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Solutions

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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Solutions

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  1. 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

  2. Gases in Solution • Temperature affects the solubility of gases. • The higher the temperature, the lower the solubility of a gas 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

  3. Pressure & 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

  4. Henry’s Law new pressure solubility × = new solubility old pressure 1150 torr 0.00414 g/100 mL × = 0.00626 g/100 mL 760 torr • We can calculate the solubility of a gas at a new pressure using Henry’s law. • 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? Chapter 14

  5. 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

  6. 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

  7. 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

  8. Miscible & 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

  9. 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

  10. 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

  11. Chemistry Connection: Colloids • Why is the flashlight beam visible in only one container? • The solution at the right 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

  12. 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

  13. Dissolving of Ionic Compounds • When a sodium chloride crystal is place 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

  14. 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

  15. 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

  16. 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

  17. 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

  18. 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

  19. 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

  20. Mass Percent Concentration mass of solute g solute × 100% = m/m % mass of solution × 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

  21. Calculating Mass/Mass Percent 5.50 g NaCl × 100% = m/m % 5.00 g NaCl + 97.0 g H2O 5.00 g NaCl × 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

  22. 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

  23. Mass Percent Calculation 100 g solution 25.0 g dextrose × 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

  24. 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

  25. Calculating Molarity 1 mol NaOH 24.0 g NaOH × = 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

  26. 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

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

  28. Molar Concentration Problem 1 mol HCl 1000 mL solution 9.15 g HCl × × 36.46 g HCl 12.0 mol HCl • 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. = 20.9 mL solution Chapter 14

  29. 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

  30. 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 × V1 = M2 × V2. • M1 and V1 are the initial molarity and volume, and M2 and V2 are the new molarity and volume. Chapter 14

  31. Dilution Problem (0.10 M) × (5.00 L) V1 = = 0.083 L 6.0 M • What volume of 6.0 M NaOH needs to be diluted to prepare 5.00 L if 0.10 M NaOH? • We want final volume and we have our final volume and concentration. M1 × V1 = M2 × V2 (6.0 M) × V1 = (0.10 M) × (5.00 L) Chapter 14

  32. 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

  33. Solution Stoichiometry Problem 187.77 g AgBr 0.100 mol AlBr3 3 mol AgBr 37.5 mL soln × × × 1 mol AgBr 1000 mL soln 1 mol AlBr3 • 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. = 2.11 g AgBr Chapter 14

  34. 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

  35. 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

  36. Chapter Summary, continued moles of solute = M liters of solution mass of solute × 100% = m/m % mass of solution • The mass/mass percent concentration is the mass of solute per 100 grams of solution: • The molarity of a solution is the moles of solute per liter of solution. Chapter 14

  37. Chapter Summary, continued • You can make a solution by diluting a more concentrated solution: M1 × V1 = M2 × 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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