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Solution Chemistry: The Hydrosphere

4. Solution Chemistry: The Hydrosphere. Chapter Outline. 4.1 Solutions on Earth and Other Places Solutions: Solute vs. Solvent 4.2 Concentration Units 4.3 Dilutions 4.4 Electrolytes and Nonelectrolytes 4.5 Acid–Base Reactions: Proton Transfer 4.6 Titrations

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Solution Chemistry: The Hydrosphere

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  1. 4 Solution Chemistry: The Hydrosphere

  2. Chapter Outline • 4.1 Solutions on Earth and Other Places • Solutions: Solute vs. Solvent • 4.2 Concentration Units • 4.3 Dilutions • 4.4 Electrolytes and Nonelectrolytes • 4.5 Acid–Base Reactions: Proton Transfer • 4.6 Titrations • 4.7 Precipitation Reactions • 4.8 Ion Exchange • 4.9 Oxidation–Reduction Reactions: Electron Transfer

  3. Earth: The Water Planet • Earth’s surface: • About 75% covered by water • Depressions in Earth’s crust filled with 1.5 × 1021 L of H2O(ℓ) • Properties of water are responsible for life on Earth and many geographical features. • All natural waters have ionic and molecular compounds dissolved in them.

  4. Solutions • Solutions: • Homogeneous mixtures of two or more substances • solvent = a component of a solution that is present in the greatest amount. • solute = any component in a solution other than the solvent (i.e., the other ingredients in the mixture). • Aqueous solutions → water is the solvent

  5. Solutions (cont.) Water is the solvent; sugar is the solute.

  6. Chapter Outline • 4.1 Solutions on Earth and Other Places • 4.2 Concentration Units • Defining the Amount of Solute in Solution • Mass-based Units: ppm and ppb • Mole-based Units: Molarity • 4.3 Dilutions • 4.4 Electrolytes and Nonelectrolytes • 4.5 Acid–Base Reactions: Proton Transfer • 4.6 Titrations • 4.7 Precipitation Reactions • 4.8 Ion Exchange • 4.9 Oxidation–Reduction Reactions: Electron Transfer

  7. Concentration Units • Concentration can be described qualitatively or quantitatively, and define the amount of solute in a solution: • Quantitative concentration units based on: • Mass of solute • Moles of solute

  8. Concentration Units (cont.) • Parts per million (ppm): • Parts per billion (ppb): • Useful for very small amounts of solute

  9. Molarity • Molarity (M) • As a conversion factor  g of solute • Small concentrations: mM (10–3 M), M (10–6 M)

  10. Practice: Calculating Molarity What is the molarity of an aqueous solution prepared by adding 36.5 g of barium chloride to enough water to make 750.0 mL of solution? - Collect and Organize: We have a mass of solute (36.5 g BaCl2), and a final volume of solution (750.0 mL). We can calculate the molar mass of barium chloride from the chemical formula (208.2 g/mol). We want to determine the molarity of the BaCl2 solution.

  11. Practice: Calculating Molarity What is the molarity of an aqueous solution prepared by adding 36.5 g of barium chloride to enough water to make 750.0 mL of solution? - Analyze: From the units for molarity, we need moles of solute (BaCl2) per liter of solution. With the mass of solute (36.5 g) and the molar mass (208.2 g/mole) we can calculate the moles of solute. The volume of solution will need to be converted from mL to L.

  12. Practice: Calculating Molarity What is the molarity of an aqueous solution prepared by adding 36.5 g of barium chloride to enough water to make 750.0 mL of solution? - Solve: (36.5 g/208.2 g/mol)/(0.7500L) = 0.234 M

  13. Practice: Calculating Molarity What is the molarity of an aqueous solution prepared by adding 36.5 g of barium chloride to enough water to make 750.0 mL of solution? - Think About It: The mass of solute represents a little less than 20% of one mole (.2 moles?). Since the volume is less than 1 liter, the overall molarity should be a little more than 0.2 M, which matches our calculated result.

  14. Practice: Mass of Solute How many grams of aluminum nitrate are required to make 500.0 mL of a 0.0525 M aqueous solution? - Collect and Organize: We know the molarity (0.0525 M) and volume (500.0 mL) of the solution. We want to calculate the mass of solute needed to prepare the solution

  15. Practice: Mass of Solute How many grams of aluminum nitrate are required to make 500.0 mL of a 0.0525 M aqueous solution? - Analyze: We can use molarity and volume to calculate the number of moles of solute required. We can then use the formula mass for the solute, (Al(NO3)3 at 213.0 g/mol)) to convert from moles to mass.

  16. Practice: Mass of Solute How many grams of aluminum nitrate are required to make 500.0 mL of a 0.0525 M aqueous solution? -Solve: (0.0525 mol/L) × (0.500 L) = 0.02625 mols (0.02625 mol) × 213.0 g/mol) = 5.59 g

  17. Practice: Molarity from Density If the density of ocean water at a depth of 10,000 m is 1.071 g/mL and if 25.0 g of water at that depth contains 190 mg of KCl, what is the molarity of KCl? - Collect and Organize: We know the mass of solute (KCl, 190 mg), the mass of water (25.0 g), and the density of water at the specified depth (1.071 g/mL). We want to determine the molarity of KCl in seawater at the specified depth.

  18. Practice: Molarity from Density If the density of ocean water at a depth of 10,000 m is 1.071 g/mL and if 25.0 g of water at that depth contains 190 mg of KCl, what is the molarity of KCl? - Analyze: We can convert the mass of solute to moles using the molar mass (74.55 g/mol). We can convert the mass of water to a volume using density. With the moles of solute and volume of water (in L) we can calculate molarity.

  19. Practice: Molarity from Density If the density of ocean water at a depth of 10,000 m is 1.071 g/mL and if 25.0 g of water at that depth contains 190 mg of KCl, what is the molarity of KCl? -Solve: (190 mg) × (1 g/1000 mg) × (1 mol/74.55 g) = 0.002549 mol KCl (25.0 g) × ( 1 mL/1.071 g) × (1 L/1000 mL) = 0.0233 L Molarity = (0.002549 mol/0.0233 L) = 0.109 M

  20. Chapter Outline • 4.1 Solutions on Earth and Other Places • 4.2 Concentration Units • 4.3 Dilutions • Concentrated (Stock) vs. Dilute Solutions • Preparation of Dilute Solutions • Measuring Concentration • 4.4 Electrolytes and Nonelectrolytes • 4.5 Acid–Base Reactions: Proton Transfer • 4.6 Titrations • 4.7 Precipitation Reactions • 4.8 Ion Exchange • 4.9 Oxidation–Reduction Reactions: Electron Transfer

  21. Dilutions Stock solution – a concentrated solution of a substance used to prepare solutions of lower concentration. Dilution – the process of lowering the concentration of a solution by adding more solvent. (mol solute)stock = (mol solute)dilute Minitial × Vinitial = Mdilute × Vdilute

  22. Dilutions (cont.)

  23. Practice: Diluting Stock Solutions Hydrochloric acid is obtained in 12.0 M stock solution. What volume of stock solution is required to make 500.0 mL of a 0.145 M dilute solution? - Collect and Organize: We know the initial molarity of the stock solution (12.0 M HCl), and the desired molarity and volume of the dilute solution (0.145 M and 500.0 mL, respectively). We want to know the volume of stock solution required to prepare the solution.

  24. Practice: Diluting Stock Solutions Hydrochloric acid is obtained in 12.0 M stock solution. What volume of stock solution is required to make 500.0 mL of a 0.145 M dilute solution? - Analyze: Knowing three of the four variables in the dilution equation, we can rearrange the equation to solve for Vinitial.

  25. Practice: Diluting Stock Solutions Hydrochloric acid is obtained in 12.0 M stock solution. What volume of stock solution is required to make 500.0 mL of a 0.145 M dilute solution? - Solve: Vinitial = (Mfinal ×Vfinal)/(Minitial)  (0.145 M)( 500 mL)/(12.0 M) = 6.04 mL

  26. Practice: Diluting Stock Solutions Hydrochloric acid is obtained in 12.0 M stock solution. What volume of stock solution is required to make 500.0 mL of a 0.145 M dilute solution? - Think About It: Since the final molarity is about 1/10th of the initial volume, we would expect the initial volume to be about 1/10th of the final volume, or about 5 mL. Our calculated result is close and of the right order of magnitude.

  27. Measuring Concentration • Intensity of color can be used to measure concentrations • Beer’s Law: A = ε·b·c • Where: • A = absorbance (amount of light absorbed by sample) • ε = molar absorptivity • b = path length • c = concentration of absorbing species

  28. Measuring Concentration (cont.) (a) (b)

  29. Chapter Outline • 4.1 Solutions on Earth and Other Places • 4.2 Concentration Units • 4.3 Dilutions • 4.4 Electrolytes and Nonelectrolytes • Strong, Weak, and Nonelectrolytes • 4.5 Acid–Base Reactions: Proton Transfer • 4.6 Titrations • 4.7 Precipitation Reactions • 4.8 Ion Exchange • 4.9 Oxidation–Reduction Reactions: Electron Transfer

  30. Electrolytes • Strong electrolytes: • Nearly 100% dissociated into ions • Conducts current efficiently • Examples: solutions of NaCl, HNO3, HCl • NaCl(s) Na+(aq) + Cl– (aq)

  31. Electrolytes (cont.) • Weak electrolytes: • Only partially dissociate into ions • Slightly conductive • Examples: vinegar (aqueous solution of acetic acid); tap water • CH3CO2H(aq)⇄ CH3CO2–(aq) + H+(aq)

  32. Nonelectrolytes Substances in which no ionization occurs; no conduction of electrical current Examples: aqueous solutions of sugar, ethanol, ethylene glycol

  33. Chapter Outline • 4.1 Solutions on Earth and Other Places • 4.2 Concentration Units • 4.3 Dilutions • 4.4 Electrolytes and Nonelectrolytes • 4.5 Acid–Base Reactions: Proton Transfer • Definition of Acids and Bases • Neutralization Reactions • Molecular vs. Ionic Equations: Net Ionic Equations • 4.6 Titrations • 4.7 Precipitation Reactions • 4.8 Ion Exchange • 4.9 Oxidation–Reduction Reactions: Electron Transfer

  34. Acid–Base Reactions Proton donor (acid) Proton acceptor (base) • Brønsted–Lowry definitions: • Acids = proton (H+) donors • Bases = proton acceptors • HCl (aq) + H2O(ℓ) → H3O+(aq) + Cl–(aq) • H+ ions strongly associated with water molecules hydronium ions (H3O+)

  35. Acid–Base Reactions (cont.) • Neutralization – reaction that takes place when an acid reacts with a base and produces a solution of a salt in water. • Salt • Product of a neutralization reaction • Made up of the cationof the base plus the anion of the acid. • Example: HCl + NaOH → NaCl + H2O

  36. Types of Equations • Molecular equations: • Reactants/products are written as undissociated molecules. • HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(ℓ) • Overall ionic equations: • Distinguish between molecular and ionic substances • Ionic species are represented as dissolved ions. • H+(aq) + Cl–(aq) + Na+(aq) + OH–(aq) → Na+(aq) + Cl–(aq) + H2O(ℓ)

  37. Types of Equations (cont.) • Net ionic equations: • Remove spectator ions (ions present in same form on both reactants and products side of chemical equation). • H+(aq) + Cl–(aq) + Na+(aq) + OH–(aq) → Na+(aq) + Cl–(aq) + H2O(ℓ) • Net ionic equation: H+(aq) + OH–(aq) → H2O(ℓ)

  38. Nonmetal Oxides and Acids • Nonmetal oxides • Form acids in hydrolysis reaction • Example: • SO3(g) + H2O(ℓ)  H2SO4(aq)

  39. Strong and Weak Acids • Strong acids/bases: • Dissociate completely in aqueous solution (i.e., strong electrolytes) • Examples: • H2SO4(aq) → H+(aq) + HSO4–(aq) • HSO4–(aq) ⇄ H+(aq) + SO42–(aq)

  40. Strong and Weak Bases • Strong bases • 1A, 2A hydroxides • NaOH(aq) Na+(aq) + OH–(aq) • Weak bases • NH3(aq) + H2O(ℓ) ⇄ NH4+(aq) + OH–(aq)

  41. Water: Acid or Base? Proton donor (acid) Proton acceptor (base) Proton acceptor (base) Proton donor (acid) • Water as base: • HCl(aq) + H2O(ℓ) → H3O+(aq) + Cl– (aq) • Water as acid: • NH3(aq) + H2O(ℓ) ⇄ NH4+(aq) + OH–(aq) • Amphiprotic: acts as acid or base

  42. Chapter Outline • 4.1 Solutions on Earth and Other Places • 4.2 Concentration Units • 4.3 Dilutions • 4.4 Electrolytes and Nonelectrolytes • 4.5 Acid–Base Reactions: Proton Transfer • 4.6 Titrations • Terms: Standard Solutions (Titrants), Equivalence Points, and End Points • Using Titrations to Analyze Unknown Solutions • 4.7 Precipitation Reactions • 4.8 Ion Exchange • 4.9 Oxidation–Reduction Reactions: Electron Transfer

  43. Key Titration Terms Titration – an analytical method to determine the concentration of a solute in a sample by reacting it with a standard solution. Standard solution – a solution of known concentration (also called the titrant.) Equivalence point – point when moles of added titrant is stoichiometrically equivalent to moles of substance being analyzed. End point – point reached when the indicator changes color.

  44. Titration Example H2SO4 + 2 NaOH Na2SO4 + 2 H2O (titrant) H2SO4 (unknown) End point

  45. Stoichiometry Calculations H2SO4 + 2 NaOH → Na2SO4 + 2 H2O At the equivalence point: # moles H2SO4 = (# moles NaOH)/2 Macid·Vacid = (Mbase·Vbase)/(2) Rearrange to find Macid: Macid = (Mbase·Vbase)/(2)(Vacid)

  46. Practice: Acid–Base Titration #1 What is the concentration of sulfuric acid if 15.00 mL of it reacts with 18.45 mL of a 0.0973 M NaOH solution? - Collect and Organize: We know Vacid = 15.00 mL, Vbase= 18.45.mL, and Mbase= 0.0973 M. We need to determine Macidfor the original sulfuric acid sample.

  47. Practice: Acid–Base Titration #1 What is the concentration of sulfuric acid if 15.00 mL of it reacts with 18.45 mL of a 0.0973 M NaOH solution? - Analyze: Knowing the stoichiometry of the reaction, we can rearrange the equation to solve for Macid.

  48. Practice: Acid–Base Titration #1 What is the concentration of sulfuric acid if 15.00 mL of it reacts with 18.45 mL of a 0.0973 M NaOH solution? - Solve: Macid = (Vbase)(Mbase)/(2)(Vacid) = 0.0598 M H2SO4

  49. Practice: Acid–Base Titration #1 What is the concentration of sulfuric acid if 15.00 mL of it reacts with 18.45 mL of a 0.0973 M NaOH solution? - Think About It: The volume of base is only slightly more than the volume of the original acid sample. If the stoichiometry was 1:1, we would expect the molarity of the acid to be similar to the molarity of the base. However, the stoichiometry is 2:1—it takes twice as much base to neutralize a given amount of the sulfuric acid, so we would expect the molarity of acid to be a little more than one half the molarity of the base. Our calculated result is consistent with this estimate.

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