cause Nonspontaneous Process - Requires something to be applied in order for it to occur (usually in the form of energy)

  40. SOLUBILITY OF IONIC COMPOUNDS - Strong electrostatic attractions between oppositely charged ions hold ionic solids together - For soluble ionic compounds the enthalpy of attraction between solvent molecules and ions must be comparable to the crystal lattice enthalpy in the solid Example - NaCl solution contains Na+ and Cl- ions - Each ion is surrounded by water molecules - Good conductor of electricity

  41. SOLUBILITY OF IONIC COMPOUNDS • Solvation Process (Hydration) • - Ions in aqueous solution are surrounded by the H2O molecules • The O atom in each H2O molecule has partial negative charge • and attract cations • - The H atoms have partial positive charge • and attract anions • - Cations and anions are prevented from recombining • - About 4 to 10 water molecules surround each cation

  42. SOLUBILITY OF IONIC COMPOUNDS - There is an increase in disorder of the solute as it separates into ions - There is an increase or decrease in disorder of the solvent depending on the solute - Solubilities are difficult to predict due to these several contributing factors

  43. SOLUBILITY OF MOLECULAR COMPOUNDS - Most molecular compounds do not form ions in solution - The molecules disperse throughout the solution Example - Sucrose in water solution contains neutral sucrose molecules - Each molecule is surrounded by water molecules - Poor conductor of electricity - A few molecular compounds form ions in aqueous solution - HCl dissociates into H+(aq) and Cl-(aq) - HNO3 dissociates into H+(aq) and NO3-(aq)

  44. SOLUBILITY OF MOLECULAR COMPOUNDS Consider mixing hydrocarbons such as C6H14 and C7H16 - London dispersion forces dominate within the molecules - Attraction between C6H14 and C7H16 molecules are also due to London dispersion forces - These two substances mix because the attractions are close in energy - Increase in disorder is the controlling factor

  45. SOLUBILITY OF MOLECULAR COMPOUNDS Consider mixing water and a hydrocarbon - Strong hydrogen bonding dominates intermolecular attractions between water molecules - London dispersion forces dominate intermolecular attractions between hydrocarbon molecules - Attraction between water and hydrocarbon molecules are due to weak London dispersion forces - Increase in disorder is not sufficient to overcome the unfavorable enthalpy change hence very low solubility results

  46. EFFECT OF PRESSURE ON SOLUBILITY - Solubilities of gases in liquids are sensitive to pressure changes - Increase in pressure increases solubility of gases - An increase in pressure of a saturated solution results in dissolving more gas molecules - Solubilities of liquids and solids change very little with pressure due to very little change in volume

  47. EFFECT OF PRESSURE ON SOLUBILITY Henry’s Law - The solubility of a gas is directly proportional to its partial pressure at any given temperature C = kP C = concentration of the gaseous solute k = proportionality constant (units depend on units of C) P = partial pressure of gaseous solute

  48. EFFECT OF TEMPERATURE ON SOLUBILITY - Effect of temperature depends on the sign of the enthalpy change - Solubility increases with increasing temperature when the enthalpy change is positive (+∆H, endothermic process) - Solubility decreases with increasing temperature when the enthalpy change is negative (-∆H, exothermic process) Generally - The more positive the ∆H the greater the change in solubility with temperature

  49. EFFECT OF TEMPERATURE ON SOLUBILITY - Solubility of most ionic solids increase with increase in temperature - Solubility of most gases decrease with increase in temperature - Enthalpy of solution of most gases in water is negative - There is little or no attraction between gas molecules but there are attractions between solvent and solute molecules - Hence the negative enthalpy change

  50. COLLIGATIVE PROPERTIES - The physical properties of a solution differ from those of the pure solvent - The physical properties of a solvent change when a solute is added to a solvent - Four physical properties change based on the amount of solute added but not the solute’s chemical identity