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Water: The Universal Solvent

One of the most valuable properties of water is its ability to dissolve. An individual water molecule has a bent shape with a H-O-H bond angle of approximately 105 degrees. Water is polar thus having positive & negative partial charges on its ends. Water: The Universal Solvent.

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Water: The Universal Solvent

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  1. One of the most valuable properties of water is its ability to dissolve. An individual water molecule has a bent shape with a H-O-H bond angle of approximately 105 degrees. Water is polar thus having positive & negative partial charges on its ends. Water: The Universal Solvent

  2. Ionic Compounds in Water • The positive ends of a water molecule are attracted to negative cations and the negative ends are attracted to positive cations in an ionic compound – this is called hydration. • The ions become hydrated & move around independently.

  3. Covalent Compounds in Water • Water also dissolves many nonionic substances such as ethanol (C2H5OH). • The reason for this is that ethanol is also polar. • Polar dissolves polar – “like dissolves like”. • This is the reason water will not dissolve oil.

  4. Solutions as Electrolytes • Strong conductor = strong electrolyte • Weak conductor = weak electrolyte • No conductor - nonelectrolyte

  5. Solution Concentration • Solution can be expressed in a variety of ways but the most common method is molarity: M = moles solute volume (L) solution • Example: calculate the molarity of a solution prepped by dissolving 1.56 g of gaseous hydrochloric acid in enough water to make 26.8 mL of solution. • 1.60 M HCl

  6. Concentration of Ions in Solution • When an ionic salt dissolves in water ions are form in solution. The moles of ions formed must be considered in concentration. • Example: what is the concentration each ion in (a) 0.50M cobalt II nitrate (b) 2.0M iron III perchlorate. • Co(NO3)2(s) + H2O  Co2+(aq) + 2NO3- (aq) [Co2+] = 1 x 0.50M = 0.50M [NO3-] = 2 x 0.50M = 1.0M • Fe(ClO4)3(s) + H2O  Fe3+(aq) + 3ClO4-(aq) [Fe3+] = 2.0M [ClO4-] = 6.0M

  7. Concentration of Ions in Solution • Example: Calculate the number of moles of chloride ions in 1.75 L of 1.0 x 10-3M zinc chloride. (step 1) ZnCl2(s) Zn2+(aq) + 2Cl-(aq) (step 2) [Cl-] = 2 x 1.0 x 10-3 = 2.0 x 10-3M (step 3) 1.75 L x 2.0 x 10-3 mole Cl- = 1 L 3.5 x 10-3 moles of chloride ions

  8. Example: Concentration & Volume • Typical blood serum is about 0.14M sodium chloride. What volume of blood contains 1.0 mg of sodium chloride? • 0.12 mL of blood • To analyze the alcohol content of a certain wine, a chemist needs 50.00 mL of an aqueous 0.200M potassium dichromate solution. How much solid potassium dichromate must be weighed out to make this solution? • 2.94 g K2Cr2O7

  9. Dilution Formula • This formula allows a chemist to prepare a diluted solution from a concentrated one. M1V1 = M2V2 or McVc = MdVd • Example: what volume of 16 M sulfuric acid must be used to prepare 1.5 L of a 0.10 M sulfuric acid solution? • 9.4 mL of H2SO4 must be diluted with 1.5 L of water.

  10. Types of Chemical Reactions

  11. Precipitation Reactions • When 2 aqueous solutions are mixed an insoluble precipitate sometimes forms – also known as double replacement or metathesis reactions. • It is important to remember that some ions are the key players and some are just spectators: • The formula equation gives the overall reaction. • The complete ionic equation represents all ions involved in the reaction. • The net ionic equation includes only those solution components undergoing a change, spectator ions are not included.

  12. The reaction of Pb(NO3)2 & NaI.Write the formula equation, ionic & net ionic equations:

  13. Stoichiometry of Precipitation Reactions • Calculate the mass of solid sodium chloride that must be added to 1.50 L of a 0.100 M silver nitrate solution to precipitate all the silver ions in the form of silver chloride. • First find the # of moles Na necessary for the # of moles of silver ions already present, then convert to grams. • 8.77 g NaCl

  14. Example: Determine Mass of Product Formed • When aqueous solutions of sodium sulfate and lead II nitrate are mixed a precipitate forms. Give the net ionic equation for the reactions and calculate the mass of this precipitate when 1.25 L of 0.0500 M lead II nitrate and 2.00 L of 0.0250 M sodium sulfate. 15.2 g PbSO4

  15. Acid-Base Reactions • Anacidis a substance that producesH+ ions when dissolved in water. • Abaseis a substance that producesOH− ions when dissolved in water.

  16. Acids and bases as Electrolytes A, Strong acids and bases are strong electrolytes, as indicated by the brightly lit bulb. B, Weak acids and bases are weak electrolytes.

  17. Neutralization Reactions

  18. Strong Acid—Strong Base: Because both ionize completely, the H+ ions and OH- ions react with each other to form water molecules. Basic: HNO3(aq) + NaOH (aq) Ionic: H+ (aq) + NO3-(aq) + Na+ (aq) + OH-(aq) Net Ionic: H+ (aq) + OH-(aq)  H2O (l)

  19. Weak Acid – Strong Base: A two-step process occurs: • 1. The ionization of the weak acid HB (aq)  H+ (aq) + B- (aq) • 2. The neutralization of the H+ ion by the OH- from the strong base H+ (aq) + OH- (aq)  H2O (l) Net Ionic:HX + OH-  X- + H2O

  20. Strong Acid – Weak Base: This is also a two-step process: • 1.The first step occurs when the weak base reacts with water to produce OH- ions. NH3 + H2O  NH4+ + OH- • 2. In the second step the H+ ions from the strong acid neutralize the OH- ions to form water. H+ + OH-  H2O • Net Ionic: H+ + X  XH+

  21. Example: Neutralization Reactions • 1)What volume of a 0.100 M HCl is needed to neutralize 25.0 mL of 0.350 M NaOH? 8.75 x 10-2 L • 2)In a certain experiment, 28.0 mL of 0.250 M nitric acid and 53.0 mL of 0.320 M potassium hydroxide are mixed. Calculate the moles of water formed in the resulting reaction. What is the [H+] and [OH-] after the reaction goes to completion? 0.024 moles H2O, [H+] = 0, [OH-] = 0.123 M

  22. Acid-Base Titration • This is a volumetric analysis technique for determining the amount (usually concentration) of a substance. • This involves the delivery (from a buret) of a measured volume of a solution of known concentration (the titrant) into a solution containing the substance being analyzed (the analyte). • The neutralization point is known as the equivalence point. This point is marked by an indicator. • The point when the indicator changes color is known as the end point. • The goal is to choose an indicator which has a similar endpoint as the equivalence point your reaction.

  23. Titration Example #1 You perform an acid-base titration to standardize an HCl solution by placing 50.00 mL of HCl in a flask with a few drops of indicator solution. You put 0.1524 M NaOH into the buret, and the initial reading is 0.55 mL. At the end point, the buret reading is 33.87 mL. What is the concentration of the HCl solution?

  24. Titration Example #2 In a titration, it is found that 25.0 mL of 0.500 M NaOH is required to react with • (a) a 15.0-mL sample of HCl. What is the molarity of HCl? 0.833 M • (b) a 15.0-mL sample of a weak acid, H2A. What is the molarity of H2A, assuming the reaction to be H2A(aq) + 2OH-(aq)  2H2O + A2-(aq)? 0.417 M • (c) an aspirin tablet weighing 2.50 g. What is the percentage of acetylsalicylic acid, HC9H7O4, in the aspirin tablet? The reaction is HC9H7O4 (s) + OH- (aq)  H2O + C9H7O4 - (aq) 90.0%

  25. Oxidation-Reduction (Redox) • Oxidation means the losing of electrons (an increase in the oxidation #) and reduction means the gaining of electrons (a decrease in the oxidation #). The 2 occur together, they are opposite sides of the same coin. • A good way to remember LEO the lion goes GER – losing electrons oxidation…..gaining electrons reduction OR OIL RIG – oxidation is losing, reduction is gaining.

  26. Oxidizing & Reducing Agents • An oxidizing agent is the species that accepts the electrons i.e. the H+ ion above. • Non-metals tend to be oxidizing agents. • A reducing agent is the species that donates the electrons i.e. the Zn above. • Metals tend to be reducing agents.

  27. Oxidation Numbers • The first step to balancing any redox is assigning oxidation numbers to reactants and products in the equation – please reference pg. 89 in text: • The oxidation # of an element in its elemental state is 0. • The oxidation # of an element in a monatomic ion is equal to the charge of that ion. • Certain elements have the same oxidation in all or almost all their compounds. • The sum of the oxidation numbers in a neutral species is 0; in a polyatomic ion, it is equal to the charge of that ion.

  28. Assigning Oxidation #’s Practice • What is the oxidation number of phosphorus • in sodium phosphate? • P = +5 • In the dihydrogen phosphate ion? • P = +5

  29. Balancing Redox Reactions • Assign oxidation numbers (Rules – Pg. 89) • Identify the oxidation and reduction reactions. Split in 2 half reactions. • Balance the element being oxidized and reduced. • Balance the elements (that is being reduced or oxidized) oxidation # by adding electrons. Oxidation adds to the right, reduction adds to the left. • Balance the oxygens by adding water molecules. • Balance the hydrogens by adding H+ ions. • If the electrons on both sides are not the same you must find the least common multiple between the 2 electrons. Multiply each reaction to get the same number of electrons on both sides. • It is important to check that all atoms and charges balance at this point. THE MISTAKES ARE LIKELY TO HAPPEN HERE!!! • Combine the reactions and simplify if necessary. • If in basic solution add OH- ions to both sides to produce water molecules on the side with H+ ions. Simplify the water molecules if necessary.

  30. Zn (s) + 2HCl (aq) ZnCl2(aq) + H2(g) • Oxidation: Zn  Zn+2 + 2e- • Reduction: 2H+ + 2e-  H2 Net ionic equation: Zn (s) + 2H+ (aq)  H2 (g) + Zn+2 (aq)

  31. Balance the following redox… • (a) Fe2+(aq)+ MnO4- (aq)Fe3+(aq) Mn2+(aq) (acidic solution) • (b) Cl2(g) Cr(OH)3(s) Cl-(aq) CrO42- (aq) (basic solution)

  32. MC #1 • When 70. milliliter of 3.0-molar Na2CO3 is added to 30. milliliters of 1.0-molar NaHCO3 the resulting concentration of Na+ is (A) 2.0 M(B) 2.4 M(C) 4.0 M(D) 4.5 M (E) 7.0 M

  33. MC #2 • A student wishes to prepare 2.00 liters of 0.100-molar KIO3 (molecular weight 214 g/mol). The proper procedure is to weigh out (A) 42.8 grams of KIO3 and add 2.00 kilograms of H2O(B) 42.8 grams of KIO3 and add H2O until the final homogeneous solution has a volume of 2.00 liters(C) 21.4 grams of KIO3 and add H2O until the final homogeneous solution has a volume of 2.00 liters(D) 42.8 grams of KIO3 and add 2.00 liters of H2O(E) 21.4 grams fo KIO3 and add 2.00 liters of H2O

  34. MC #3 • A 20.0-milliliter sample of 0.200-molar K2CO3 solution is added to 30.0 milliliters of 0.400-molar Ba(NO3)2 solution. Barium carbonate precipitates. The concentration of barium ion, Ba2+, in solution after reaction is (A) 0.150 M(B) 0.160 M(C) 0.200 M(D) 0.240 M(E) 0.267 M

  35. MC #4 • The weight of H2SO4 (molecular weight 98.1 g/mol) in 50.0 milliliters of a 6.00-molar solution is (A) 3.10 grams(B) 12.0 grams(C) 29.4 grams(D) 294 grams(E) 300. grams

  36. MC #5 • Given that a solution is 5 percent sucrose by mass, what additional information is necessary to calculate the molarity of the solution? I. The density of waterII. The density of the solutionIII. The molar mass of sucrose (A) I only(B) II only(C) III only(D) I and III(E) II and III

  37. MC #6 • A yellow precipitate forms when 0.5 M NaI(aq) is added to a 0.5 M solution of which of the following ions? A) Pb2+ (aq)B) Zn2+(aq)C) CrO42¯ (aq)D) SO42¯ (aq)E) OH¯ (aq)

  38. MC #7 • When 100 mL of 1.0 M Na3PO4 is mixed with 100 mL of 1.0 M AgNO3, a yellow precipitate forms and [Ag+] becomes negligibly small. Which of the following is a correct listing of the ions remaining in solution in order of increasing concentration? A) [PO43¯] < [NO3¯] < [Na+]B) [PO43¯] < [Na+] < [NO3¯]C) [NO3¯] < [PO43¯] < [Na+]D) [Na+] < [NO3¯] < [PO43¯]E) [Na+] < [PO43¯] < [NO3¯]

  39. MC #8 • The volume of distilled water that should be added to 10.0 mL of 6.00 M HCl(aq) in order to prepare a 0.500 M HCl(aq) solution is approximately A) 50.0 mLB) 60.0 mLC) 100. mLD) 110. mLE) 120. mL

  40. MC #9 • The net ionic equation for the reaction between silver carbonate and hydrochloric acid is (A) Ag2CO3 + 2H+ + 2 Cl¯ ---> 2 AgCl + H2O + CO2 (B) 2Ag+ + CO32¯ + 2 H+ + 2 Cl¯ ---> 2 AgCl + H2O + CO2(C) CO32¯ + 2 H+ ---> H2O + CO2(D) Ag+ + Cl¯ ---> AgCl(E) Ag2CO3 + 2H+ ---> 2Ag+ + H2CO3

  41. MC #10 5 Fe2+ + MnO4¯ + 8 H+ <===> 5 Fe3+ + Mn2+ + 4H2O • In a titration experiment based on the equation above, 25.0 milliliters of an acidified Fe2+ solution requires 14.0 milliliters of standard 0.050-molar MnO4¯ solution to reach the equivalence point. The concentration of Fe2+ in the original solution is (A) 0.0010 M(B) 0.0056 M(C) 0.028 M(D) 0.090 M(E) 0.14 M

  42. FRQ #1 • A 1.2516 gram sample of a mixture of CaCO3 and Na2SO4 was analyzed by dissolving the sample and completely precipitating the Ca2+ as CaC2O4. The CaC2O4 was dissolved in sulfuric acid and the resulting H2C2O4 was titrated with a standard KMnO4 solution. (a) Write the balanced equation for the titration reaction, shown unbalanced below: MnO4- + H2C2O4 + H+  Mn2+ + CO2 + H2O (i) Indicate which substance is the oxidizing agent and which substance is the reducing agent. (b) The titration of the H2C2O4 obtained required 35.62 milliliters of 0.1092 molar MnO4- solution. Calculate the number of moles of H2C2O4 that reacted with the MnO4- (c) Calculate the number of moles of CaCO3 in the original sample. (d) Calculate the percentage by weight of CaCO3 in the original sample.

  43. FRQ #2 • Permanganate ion, MnO4-, oxidizes sulfite ions to sulfate ion. The manganese product depends upon the pH of the reaction mixture. The mole ratio of oxidizing to reducing agent is two to five at pH 1 (acidic), and is two to one at pH 13 (basic). For each of these cases, write a balanced equation for the reaction, and indicate the oxidation state of the manganese in the product containing manganese.

  44. FRQ #3 • A 0.150 g sample of solid lead(II) nitrate is added to 125 mL of 0.100 M sodium iodide solution. Assume no change in volume of the solution. The chemical reaction that takes place is represented by the following equation: Pb(NO3)2(s) + 2 NaI(aq) PbI2(s) + 2NaNO3(aq) (a) List an appropriate observation that provides evidence of a chemical reaction between the two compounds. (b) Calculate the number of moles of each reactant. (c) Identify the limiting reactant. Show calculations to support your identification. (d) Calculate the molar concentration of NO3–(aq) in the mixture after the reaction is complete.

  45. FRQ #4 • Answer the following questions about acetylsalicylic acid, the active ingredient in aspirin. (a) The amount of acetylsalicylic acid in a single aspirin tablet is 325 mg, yet the tablet has a mass of 2.00 g. Calculate the mass percent of acetylsalicylic acid in the tablet. (b) A student dissolved 1.625 g of pure acetylsalicylic acid in distilled water and titrated the resulting solution to the equivalence point using 88.43 mL of 0.102 M NaOH(aq). Assuming that acetylsalicylic acid has only one ionizable hydrogen, calculate the molar mass of the acid.

  46. FRQ #5 (part I) 5 Fe2+(aq) + MnO4–(aq) + 8 H+(aq) 5 Fe3+(aq) + Mn2+(aq) + 4H2O(l) • The mass percent of iron in a soluble iron(II) compound is measured using a titration based on the balanced equation above. (a) What is the oxidation number of manganese in the permanganate ion, MnO4–(aq)? (b) Identify the reducing agent in the reaction represented above. Explain your reasoning. • The mass of a sample of the iron(II) compound is carefully measured before the sample is dissolved in distilled water. The resulting solution is acidified with H2SO4(aq). The solution is then titrated with MnO4–(aq) until the end point is reached. (c) Describe the color change that occurs in the flask when the end point of the titration has been reached. Explain why the color of the solution changes at the end point.

  47. FRQ #5 (part II) (d) Let the variables g, M, and V be defined as follows: g = the mass, in grams, of the sample of the iron(II) compound M = the molarity of the MnO4–(aq) used as the titrant V = the volume, in liters, of MnO4–(aq) added to reach the end point • In terms of these variables, the number of moles of MnO4–(aq)added to reach the end point of the titration is expressed as M x V. Using the variables defined above, the molar mass of iron (55.85 g mol-1), and the coefficients in the balanced chemical equation, write the expression for each of the following quantities (i) The number of moles of iron in the sample (ii) The mass of iron in the sample, in grams (iii) The mass percent of iron in the compound (e) What effect will adding too much titrant have on the experimentally determined value of the mass percent of iron in the compound? Justify your answer.

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