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1 11 Reactions in Aqueous Solutions II: Calculations

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2 Chapter Goals Aqueous Acid-Base Reactions 1.Calculations Involving Molarity 2.Titrations 3.Calculations for Acid-Base Titrations Oxidation-Reduction Reactions 4.Balancing Redox Equations 5.Adding in H +, OH -, or H 2 O to Balance Oxygen or Hydrogen 6.Calculations for Redox Titrations

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3 Calculations Involving Molarity Example 11-1: If 100.0 mL of 1.00 M NaOH and 100.0 mL of 0.500 M H2SO4 solutions are mixed, what will the concentration of the resulting solution be? What is the balanced reaction? –It is very important that we always use a balanced chemical reaction when doing stoichiometric calculations.

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4 Calculations Involving Molarity Reaction Ratio: 2 mmol1 mmol 2 mmol Before Reaction: 100 mmol 50 mmol0 mmol After Reaction: 0 mmol 50 mmol 100 mmol

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5 Calculations Involving Molarity What is the total volume of solution? 100.0 mL + 100.0 mL = 200.0 mL What is the sodium sulfate amount, in mmol? 50.0 mmol What is the molarity of the solution? M = 50 mmol/200 mL = 0.250 M Na 2 SO 4

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6 Calculations Involving Molarity Example 11-2: If 130.0 mL of 1.00 M KOH and 100.0 mL of 0.500 M H 2 SO 4 solutions are mixed, what will be the concentration of KOH and K 2 SO 4 in the resulting solution? What is the balanced reaction?

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7 Calculations Involving Molarity Reaction Ratio:2 mmol1 mmol 2 mmol Before Reaction:130 mmol 50 mmol0 mmol After Reaction:30 mmol 0 mmol50 mmol100 mmol

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8 Calculations Involving Molarity What is the total volume of solution? 130.0 mL + 100.0 mL = 230.0 mL What are the potassium hydroxide and potassium sulfate amounts? 30.0 mmol & 50.0 mmol What is the molarity of the solution? M = 30.0 mmol/230.0 mL = 0.130 M KOH M = 50.0 mmol/230.0 mL = 0.217 M K 2 SO 4

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9 Calculations Involving Molarity Example 11-3: What volume of 0.750 M NaOH solution would be required to completely neutralize 100 mL of 0.250 M H 3 PO 4 ? You do it!

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10 Calculations Involving Molarity

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11 Titrations Acid-base Titration Terminology Titration – A method of determining the concentration of one solution by reacting it with a solution of known concentration. Primary standard – A chemical compound which can be used to accurately determine the concentration of another solution. Examples include KHP and sodium carbonate. Standard solution – A solution whose concentration has been determined using a primary standard. Standardization – The process in which the concentration of a solution is determined by accurately measuring the volume of the solution required to react with a known amount of a primary standard.

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12 Titrations Acid-base Titration Terminology Indicator – A substance that exists in different forms with different colors depending on the concentration of the H + in solution. Examples are phenolphthalein and bromothymol blue. Equivalence point – The point at which stoichiometrically equivalent amounts of the acid and base have reacted. End point – The point at which the indicator changes color and the titration is stopped.

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13 Titrations Acid-base Titration Terminology

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14 Calculations for Acid-Base Titrations Potassium hydrogen phthalate is a very good primary standard. –It is often given the acronym, KHP. –KHP has a molar mass of 204.2 g/mol.

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15 Calculations for Acid-Base Titrations Example 11-4: Calculate the molarity of a NaOH solution if 27.3 mL of it reacts with 0.4084 g of KHP.

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16 Calculations for Acid-Base Titrations Example 11-5: Calculate the molarity of a sulfuric acid solution if 23.2 mL of it reacts with 0.212 g of Na 2 CO 3.

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17 Calculations for Acid-Base Titrations Example 11-6: An impure sample of potassium hydrogen phthalate, KHP, had a mass of 0.884 g. It was dissolved in water and titrated with 31.5 mL of 0.100 M NaOH solution. Calculate the percent purity of the KHP sample. Molar mass of KHP is 204.2 g/mol. NaOH + KHP NaKP + H 2 O You do it!

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18 Calculations for Acid-Base Titrations

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19 Oxidation-Reduction Reactions We have previously gone over the basic concepts of oxidation & reduction in Chapter 4. Rules for assigning oxidation numbers were also introduced in Chapter 4. –Refresh your memory as necessary. We shall learn to balance redox reactions using the half-reaction method.

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20 Balancing Redox Reactions Half reaction method rules: 1.Write the unbalanced reaction. 2.Break the reaction into 2 half reactions: One oxidation half-reaction and One reduction half-reaction Each reaction must have complete formulas for molecules and ions. 3.Mass balance each half reaction by adding appropriate stoichiometric coefficients. To balance H and O we can add: H + or H 2 O in acidic solutions. OH - or H 2 O in basic solutions.

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21 Balancing Redox Reactions 4.Charge balance the half reactions by adding appropriate numbers of electrons. Electrons will be products in the oxidation half- reaction. Electrons will be reactants in the reduction half- reaction. 5.Multiply each half reaction by a number to make the number of electrons in the oxidation half- reaction equal to the number of electrons reduction half-reaction. 6.Add the two half reactions. 7.Eliminate any common terms and reduce coefficients to smallest whole numbers.

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22 Balancing Redox Reactions Example 11-12: Tin (II) ions are oxidized to tin (IV) by bromine. Use the half reaction method to write and balance the net ionic equation.

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23 Balancing Redox Reactions

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24 Balancing Redox Reactions

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25 Balancing Redox Reactions Example 11-13: Dichromate ions oxidize iron (II) ions to iron (III) ions and are reduced to chromium (III) ions in acidic solution. Write and balance the net ionic equation for the reaction.

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26 Balancing Redox Reactions

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27 Balancing Redox Reactions

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28 Balancing Redox Reactions Example 11-15: When chlorine is bubbled into basic solution, it forms hypochlorite ions and chloride ions. Write and balance the net ionic equation. You do it! This is a disproportionation redox reaction. The same species, in this case Cl 2, is both reduced and oxidized.

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Balancing Redox Reactions 29

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30 Calculations for Redox Titrations Just as we have done stoichiometry with acid-base reactions, it can also be done with redox reactions. Example 11-16: What volume of 0.200 M KMnO 4 is required to oxidize 35.0 mL of 0.150 M HCl? The balanced reaction is: You do it!

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31 Calculations for Redox Titrations

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32 Calculations for Redox Titrations Example 11-17: A volume of 40.0 mL of iron (II) sulfate is oxidized to iron (III) by 20.0 mL of 0.100 M potassium dichromate solution. What is the concentration of the iron (II) sulfate solution? From Example 11-19 the balanced equation is: You do it!

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33 Calculations for Redox Titrations

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34 Synthesis Question A 0.7500 g sample of an impure FeSO 4 sample is titrated with 26.25 mL of 0.0200 M KMnO 4 to an endpoint. What is the % purity of the FeSO 4 sample?

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35 Synthesis Question

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36 11 Reactions in Aqueous Solutions II: Calculations

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