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Chapter 14

Chapter 14. Solutions by Christopher Hamaker. 1. Chapter 14. Solutions. A solution is a homogeneous mixture. A solution is composed of a solute dissolved in a solvent . Solutions exist in all three physical states:. Gases in Solution. Temperature affects the solubility of gases.

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Chapter 14

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  1. Chapter 14 Solutions by Christopher Hamaker 1 Chapter 14 Chapter 14

  2. Solutions A solution is a homogeneous mixture. A solution is composed of a solute dissolved in a solvent. Solutions exist in all three physical states: Chapter 14

  3. Gases in Solution • Temperature affects the solubility of gases. • The higher the temperature is, the lower the solubility of a gas is in solution. • An example is carbon dioxide in soda: • Less CO2 escapes when you open a cold soda than when you open a warm soda. Chapter 14

  4. Pressure and Gas Solubility Pressure also influences the solubility of gases. According to Henry’s law, the solubility of a gas is directly proportional to the partial pressure of the gas above the liquid. If we double the partial pressure of a gas, we double the solubility. Chapter 14

  5. Henry’s Law new pressure solubility x = new solubility old pressure 1150 torr 0.00414 g/100 mL x = 0.00626 g/100 mL 760 torr • What is the solubility of oxygen gas at 25 C and a partial pressure of 1150 torr if the solubility of oxygen is 0.00414 g/100 mL at 25 C and 760 torr? We can calculate the solubility of a gas at a new pressure using Henry’s law. Chapter 14

  6. Polar Molecules When two liquids make a solution, the solute is the lesser quantity, and the solvent is the greater quantity. Recall that a net dipole is present in a polar molecule. Water is a polar molecule. Chapter 14

  7. Polar and Nonpolar Solvents A liquid composed of polar molecules is a polar solvent. Water and ethanol are polar solvents. A liquid composed of nonpolar molecules is a nonpolar solvent. Hexane is a nonpolar solvent. Chapter 14

  8. Like Dissolves Like Polar solvents dissolve in one another. Nonpolar solvents dissolve in one another. This is the like dissolves like rule. Methanol dissolves in water, but hexane does not dissolve in water. Hexane dissolves in toluene, but water does not dissolve in toluene. Chapter 14

  9. Miscible and Immiscible Two liquids that completely dissolve in each other are miscible liquids. Two liquids that are not miscible in each other are immiscible liquids. Polar water and nonpolar oil are immiscible liquids and do not mix to form a solution. Chapter 14

  10. Solids in Solution When a solid substance dissolves in a liquid, the solute particles are attracted to the solvent particles. When a solution forms, the solute particles are more strongly attracted to the solvent particles than other solute particles. We can also predict whether a solid will dissolve in a liquid by applying the like dissolves like rule. Chapter 14

  11. Like Dissolves Like for Solids Ionic compounds, like sodium chloride, are soluble in polar solvents and insoluble in nonpolar solvents. Polar compounds, like table sugar (C12H22O11), are soluble in polar solvents and insoluble in nonpolar solvents. Nonpolar compounds, like naphthalene (C10H8), are soluble in nonpolar solvents and insoluble in polar solvents. Chapter 14

  12. Chemistry Connection: Colloids Why is the flashlight beam visible in only one container? The solution, at the right of this slide, is a colloid. A colloid is a solution with large solute particles (ranging from 1 to 100 nm). The solute particles in a colloid are large enough to scatter light via a phenomenon known as the Tyndall effect. Chapter 14

  13. The Dissolving Process When a soluble crystal is placed into a solvent, it begins to dissolve. • When a sugar crystal is placed in water, the water molecules attack the crystal and begin pulling part of it away and into solution. • The sugar molecules are held within a cluster of water molecules called a solvent cage. Chapter 14

  14. Dissolving of Ionic Compounds When a sodium chloride crystal is placed in water, the water molecules attack the edge of the crystal. • In an ionic compound, the water molecules pull individual ions off of the crystal. • The anions are surrounded by the positively charged hydrogens on water. • The cations are surrounded by the negatively charged oxygen on water. Chapter 14

  15. Rate of Dissolving • There are three ways we can speed up the rate of dissolving for a solid compound: • Heating the solution • This increases the kinetic energy of the solvent, and the solute is attacked faster by the solvent molecules. • Stirring the solution • This increases the interaction between solvent and solute molecules. • Grinding the solid solute • There is more surface area for the solvent to attack. Chapter 14

  16. Solubility of Solids and Temperature The solubility of a compound is the maximum amount of solute that can dissolve in 100 g of water at a given temperature. • In general, a compound becomes more soluble as the temperature increases. Chapter 14

  17. Saturated Solutions A solution containing exactly the maximum amount of solute at a given temperature is a saturated solution. A solution that contains less than the maximum amount of solute is an unsaturated solution. Under certain conditions, it is possible to exceed the maximum solubility of a compound. A solution with greater than the maximum amount of solute is a supersaturated solution. Chapter 14

  18. Supersaturated Solutions At 55 C, the solubility of NaC2H3O2 is 100 g per 100 g water. If a saturated solution at 55 C is cooled to 20 C, the solution is supersaturated. • Supersaturated solutions are unstable. The excess solute can readily be precipitated. Chapter 14

  19. Supersaturation A single crystal of sodium acetate added to a supersaturated solution of sodium acetate in water causes the excess solute to rapidly crystallize from the solution. Chapter 14

  20. Concentration of Solutions • The concentration of a solution tells us how much solute is dissolved in a given quantity of solution. • We often hear imprecise terms such as a “dilute solution” or a “concentrated solution.” • There are two precise ways to express the concentration of a solution: • Mass/mass percent • Molarity Chapter 14

  21. Mass Percent Concentration mass of solute g solute x 100% = m/m % mass of solution x 100% = m/m % g solute + g solvent Mass percent concentration compares the mass of solute to the mass of solvent. The mass/mass percent (m/m %) concentration is the mass of solute dissolved in 100 g of solution. Chapter 14

  22. Calculating Mass/Mass Percent 5.50 g NaCl x 100% = m/m % 5.00 g NaCl + 97.0 g H2O 5.00 g NaCl x 100% = 4.90 % 102 g solution A student prepares a solution from 5.00 g NaCl dissolved in 97.0 g of water. What is the concentration in m/m %? Chapter 14

  23. Mass Percent Unit Factors 100 g solution 100 g solution 4.90 g NaCl 95.1 g water 100 g solution 100 g solution 4.90 g NaCl 95.1 g water 95.1 g water 4.90 g NaCl 95.1 g water 4.90 g NaCl We can write several unit factors based on the concentration 4.90 m/m % NaCl: Chapter 14

  24. Mass Percent Calculation 100 g solution 25.0 g dextrose x 5.00 g dextrose = 500 g solution What mass of a 5.00 m/m % solution of dextrose contains 25.0 grams of dextrose? We want grams solution; we have grams dextrose. Chapter 14

  25. Molar Concentration moles of solute = M liters of solution The molar concentration, or molarity (M), is the number of moles of solute per liter of solution, and is expressed as moles/liter. Molarity is the most commonly used unit of concentration. Chapter 14

  26. Calculating Molarity 1 mol NaOH 24.0 g NaOH x = 6.00 M NaOH 40.00 g NaOH 0.100 L solution What is the molarity of a solution containing 24.0 g of NaOH in 0.100 L of solution? We also need to convert grams NaOH to moles NaOH (M = 40.00 g/mol). Chapter 14

  27. Molarity Unit Factors 1 L solution 6.00 mol NaOH 1 L solution 6.00 mol NaOH 1000 mL solution 6.00 mol NaOH 1000 mL solution 6.00 mol NaOH We can write several unit factors based on the concentration 6.00 M NaOH: Chapter 14

  28. Molar Concentration Problem 0.100 mol K2Cr2O7 294.2 g K2Cr2O7 x 250.0 mL solution x 1000 mL solution 1 mol K2Cr2O7 = 7.36 g K2Cr2O7 How many grams of K2Cr2O7 are in 250.0 mL of 0.100 M K2Cr2O7? We want mass K2Cr2O7; we have mL solution. Chapter 14

  29. Molar Concentration Problem, Continued 1 mol HCl 1000 mL solution 9.15 g HClx x 36.46 g HCl 12.0 mol HCl = 20.9 mL solution What volume of 12.0 M HCl contains 9.15 g of HCl solute (M = 36.46 g/mol)? We want volume; we have grams HCl. Chapter 14

  30. Critical Thinking: Water Fluoridation Cities often add fluoride to drinking water. Tooth enamel is made mostly of the mineral hydroxyapatite, Ca10(PO4)6(OH)2. Fluoride prevents tooth decay by converting some of the hydroxyapatite to Ca10(PO4)6F2, which is more resistant to acid. Typically, fluoridation levels are less than 1 mg/L. Chapter 14

  31. Dilution of a Solution • Rather than prepare a solution by dissolving a solid in water, we can prepare a solution by diluting a more concentrated solution. • When performing a dilution, the amount of solute does not change, only the amount of solvent. • The equation we use is: M1 xV1 = M2 xV2. • M1 and V1 are the initial molarity and volume, and M2 and V2 are the new molarity and volume. Chapter 14

  32. Dilution Problem (0.10 M) x (5.00 L) V1 = = 0.083 L 6.0 M • What volume of 6.0 MNaOH needs to be diluted to prepare 5.00 L if 0.10 MNaOH? • We want final volume and we have our final volume and concentration. M1 xV1 = M2 xV2 (6.0 M) xV1 = (0.10 M) x(5.00 L) Chapter 14

  33. Solution Stoichiometry balanced equation solution concentration molar mass • In Chapter 10, we performed mole calculations involving chemical equations: stoichiometry problems. • We can also apply stoichiometry calculations to solutions. molarity known  moles known  moles unknown  mass unknown Chapter 14

  34. Solution Stoichiometry Problem 187.77 g AgBr 0.100 mol AlBr3 3 mol AgBr 37.5 mL solnx x x 1 mol AgBr 1000 mL soln 1 mol AlBr3 = 2.11 g AgBr • What mass of silver bromide is produced from the reaction of 37.5 mL of 0.100 M aluminum bromide with excess silver nitrate solution? AlBr3(aq) + 3 AgNO3(aq) → 3 AgBr(s) + Al(NO3)3(aq) • We want g AgBr; we have volume of AlBr3. Chapter 14

  35. Chapter Summary • Gas solubility decreases as the temperature increases. • Gas solubility increases as the pressure increases. • When determining whether a substance will be soluble in a given solvent, apply the like dissolves likerule. • Polar molecules dissolve in polar solvents. • Nonpolar molecules dissolve in nonpolar solvents. Chapter 14

  36. Chapter Summary, Continued • Three factors can increase the rate of dissolving for a solute: • Heating the solution • Stirring the solution • Grinding the solid solute • In general, the solubility of a solid solute increases as the temperature increases. • A saturated solution contains the maximum amount of solute at a given temperature. Chapter 14

  37. Chapter Summary, Continued moles of solute = M liters of solution mass of solute x 100% = m/m % mass of solution The mass/mass percent concentration is the mass of solute per 100 grams of solution. The molarityof a solution is the moles of solute per liter of solution. Chapter 14

  38. Chapter Summary, Continued • You can make a solution by diluting a more concentrated solution. M1 x V1 = M2 x V2 • We can apply stoichiometry to reactions involving solutions using the molarity as a unit factor to convert between moles and volume. Chapter 14

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