Chapter 4 Reactions in Aqueous Solution

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Chapter 4 Reactions in Aqueous Solution Homework Chapter 4 11, 15, 21, 23, 27, 29, 35, 41, 45, 47, 51, 55, 57, 61, 63, 73, 75, 81, 85 1 2 Chapter Objectives Solution To understand the nature of ionic substances when dissolved in water To be able to predict the water solubility of ionic compounds To show that only a few general types of reactions occur in aqueous solution To learn how to handle quantitative aspects of reactions in aqueous solution A solution is a homogenous mixture of two or more substances, in which one is generally considered the solvent, the medium in which another substance - the solute is dissolved. 3 4 Some Properties of Water Water is bent or V-shaped. The O-H bonds are covalent. Water is a polar molecule. Hydration occurs when salts dissolve in water. A Solute dissolves in water (or other solvent ) changes phase (if different from the solvent) is present in lesser amount (if the same phase as the solvent) 5 6

A Solvent retains its phase (if different from the solute) is present in greater amount (if the same phase as the solute) Electrolytes A substance that ionizes in water or on melting to form an electrically conducting solution. There can be strong or weak electrolytes depending on the concentration of the electrically conducting ions. 7 8 Non-electrolytes A non-electrolyte is something that dissolves in water, but does not conduct electricity. 9 10 Acids Strong acids - dissociate completely to produce H + in solution hydrochloric and sulfuric acid Weak acids - dissociate to a slight extent to give H + in solution acetic and formic acid Bases Strong bases - react completely with water to give OH ions. sodium hydroxide Weak bases - react only slightly with water to give OH ions. ammonia 11 12

Molarity Molarity (M) = moles of solute per volume of solution in liters: M = molarity = moles of solute liters of solution 6 moles of HCl 3 M HCl = 2 liters of solution Common Terms of Solution Concentration Stock - routinely used solutions prepared in concentrated form. Concentrated - relatively large ratio of solute to solvent. (5.0 M NaCl) Dilute - relatively small ratio of solute to solvent. (0.01 M NaCl) 13 14 3 Basic Type of Aqueous Chemical Reactions Precipitation (formation of a solid) Acid / Base Neutralization Gas-forming Oxidation / Reduction (REDOX) 15 16 Precipitation Reactions A precipitation reaction produces an insoluble product called a precipitate. Figure 4.13: When yellow aqueous potassium chromate is added to a colorless barium nitrate solution, yellow barium chromate precipitates. 17

Figure 4.17: The reaction of KCl(aq) with AgNO 3 to form AgCl(s). Example AgNO 3 (aq) + KCl(aq) AgCl(s) + KNO 3 (aq) Reactants Ag + (aq) NO 3- (aq) K + (aq) Cl - (aq) Products AgCl(s) K + (aq) NO 3- (aq) NO 3- (aq) 19 20 Net Ionic Equations Last Example for PPT RXN The balanced equation that results from leaving out the spectator ions. Spectator ions: the ions that are not involved in the creation of the solid. Example: AgNO 3 (aq) + KCl(aq) AgCl(s) + KNO 3 (aq) Ag + (aq) + Cl - (aq) AgCl(s) Write a balanced, net ionic equation for the reaction of AgNO 3 and CaCl 2 to give AgCl and Ca(NO 3 ) 2. AgNO 3 + CaCl 2 AgCl + Ca(NO 3 ) 2 2AgNO 3 (aq) + CaCl 2 (aq) 2 AgCl(s) + Ca(NO 3 ) 2 (aq) 2 Ag + (aq) + 2 Cl - (aq) 2 AgCl(s) 21 22 Acid / Base Reactions Strong Acids and Bases There are 3 major ways of describing acids & bases. We will only discuss one in this chapter. Acid An substance that, when dissolved in pure water, increases the concentration of hydrogen ions (H + ). Base An substance the, when dissolved in pure water, increases the concentration of hydroxide ions (OH - ). 23 A strong acid or base is a compound that fully dissociates in water. Common Strong Acids HCl HBr HI HNO 3 H 2 SO 4 (polyprotic acid) Common Strong Bases LiOH NaOH KOH 24

Weak Acid and Bases A weak acid or base is a compound that does NOT fully dissociates in water. Common Weak Acids H 2 PO 4 CH 3 CO 2 H H 2 CO 3 Common Weak Base NH 3 Performing Calculations for Acid-Base Reactions 1. List initial species and predict reaction. 2. Write balanced net ionic reaction. 3. Calculate moles of reactants. 4. Determine limiting reactant. 5. Calculate moles of required reactant/product. 6. Convert to grams or volume, as required. 25 26 Key Titration Terms Titrant - solution of known concentration used in titration Analyte - substance being analyzed Equivalence point - enough titrant added to react exactly with the analyte Endpoint - the indicator changes color so you can tell the equivalence point has been reached. Figure 4.18a: The titration of an acid with a base. 27 Figure 4.18b: The titration of an acid with a base. Figure 4.18c: The titration of an acid with a base.

Oxides of Nonmetal and Metals There are oxides that when they react with water don t obviously increase the [H + ] or [OH - ] When oxides of nonmetals react with water it produces H + ions. Example: CO 2 reacts with water to create carbonic acid (H 2 CO 3 ) which can dissociate to produce H +. When metal oxides react with water it produces OH - ions. Example: CaO reacts with water to create Ca(OH) 2 which can dissociate to produce OH - Acid / Base Reactions There are two major types of acid / base reactions that we discuss in this chapter. Neutralization A strong acid and base react with each other to create water and a common salt. Gas Producing In reality a gas producing acid / base reaction is still a type of neutralization reaction, but just produces gas. 31 32 Neutralization Example Write the balanced overall equation and the net ionic equation for the reaction of calcium hydroxide with acetic acid. Ca(OH) 2 + CH 3 CO 2 H Ca(CH 3 CO 2 ) 2 + H 2 O Ca(OH) 2 + 2 CH 3 CO 2 H Ca(CH 3 CO 2 ) 2 + 2 H 2 O Gas-Forming Reactions Are reactions with an acid and some other compound that form gases. 33 34 Example: CaCO 3 (s) + HCl(aq) CaCl 2 (aq) + H 2 CO 3 (aq) H 2 CO 3 (aq) H 2 O(l) + CO 2 (g) Exchange Reactions Oxidation Numbers All of the three different types of reactions we have talked about so far can be examined by using exchange reactions. In other words, the ions of the reactants changed partners Example A + B - + C + D - A + D - + C + B - 35 The oxidation number of an atom in a molecule or ion is defined as the electric charge an atom has, or appears to have. Example: Al 2 O 3 We know that the charge on O is 2-, based upon that we can determine the oxidation number on Al. 2-3 = 6- Since we have 2 Al, the charge must be +3 then. 36

Guidelines for Determine Oxidation Numbers (p. 171) 1. Each atom in a pure element has the oxidation number of zero. 2. For ions consisting of a single atom, the oxidation number is equal to the charge on the ion. 3. Fluorine is always -1 in compounds with other elements. 4. Cl, Br, and I are always -1 in compounds except when combined with oxygen and fluorine. 5. The oxidation number of H is +1 and O is -2 in most compounds. 6. The algebraic sum of the oxidation numbers in a neutral compound must be zero, in a polyatomic ion, the sum must be equal to the ion charge. 37 Example Oxidation Number Determinations What are the oxidation numbers on all of the elements in the following compounds? H 2 SO 4 H is +1 O is -2 So, 2(+1) + 4(-2) = -6 Therefore, S must be +6 MnO - 4 O is -2 So, 4(-2) = -8 Therefore, Mn must be +7 because of the overall charge of -1 on the ployatomic ion. Cr 2 O 2-7 O is -2 So, 7(-2) = -14 Therefore, Cr must be +6 because the overall charge is -2 and that there are two atoms of Cr. 38 Oxidation Reduction Reactions A reaction involving the transfer of one or more electrons from one species to another. This type of reaction can take place in acidic and basic solutions. Example: Cu(s) + 2NO 3- (aq) + 4H + (aq) Cu 2+ (aq) + 2NO 2 (g) + 2H 2 O(l) The problem with REDOX reactions is trying to identify and balance them. Right now we are just going to learn how to identify REDOX reactions. 39 Identifying REDOX Reactions The easiest way to determine if a REDOX reaction is occurring is to check the oxidation state of the easiest identifiable metal. Example: Fe 2 O 3 (s) + 2 Al(s) 2 Fe(s) + Al 2 O 3 (s) Fe +3 on the reactant side. Fe 0 on the product side. Therefore it is being reduced and is the oxidizing agent 40 What?! Something is Reduced When the oxidation state of that compound is lowered. That compound is then the oxidizing agent. Something is Oxidized When the oxidation number of that compound is increased. That compound is then called the reducing agent. 41 42

Back to the Example Another Example Fe 2 O 3 (s) + 2 Al(s) 2 Fe(s) + Al 2 O 3 (s) Fe is the most identifiable metal. Figure out the oxidation states of iron for both iron containing compounds. Compare oxidation state of Fe. If the oxidation state of Fe lowers it is being reduced. If the oxidation state of Fe raises it is being oxidized. Identify the other element being either reduced of oxidized. Al 0 Al +3 + 3 e - 43 So Al is being oxidized and is the reducing agent. 2 K(s) + 2 H 2 O(l) 2 KOH + H 2 (g) Is this a PPT RXN? Is this an acid/base RXN? Check the oxidation state of the most identifiable metal. K 0 on the reactant side. K + on the product side. So, K is being oxidized and is the reducing agent. H + on the reactant side. H 0 on the product side So, H is being reduced and is the oxidizing agent. 44 Balancing REDOX Reactions Acid REDOX reactions Identify that it IS a REDOX reaction. Break up in to two half reactions. Balance all non-redox components. Balance the oxygen with water. Balance the hydrogen with H +. Balance the charge with electrons (e - ). Find the lowest common multiple between the two half reactions with respect to electrons and multiply each half reaction accordingly. Cancel certain components. Add the two half reactions up. 45 Balancing REDOX Reactions 46 Base REDOX reaction Follow the same procedure to balance the Acid REDOX reaction. Add OH - to both sides of the equation in order to cancel out the H + to create water. Molarity Example Molarity is a unit of concentration of a solution. Mols of solute Molarity = Liters of solution If I have 23 g of NaCl in a 500 ml solution what is the molarity (M) of that solution? 23 g of NaCl = 0.397 mols NaCl 0.397 mols NaCl / 0.500 L = 0.79 M solution of NaCl 47 48