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Unit: 11: Solution Chemistry

Unit: 11: Solution Chemistry. Solutions. Mixture. A combination of 2 or more kinds of matter, each of which retains its own composition and properties To distinguish the components of a mixture one would have to inspect the mixture molecule by molecule

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Unit: 11: Solution Chemistry

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  1. Unit: 11: Solution Chemistry Solutions

  2. Mixture • A combination of 2 or more kinds of matter, each of which retains its own composition and properties • To distinguish the components of a mixture one would have to inspect the mixture molecule by molecule • Mixtures differ according to the size of the particles in the mixture

  3. Mixtures: • Are classified as either suspensions, colloids or solutions based on size of particles • Solutions contains smallest particles and suspension contains the largest particles

  4. Types of Solution • Solution – homogeneous mixture of two or more substances of ions or molecules. E.g. NaCl (aq) • Solvent = component which is the component in greater amount. • Solute = component which is present in the smaller amount. • Soluble: means it can be dissolved

  5. Gaseous = gases are completely miscible in each other. • Liquid = gas, liquid or solid solute dissolved in solute. • Solid = mixture of two solids that are miscible in each other to form a single phase.

  6. Colloid – appears to be a homogeneous mixture, but particles are much bigger, but not filterable. E.g. Fog, smoke, whipped cream, mayonnaise, etc. • Suspension: larger particle sizes, filterable. E.g. mud, freshly squeezed orange juice.

  7. Solubility and the Solution Process • The solid dissolves rapidly at first but as the solution approaches saturation the net rate of dissolution decreases since the process is in dynamic equilibrium. • When the solution has reached equilibrium the amount of solute does not change with time; • At equilibrium: the rate of dissolution = rate of solution Fig. 12.2 Solubility Equilibrium

  8. 3 Factors that influence the rate of dissolving • Increase the surface area of the solute • Agitate the solution (stir or shake) • Heat the solvent

  9. Solubility Solubility is • the maximum amount of solute that dissolves in a specific amount of solvent. • expressed as grams of solute in 100 grams of solvent water. g of solute 100 g water

  10. Learning Check At 40C, the solubility of KBr is 80 g/100 g H2O. Identify the following solutions as either 1) saturated or (2) unsaturated. Explain. A. 60 g KBr added to 100 g of water at 40C. B. 200 g KBr added to 200 g of water at 40C. C. 25 g KBr added to 50 g of water at 40C.

  11. Solution A. 2 Amount of 60 g KBr/100 g water is less than the solubility of 80 g KBr/100 g water. B. 1 In 100 g of water, 100 g KBr exceeds the solubility of 80 g KBr water at 40C. C. 2 This is the same as 50 g KBr in 100 g of water, which is less than the solubility of 80 g KBr/100 g water at 40C.

  12. Solubility and the Solution Process • Saturated solution: maximum amount of solute is dissolved in solvent. Trying to dissolve more results in undissolved solute in container. • Solubility: Amount of solute that dissolves in a solvent to produce a saturated solution. (Solubility often expressed in g/100 mL.) E.g. 0.30 g of I2 dissolved in 1000 g of H2O.

  13. Unsaturated solution: less than max. amount of solute is dissolved in solvent. E.g. 0.20 g of I2 dissolved in 1000 g of H2O. • Supersaturation = more solute in solution than normally allowed; we call this a supersaturated solution.

  14. Mixing of Gas Molecules Factors Affecting Solubility • “like dissolves like” = substances with similar molecular structure are usually soluble in each other. • Gases = generally completely soluble in each other because of entropy • Molecules in gas phase are far apart from each other and not interacting strongly with each other in solution.

  15. Solvent Solute Solution Solvent – solvent Solute – solute Solute – solvent Energy Changes and the Solution Process • Intermolecular forces are also important in determining the solubility of a substance. • “like” intermolecular forces for solute and solvent will make the solute soluble in the solvent. • Hsoln is sometimes negative and sometimes positive. +

  16. Solvent – solvent interactions: energy required to break weak bonds between solvent molecules. • Solute – solute interactions: energy required to break intermolecular bonds between the solute molecules. • Solute – solvent interactions: H is negative since bonds are formed between them.

  17. Molecular Solutions • Molecular compounds with similar chemical structures and polarities tend to be miscible.(not soluble in each other) • Homologous alcohol series have polar and non-polar ends.

  18. Ionic Solutions • Solubility affected by: • Energy of attraction (due Ion-dipole force) affects the solubility. Also called hydration energy, • Lattice energy (energy holding the ions together in the lattice. Related • to the charge on ions; larger charge means higher lattice energy. • Inversely proportional to the size of the ion; large ions mean smaller lattice energy.

  19. Solubility increases with increasing ion size, due to decreasing lattice energy; Mg(OH)2(least soluble), Ca(OH)2, Sr(OH)2, Ba(OH)2(most soluble) (lattice energy changes dominant). • Energy of hydration increases with for smaller ions than bigger ones; thus ion size. MgSO4(most soluble),... BaSO4 (least soluble.) Hydration energy dominant.

  20. Solubility: Temperature Dependence • All solubilities are temperature dependent; must report temperatures with solubilities. • Most solids are more soluble at higher temperatures. Exceptions exist. • All gases are less soluble at higher temperatures.

  21. Effect of Temperature on Solubility Solubility • Depends on temperature. • Of most solids increases as temperature increases. • Of gases decreases as temperature increases.

  22. Temperature related to sign of Hsoln; • negative means less soluble at high temperatures • positive means more soluble (Le Chatelier’s principle). • E.g. Predict the temperature dependence of the solubility of Li2SO4, Na2SO4 and K2SO4 if their Hsoln are 29.8 kJ/mol, 2.4 kJ/mol and +23.8 kJ/mol, respectively.

  23. Learning Check A.Why do fish die in water that is too warm?

  24. Solution Because O2 gas is less soluble in warm water, fish cannot obtain the amount of O2 required for their survival.

  25. Solubility: Pressure Dependence • Pressure has little effect on the solubility of a liquid or solid, but has dramatic effect on gas solubility in a liquid. • Gas solubilities in liquids always increases with increase pressure • Henry’s law S = kHP. Allows us to predict the solubility of a gas at any pressure. E.g. At 25C P(O2 in air) = 0.21 atm. Its solubility in water is 3.2x10-4M. Determine its solubility when pressure of O2 = 1.00 atm.

  26. Solubility and Pressure Henry’s Law states • the solubility of a gas in a liquid is directly related to the pressure of that gas above the liquid. • at higher pressures, more gas molecules dissolve in the liquid. • (increase pressure = increase solubility of gas)

  27. Example Problem Henry’s Law • The solubility of methane, the major component of natural gas, in water at 20C and 1.00atm pressure is 0.026g/L. If the temperature remains constant, what will the solubility of this gas be at the following pressure: .6 atm • Solution: • Given 1 L = 0.26 g K = equilibrium constant • [CH4] = KP[CH4] • .026g/L =K (1atm) • K = .026g/atm solubility = (.026g atm)(.6 atm)= .0156g

  28. Example Henry’s Law • To increase the solubility of a gas at constant temperature from 0.85g/ml (1 atm) to 5.1 g/ml, the pressure would have to be increased to what? • Solution: • Given .85g = 1 ml • [gas] = KP[gas] • .85g/L =K (1atm) • K = .85g/atm solubility: 5.1g/ml = (.85g/atm) (x) • x= 5.96 atm

  29. Example Problem Henry’s Law • The solubility of methane, the major component of natural gas, in water at 20C and 1.00atm pressure is 0.026g/L. If the temperature remains constant, what will the solubility of this gas be at the following pressure: 1.8 atm

  30. Solution Henry’s Law • Solution: • Given 1 L = 0.26 g K = equilibrium constant • [CH4] = KP[CH4] • .026g/L =K (1atm) • K = .026g/atm solubility = (.026g atm)(1.8 atm)= .0468g

  31. Why could a bottle of carbonated drink possibly burst (explode) when it is left out in the hot sun ? A. The pressure in a bottle increases as the gas leaves solution as it becomes less soluble at high temperatures. As pressure increases, the bottle could burst.

  32. Units of Concentration • Physical properties of solutions are often related to the concentration of the solute in the solution.Molarity • Mole fraction: The same quantity we have used in fractional abundances as well as with gases (Dalton’s law). A unitless number.Weight (mass) Percent (wt%) – similar to mole fraction except use mass of each.

  33. Percent by mass • The number of grams of solute dissolved in 100g of solvent

  34. Percent by mass • A solution of sodium chloride is prepared by dissolving 5 g of slat in 550 g of water. What is the concentration of this solution given as percent by mass?

  35. Percent by mass example Determine mass % of solution made from dissolving 30.0 g H2O2 with 70.0 g H2O. 30.0 g H2O2 . x100 30.0 g H2O2 +70.0 g H2O. Answer: 30%

  36. Percent by mass example • A 7.5% by mass aqueous solution of sodium chloride has a mass of 650 g. What mass of sodium chloride is contained in the solution? What mass of water is contained in the solution? • Solution: • 650g solution x 7.5g NaCl/100g solution = 48.8g NaCl • 650 g solution x 92.5gH2O/100g solution = 601 g H2O

  37. Percent by mass example • A 60% by mass solution of H2SO4 in water is prepared using 225g of H2SO4. What mass of solution is produced? What mass of water is required? • Solution: • 225g H2SO4 x 100g solution/60g H2SO4 = 375 g solution • 225 g H2SO4 x 40gH2O/60g H2SO4 = 150 gH2O

  38. Units of Concentration • Other units: parts per million (ppm) and parts per billion (ppb) for small concentrations.

  39. Units of Concentration: Molality • Molality(m): defined as the mol of solute per kg of solvent. Unlike Molarity this unit is temperature independent.

  40. Molality example • A solution contains 17.1 g sucrose ( C12H22 O11) dissolved in 125G H2O . Find the molal concentration of the solution • Solution • Mass solute/mass solvent x 1/molar mass solute x 1000 g/1kg • 17.1g C12H22 O11 /125g H2O x 1 mol C12H22 O11 / 342g C12H22 O11 x 1000g/1kg = • .4 molality

  41. Molality example • What is the molality of a solution in which 3 moles of NaCl is dissolved in 1.5Kg of water? • Solution • 3moles/1.5Kg = 2m

  42. Molality example • What is the molality of solution in which 25g of NaCl is dissolved in 2.0 Kg of water? • Solution • Convert grams NaCl to Moles Nacl • 25g NaCl x 1 molNaCl/58g NaCl = .4274 moles NaCl • .4274NaCl / 2Kg = .2137m

  43. Molarity • The Molarity of a solution is the number of moles of solute per liter of solution.

  44. Molarity example Problem • What mass of K3PO4 is required to prepare 4.00 L of 1.5 M solution? • Solution: • 1.5M = X/4L solve for x • X= 6 moles • Convert moles to grams • 6 moles K3PO4 x 212gK3PO4 /1 molK3PO4 = 1272 g

  45. Molarity example Problem • What volume of .075M solution can be prepared using 90g of NH4Cl • Solution: • Convert grams NH4Cl to moles • 90g x 1/153g(wt PT) = 1.698 mole NH4Cl • .75M = 16.98moesl/x solve for x • X =2.26L

  46. Molarity example Problem • What is the molarity of a solution that contains 210g of Al2(SO4)3 in 2.75 liters of solution • Solution: • Convert grams Al2(SO4)3 to moles • 210g x 1mole/ 369g (wt PT) = .0569 moles Al2(SO4)3 • .0569/2.7 L = .2069M

  47. Vapor-Pressure Lowering of Solutions: Raoult’s Law • Raoult’s Law: Psoln = PsolvxXsolv • Non–volatile solute: vapor pressure decreases upon addition of solute. • Linear for dilute solutions • Vapor pressure lowering : P = Po P = Po(1Xsolv)

  48. Raoult’s law: states that the partial vapor pressure of each component of an ideal mixture of liquids is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture. Thus the total vapor pressure of the ideal solution depends only on the vapor pressure of each chemical component (as a pure liquid) and the mole fraction of the component present in the solution Is used to determine the vapor pressure of a solution when a solute has been added to it

  49. Raoult's law is based on the assumption that intermolecular forcesbetween unlike molecules are equal to those between similar molecules: the conditions of an ideal solution. This is analogous to the ideal gas law

  50. Raoult's law: Example • 25 grams of cyclohexane (Po = 80.5 torr, MM = 84.16g/mol) and 30 grams of ethanol (Po = 52.3 torr , MM = 92.14) are both volatile components present in a solution. What is the partial pressure of ethanol?

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