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Types of Chemical Reactions and Solution Stoichiometry. Molarity. The concentration of a solution measured in moles of solute per liter of solution. mol = M L. Preparation of Molar Solutions.

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Types of Chemical Reactions and Solution Stoichiometry


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    1. Types of Chemical Reactionsand Solution Stoichiometry

    2. Molarity The concentration of a solution measured in moles of solute per liter of solution. mol= M L

    3. Preparation of Molar Solutions Problem: How many grams of sodium chloride are needed to prepare 1.50 liters of 0.500 M NaCl solution? • Step #1: Ask “How Much?” (What volume to prepare?) • Step #2: Ask “How Strong?” (What molarity?) • Step #3: Ask “What mass should I use?.” (Molar mass is?) 1.500 L 0.500 mol 58.44 g = 43.8 g 1 L 1 mol

    4. Dilution Problem: What volume of stock (11.6 M) hydrochloric acid is needed to prepare 250. mL of 3.0 M HCl solution? MstockVstock = MdiluteVdilute (11.6 M)(x Liters) = (3.0 M)(0.250 Liters) x Liters = (3.0 M)(0.250 Liters) 11.6 M = 0.065 L

    5. Dilution • It is not practical to keep solutions of many different concentrations on hand, so chemists prepare more dilute solutions from a more concentrated “stock” solution.

    6. Dilution • What would the final volume of a solution be if 50.0 ml of 0.500 M NaOH is diluted to 0.125 M? • 200 ml

    7. Dilution • 300.0 ml of NaOH has a molarity of 1.25 M. The original volume was 250.0 ml. What was the original molarity? • 1.5 M

    8. Dilution • 30.0 ml of 1.25 M KCl is diluted to 0.450 M. What volume of water was added to the solution? • 53 ml

    9. LEARN THESE FOR QUIZ

    10. Precipitation (Double) Replacement Reactions The ions of two compounds exchange places in an aqueous solution to form two new compounds. AX + BY  AY + BX One of the compounds formed is usually a precipitate (an insoluble solid), an insoluble gas that bubbles out of solution, or a molecular compound, usually water.

    11. Precipitation (Double) replacement forming a precipitate… Precipitation (Double) replacement (ionic) equation Pb(NO3)2(aq) + 2KI(aq)  PbI2(s) + 2KNO3(aq) Complete ionic equation shows compounds as aqueous ions Pb2+(aq) + 2 NO3-(aq) + 2 K+(aq) +2 I-(aq)  PbI2(s) + 2K+(aq) + 2 NO3-(aq) Net ionic equation eliminates the spectator ions Pb2+(aq) + 2 I-(aq)  PbI2(s)

    12. Problem • Using your solubility rules, predict what will happen with the following pair of solutions are mixed: KNO3 (aq) and BaCL2 (aq). • Possible solid products? • K+ + Cl- and NO3- + Ba2+ • KCL and Ba(NO3)2 are soluble in water • Therefore, no precipitate forms

    13. Homework • Problems #35, 37