Reactions in Aqueous Solution

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1 Reading Assignments: Reactions in Aqueous Solution Chapter 4 Chapter 4 in R. Chang, Chemistry, 9 th Ed., McGraw-Hill, or previous editions. Or related topics in other textbooks. Consultation outside lecture room: Office Hours: Tuesday & Thursday 10 AM -12 noon, Wednesday or by appointment A solution is a homogenous mixture of 2 or more substances The is(are) the substance(s) present in the smaller amount(s) An is a substance that, when dissolved in water, results in a solution that can conduct electricity. A is a substance that, when dissolved, results in a solution that does not conduct electricity. The amount is the substance present in the larger Solution Solvent Solute Soft drink (l) Air (g) Soft Solder (s) H 2 O Sugar, CO 2 N 2 O 2, Ar, CH 4 Pb Sn 4.1 nonelectrolyte weak electrolyte strong electrolyte 4.1

2 Conduct electricity in solution? Cations (+) and Anions (-) CH 3 COOH Ionization of acetic acid CH 3 COO - (aq) + H + (aq) Strong Electrolyte 100% dissociation NaCl (s) H 2 O Na + (aq) + Cl - (aq) A reversible reaction. The reaction can occur in both directions. Weak Electrolyte not completely dissociated CH 3 COOH CH 3 COO - (aq) + H + (aq) Acetic acid is a weak electrolyte because its ionization in water is incomplete Hydration is the process in which an ion is surrounded by water molecules arranged in a specific manner. Nonelectrolyte does not conduct electricity? No cations (+) and anions (-) in solution H 2 O C 6 H 12 O 6 (s) C 6 H 12 O 6 (aq) δ δ + H 2 O

3 Precipitation Reactions Precipitation of Lead Iodide Precipitate insoluble solid that separates from solution precipitate Pb(NO 3 ) 2 (aq) + 2NaI (aq) PbI 2 (s) + 2NaNO 3 (aq) molecular equation Pb NO Na + + 2I - PbI 2 (s) + 2Na + + 2NO 3 - ionic equation PbI 2 Pb I - PbI 2 (s) net ionic equation Na + and NO 3- are spectator ions PbI Solubility is the maximum amount of solute that will dissolve in a given quantity of solvent at a specific temperature. Writing Net Ionic Equations 1. Write the balanced molecular equation. 2. Write the ionic equation showing the strong electrolytes completely dissociated into cations and anions. 3. Cancel the spectator ions on both sides of the ionic equation 4. Check that charges and number of atoms are balanced in the net ionic equation Write the net ionic equation for the reaction of silver nitrate with sodium chloride. AgNO 3 (aq) + NaCl (aq) AgCl (s) + NaNO 3 (aq) Ag + + NO 3- + Na + + Cl - AgCl (s) + Na + + NO

4 Chemistry In Action: An Undesirable Precipitation Reaction Ca 2+ (aq) + 2HCO 3 - (aq) CaCO 3 (s) + CO 2 (aq) + H 2 O (l) CO 2 (aq) CO 2 (g) Acids Have a sour taste. Vinegar owes its taste to acetic acid. Citrus fruits contain citric acid. Cause color changes in plant dyes. React with certain metals to produce hydrogen gas. 2HCl (aq) + Mg (s) MgCl 2 (aq) + H 2 (g) React with carbonates and bicarbonates to produce carbon dioxide gas 2HCl (aq) + CaCO 3 (s) CaCl 2 (aq) + CO 2 (g) + H 2 O (l) Aqueous acid solutions conduct electricity Have a bitter taste. Bases Oxidation-Reduction Reactions (electron transfer reactions) Feel slippery. Many soaps contain bases. Cause color changes in plant dyes. Aqueous base solutions conduct electricity Mg 2Mg e - O 2 + 4e - 2O 2- Oxidation half-reaction (lose e - ) Reduction half-reaction (gain e - ) 2Mg + O 2 + 4e - 2Mg O e - 2Mg + O 2 2MgO

5 Zn (s) + CuSO 4 (aq) ZnSO 4 (aq) + Cu (s) Zn Zn e - Zn is oxidized Zn is the reducing agent Cu e - Cu Cu 2+ is reduced Cu 2+ is the oxidizing agent Copper wire reacts with silver nitrate to form silver metal. What is the oxidizing agent in the reaction? Cu (s) + 2AgNO 3 (aq) Cu(NO 3 ) 2 (aq) + 2Ag (s) Cu Cu e - Ag + + 1e - Ag Ag + is reduced Ag + is the oxidizing agent Oxidation number The charge the atom would have in a molecule (or an ionic compound) if electrons were completely transferred. 1. Free elements (uncombined state) have an oxidation number of zero. Na, Be, K, Pb, H 2, O 2, P 4 2. In monatomic ions, the oxidation number is equal to the charge on the ion. Li +, Li = +1; Fe 3+, Fe = +3; O 2-, O = The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases, its oxidation number is Group IA metals are +1, IIA metals are +2 and fluorine is always The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion. 7. Oxidation numbers do not have to be integers. Oxidation number of oxygen in the superoxide ion, O 2-, is -½. Oxidation numbers of all the elements in HCO 3-? HCO 3-3. The oxidation number of oxygen is usually 2. In H 2 O 2 and O 2 2- it is 1.

6 The oxidation numbers of elements in their compounds Oxidation numbers of all the elements in the following? IF 7 NaIO 3 K 2 Cr 2 O 7 Types of Oxidation-Reduction Reactions Combination Reaction A + B C Al + 3Br 2 2AlBr 3 Decomposition Reaction C A + B KClO 3 2KCl + 3O 2 Types of Oxidation-Reduction Reactions Combustion Reaction A + O 2 B S + O 2 SO MgO 2Mg + O 2

7 Types of Oxidation-Reduction Reactions Displacement Reaction A + BC AC + B Sr + 2H 2 O Sr(OH) 2 + H TiCl 4 + 2Mg Ti + 2MgCl 2 Hydrogen Displacement Metal Displacement Types of Oxidation-Reduction Reactions Disproportionation Reaction Element is simultaneously oxidized and reduced Cl 2 + 2OH - Chlorine Chemistry ClO - + Cl - + H 2 O Cl 2 + 2KBr 2KCl + Br 2 Halogen Displacement Classify the following reactions. Chemistry in Action: Breath Analyzer +6 3CH 3 CH 2 OH + 2K 2 Cr 2 O 7 + 8H 2 SO 4 Ca 2+ + CO 3 2- CaCO 3 NH 3 + H + NH CH 3 COOH + 2Cr 2 (SO 4 ) 3 + 2K 2 SO H 2 O Zn + 2HCl ZnCl 2 + H 2 Ca + F 2 CaF 2

8 Solution Stoichiometry The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. M = = moles of solute liters of solution What mass of KI is required to make 500. ml of a 2.80 M KI solution? M KI M KI volume of KI solution moles KI grams KI 1 L 500. ml x 1000 ml 2.80 mol KI x x 1 L soln 166 g KI 1 mol KI Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution. How would you prepare 60.0 ml of M HNO 3 from a stock solution of 4.00 M HNO 3? M i V i = M f V f Dilution M i = 4.00 M f = V f = 0.06 L V i =? L Add Solvent V i = M f V f M i = x = L = 3 ml Moles of solute before dilution (i) = Moles of solute after dilution (f) 3 ml of acid + 57 ml of water M i V i = M f V f

9 Gravimetric Analysis 1. Dissolve unknown substance in water 2. React unknown with known substance to form a precipitate 3. Filter and dry precipitate 4. Weigh precipitate 5. Use chemical formula and mass of precipitate to determine amount of unknown ion Titrations In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point the point at which the reaction is complete Indicator substance that changes color at (or near) the equivalence point Slowly add base to unknown acid UNTIL the indicator changes color

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