Chapter 4 CHEMICAL REACTIONS I. Types of chemical reactions A) There are millions of reactions. We can’t remember them all so what to we do? B) Fall into several categories. C) Your text divides them into three types. 1. Precipitation reactions – (metathesis reactions - 2 compounds exchange parts - generally take place in water solutions of two ionic solids and a solid (ppt) forms.) The general form of the equation for such a reaction is: AB + CD AD + CB 2) Acid-Base reactions. An acid substance react with a substance called a base. Such reactions involve the transfer of a proton (H+) between reactants. 3) Oxidation-reduction reactions. These involve the transfer of electrons between reactants. D) Ionic Theory of Solutions and Solubility Rules 1) A solution is a homogeneous mixture of particles less than one nanometer in diameter distributed evenly throughout. 3) We will focus on solids dissolved in water, but there are other solvents than water, and solutions other than liquids - gaseous (clean air), solid solutions (alloys like steel, solder, brass, bronze. 4) Solutes which dissolve are either electrolytes or nonelectrolytes. a) electrolytes are substances which when dissolved in water result in a solution which conducts electricity. Moving charged particles must be there. These charged particles are called ions. There are strong and weak electrolytes. THE LIGHT BULBS !!! b) nonelectrolytes are substances which when dissolved in water do not conduct electricity - uncharged molecules are the particles in solution. c) Soluble Salts - substances which result from the neutralization of an acid by a base and are soluble in water are of interest - since they dissociate in water to give separate ions NaCl(aq) Na+(aq) + Cl- (aq) CaCl2(aq) Ca2+(aq) + 2 Cl- (aq) C12H22O11(aq) C12H22O11(aq) molecules stay intact - nonelectrolyte 5) The formation of an insoluble solid (a precipitate) drives a chemical reaction. 6) For these reactions we can write three kinds of equations. MOLECULAR, IONIC, AND NET IONIC EQUATIONS a) Molecular Equations-complete formulas are written for all the reactants and products, no ions are written. b) Ionic equations-all strongly soluble electrolytes are written in their dissociated (ionized) forms. c) Net Ionic equations-only involve those chemical species which are involved in a chemical reaction. All spectator ions are eliminated. Spectator ions-those ions which do not participate in the chemical reaction but are present in the reaction mixture. d) Examples Write the molecular, ionic, and net ionic equations for the reaction of an aqueous solution of CaCl2 and an aqueous solution of Na2CO3. 1) the molecular equation is: CaCl2(aq) + Na2CO3(aq) CaCO3(s)+2NaCl(aq 2) the ionic equation is: Ca2+(aq) + 2 Cl-(aq) + 2 Na+(aq) + CO32-(aq) CaCO3(s) + 2 Na+(aq) + 2 Cl-(aq) 3) the net ionic equation is: Ca2+(aq) + CO32- (aq) CaCO3(s) D) PREDICTING METATHESIS REACTIONS (PRECIPITATION REACTIONS) USING SOLUBILITY RULES The solubility rules are used to determine whether precipitation reactions occur or not. 1) you need to know whether the predicted products are soluble or insoluble in water. A precipitate is an insoluble compound formed during a chemical reaction in solution. 9)The solubility rules are given on page 128 - table 4.1 You do not have to memorize these rules, I will give them to you on a quiz or exam. You must know how to use the chart in order to predict whether or not a precipitate will occur and the formula of the precipitate. Examples: If we add a solution of KCl to a AgNO3 solution will a precipitate form? We first write a predicted metathesis reaction. KCl(aq) + AgNO3(aq) KNO3 + AgCl Then we look at the solubility table to see if any of the products are insoluble in water. We see that the table indicates that AgCl is insoluble - most chlorides are soluble except for Ag+, Hg2+, Hg22+ and Pb2+ The molecular equation then becomes: KCl(aq) + AgNO3(aq) KNO3(aq)+ AgCl(s) The net ionic equation is: Cl-(aq) + Ag+(aq) AgCl(s) If we add a solution of NaNO3 to an NH4Cl solution will a precipitate form? The predicted equation then becomes: NaNO3(aq) + NH4Cl(aq) NaCl + NH4NO3 We see that the table indicates that both compounds are soluble - most chlorides are soluble, NaCl is soluble, and nitrates are soluble as well as NH4+ compounds, so mixing these two solutions gives no precipitates, no reaction results. After the we would write N. R. for no reaction. What will result if we add a solution of Pb(NO3)2 to a solution of KI? II) ACID-BASE REACTIONS A) There are three definitions of acids and bases. 1) The Arrhenius definition - which is based on properties of these substances: in water solution, acids taste sour, turn blue litmus to red, react with active metals to give off hydrogen gas, bases have a bitter taste, turn red litmus to blue, feel soapy (slippery) and neutralize acids. The result of these observations led Arrhenius to the theory that acids ionize in water to give H+ ions and bases give OH- ions. HCl(g) dissolves in water to give H+(aq) and Cl-(aq) ions. Therefore an Arrhenius acid. NaOH(s) dissolves in water to give Na+(aq) ions and OH- (aq) ions. Therefore an Arrhenius base. The neutralization reaction then becomes: HCl(aq) + NaOH (aq) H2O(l) + NaCl(aq) Examples: Sodium hydroxide and Nitric acid Potassium hydroxide and Sulfuric acid 2) The Brønsted - Lowry Theory of acids and bases. A B-L acid is a species which donates a proton (H+ ion) to another species. A base is a species which accepts a proton from another species. You must have a reaction in order to name a species as a B-L acid or base. HCl(aq) + H2O (l) H3O+(aq) + Cl-(aq) HCl is the B-L Acid, it donates the proton to the H2O which is the B-L Base since it accepts the proton (H+ion) NH3(g) + H2O(l) NH4+(aq) + OH-(aq) H2O is the B-L Acid, it donates the proton to the NH3 which is the B-L Base since it accepts the proton (H+ion). We will discuss the third theory of acids and bases, the Lewis theory, later. B) Strong and weak acids and bases 1) A strong acid is an acid that ionizes completely in water; it is a strong electrolyte. HCl(aq) + H2O (l) H3O+(aq) + Cl-(aq) At this time you should learn 6 strong acids and bases: 2) A weak acid is an acid that only partially ionizes in water; it is a weak electrolyte. The hydrogen cyanide molecule, HCN, reacts with water to produce a small percentage of ions in solution. HCN(aq) + H2O(l) CN-(aq) + H3O+(aq) Table 4.1 lists some common weak acids, remember, if you memorize the strong ones, all others are weak. 3) A strong base is a base that is present entirely as ions, one of which is OH-. 4) Neutralization reactions involving weak acids. Since KCN is a strong electrolyte, the net ionic equation is 5) There are acid-base reactions which give a gas as a product. The net ionic equation would be: SO32-(aq) + 2H+(aq) H2O(l) + SO2(g) III) REDOX REACTIONS A) When electrons are transferred from one compound to another,an oxidation- reduction is said to occur. B) The species which loses electrons is said to be oxidized. The oxidation number of the species increases. This is an oxidation reaction. C) The species which gains electrons is said to be reduced. The oxidation number of the species decreases. This is a reduction reaction. Oxidizing agents are themselves reduced. They oxidize the other reactant. Reducing agents are therefore oxidized. They reduce the other reactant. D) OXIDATION NUMBERS Simply a bookkeeping method for studying redox reactions. E) Rules for determining oxidation numbers. 1. The oxidation number of any free element is zero, regardless of how complex the allotrope may be. 2. The oxidation number for any simple, monatomic ion is equal to the charge of the ion. 3. The sum of all the oxidation numbers of the atoms in a molecule or polyatomic ion must be equal to the charge of the substance. 4. In its compounds, fluorine always has an oxidation number of –1. 5. In its compounds, hydrogen has an oxidation number of +1 except for hydrides, where the oxidation number is -1. 6. In its compounds, oxygen has an oxidation number of -2 except for the peroxides and superoxides, which have oxidation numbers of -1 and -1/2 respectively. 7. Examples: Find the oxidation numbers of all of the species present in the following compounds. KMnO4 Na2Cr2O7 F. An example: 4 Fe(s) + 3 O2(g) 2 Fe2O3(s) Feo Fe3+ + 3 e- (loss of electrons is oxidation) 4 e + O2o 2 O2- (gain of electrons is reduction) Fe(s) is oxidized - it gives electrons - it is the reducing agent. O2(g) is reduced - it takes electrons - it is the oxidizing agent. G. Most oxidation-reduction reactions fall into one of the following simple categories: 1. Combination 2. Decomposition 3. Displacement 4. Combustion Balancing Redox equations on separate sheets. IV. Solutions A) to describe the relative amounts of solute (that which is in the smaller amount in the solution) and the solvent (greater amount) the general term used is CONCENTRATION. B) Chemists like numbers, so we have quantitative descriptions of concentrations of solutions, one of which is MOLARITY which MUST be related to moles. The important idea here is that when we measure out a certain volume of solution we know how many moles of solute we are getting. The number of grams of solute we are pouring out of the container, is related to the number of PARTICLES OF SOLUTE we have in our new container. 2) Examples: If we have 1.00 L of a 3.00 M solution of HCl, how many moles of HCl are in the liter of solution? How many grams of HCl are in 1.00 L of 3.00 M HCl? If we have 500. mL of 6.00 M HCl solution, how many moles do we have? How many moles of HCl are in 25.0 mL of 6.00 M HCl solution? 3) Preparing solutions of a desired M a) Calculate the amount of solute needed and weigh out that amount. b) Obtain the correct volumetric flask. c) Put some water into the flask, add the solute and dissolve. d) Fill the flask with solvent to the scratch on the flask and shake to homogenize. 4) How would you prepare 250.00 mL of 6.00 M aqueous solution of NaOH from solid NaOH? a) The calculation of the amount of NaOH required is found as follows: b) Obtain the 250.00 mL volumetric flask. c) Add a bit of water, add 60.0 g of NaOH, dissolve. d) Fill to scratch with water and shake to homogenize. MAKE CERTAIN THAT YOU REALIZE THAT YOU DID NOT MEASURE THE AMOUNT OF WATER ADDED. C) Many times we start with concentrated solutions to make dilute solutions. 1) We must calculate the volume of the concentrated solution required to make the requested amount of dilute solution and measure its volume. 2) Obtain the correct volumetric flask. 3) put some water in the flask and a measured amount of concentrated solution. 4) add solvent to the scratch and shake to homogenize. D) To calculate the amount of concentrated solution required, the major concept to keep in mind is that the moles of solute obtained from the concentrated solution are equal to the moles of solute in the dilute solution. Upon dilution, the moles of solute do not change only the amount of solvent changes. moles of solute in the conc. solution = moles of solute in the dilute solution. Mc X Lc or Mc X Vc Md X Ld or Md X Vd Mc X Vc = Md X Vd How would you prepare 250.0 mL of 0.100M H2SO4 from 6.00 M H2SO4? 1) first calculate the amount of concentrated sulfuric acid (6.00 M) we need. 2) obtain a 250.00 mL volumetric flask. 3) put some water in the flask and add the 4.17 mL H2SO4 from 6.00 M bottle. 4) fill to the scratch with water and homogenize.
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