CHM1045 Lecture 13 Outline of Last Lecture I Properties of Acids II Properties of Bases III Acids IV Neutralization Reaction V Neutralization Reaction Involving a Weak Electrolyte VI Neutralization Reaction Producing a Gas VII Oxidation Reduction Reactions Outline of Current Lecture I Oxidation number II Types of Oxidation Reduction Reactions III The Activity Series for Halogens IV Chemistry in Action Breath Analyzer V Solution Stoichiometry VI Gravimetric Analysis VII Titrations Current Lecture 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 H2 O2 P4 0 2 In monatomic ions the oxidation number is equal to the charge on the ion Li Li 1 Fe3 Fe 3 O2 O 2 3 The oxidation number of oxygen is usually 2 In H2O2 and O22 it is 1 4 The oxidation number of hydrogen is 1 except when it is bonded to metals in binary compounds In these cases its oxidation number is 1 5 Group IA metals are 1 IIA metals are 2 and fluorine is always 1 6 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 The oxidation number of oxygen in the superoxide ion O2 is Example 1 Assign oxidation numbers to all the elements in the following compounds and ion a Li2O b HNO3 c Answer a Lithium has an oxidation of 1 Li and oxygens oxidation number is 2 O2 b H has an oxidation number 1 The other group nitrate ion must have a net oxidation of 1 Oxygen has an oxidation number of 2 and of we use x to represent the oxidation of nitrogen then the nitrate ion can be written N x O 2 3 x 3 2 1 x 5 c The sum of oxidation numbers in the dichromate ion Cr2O2 7 must be 2 We know that the oxidation number of O is 2 so all the remains to determine the oxidation number of Cr which we call y The dichromate ion is written Cr y zO7 2 22 y 7 2 2 y 6 Types of Oxidation Reduction Reactions Combination Reaction A B C 2Al 3Br2 2AlBr3 Decomposition Reaction C A B 2KClO3 2KCl 3O2 Combustion Reaction A O2 B S O2 SO2 2Mg O2 2MgO Displacement Reaction A BC AC B Sr 2H2O Sr OH 2 H2 TiCl4 2Mg Ti 2MgCl2 Cl2 2KBr 2KCl Br2 Hydrogen Displacement Metal Displacement Halogen Displacement The Activity Series for Halogens F2 Cl2 Br2 I2 Halogen Displacement Reaction Cl2 2KBr 2KCl Br2 I2 2KBr 2KI Br2 Types of Oxidation Reduction Reactions Disproportionation Reaction The same element is simultaneously oxidized and reduced Example Reduced Cl2 2OH ClO Cl H2O Oxidized Example 2 Classify the following redox reactions and indicate changes in the oxidation numbers of the elements a b c d Answer a This is a decomposition reaction because one reactant is converted to 2 different products The oxidation of N changes from 1 to 0 while that of O changes from 2 to 0 b Combination reaction Oxidation of Li changes from 0 to 1 while that of N changes from 0 to 3 c Metal displacement reaction Ni metals replaces the Pb2 ion The oxidation of Ni increases from 0 to 2 while that of Pb decreases from 2 to 0 d Oxidation number of N is 4 in NO2 and it is 3 in HNO2 and 5 in HNO3 Because the oxidation of the same element both decreases and increases this is a disproportionation reaction Chemistry in Action Breath Analyzer 3CH3CH2OH 2K2Cr2O7 8H2SO4 3CH3COOH 2Cr2 SO4 3 2K2SO4 11H2O Solution Stoichiometry The concentration of a solution is the amount of solute present in a given quantity of solvent or solution M Molarity moles of solute Liters of solution Example 3 How many grams of potassium dichromate K2Cr2O7 are required to prepare a 250 mL solution whose concentration is 2 16 M Answer moles of solute molarity L soln Moles of K2Cr2O7 2 16 mol K2Cr2O7 1 L soln 540 mol K2Cr2O7 250 L soln The molar mass of K2Cr2O7 is 294 2 g Grams of K2Cr2O7 needed 540 mol K2Cr2O7 294 2 K2Cr2O7 1 mol K2Cr2O7 159 g K2Cr2O7 Example 4 In a biochemical assay a chemist needs to add 3 81 g of glucose to a reaction mixture Calculate the volume in milliliters of a 2 53 M glucose solution she should use for the addition Answer 3 81 g C6H12O6 1 mol C6H12O6 2 114 10 2 mol C6H12O6 180 2 g C6H12O6 V n 2 114 10 2 mol C6H12O6 1000mL soln M 2 53 mol C6H12O6 L soln 1 L soln 8 36 mL soln Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution Example 5 Describe how you would prepare 5 00 102 mL of a 1 75 M H2SO4 solution starting with an 8 61 M stock solution of H2SO4 8 61 M V1 1 75 M 5 00 102 mL V1 1 75 M 5 00 102 mL 8 61 M 102 mL 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 Example 6 A 0 5662 g sample of an ionic compound containing chloride ions and an unknown metal is dissolved in water and treated with an excess of AgNO3 If 1 0882 g of AgCl precipitate forms what is the percent by mass of Cl in the original compound Cl 35 45 g Cl 100 143 4 g AgCl 24 72 mass of Cl Cl 2472 1 0882 g 2690 g 2690 g 100 5662 47 51 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 Titrations can be used in the analysis of Acid base reactions H2SO4 2NaOH 2H2O Na2SO4 Redox reactions 5Fe2 MnO4 8H Mn2 5Fe3 4H2O Example 7 A 16 42 mL volume of 0 1327 M KMnO4 solution is needed to oxidize 25 00 mL of a FeSO4 solution in an acidic medium What is the concentration of the FeSO4 solution in molarity The net ionic equation is Answer moles of KMnO4 1327 mol KMnO4 16 42 mL 1000 mL soln 2 179 10 3 mol KMnO4 moles FeSO4 2 179 10 3 mol KMnO4 5 mol FeSO4 1 mol KMnO4 1 090 10 2 mol FeSO4 molarity of FeSO4 mol FeSO4 L soln 1 090 10 2 mol FeSO4 1000mL soln 25 00 mL soln 1 L soln 4360 M
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