Chapter 4 Types of Chemical Reactions and Solution Stoichiometry Water is the dissolving medium of the common solvent: Some properties Water is “bent” or v-shaped The O—H bonds are covalent Water is a polar molecule Hydration occurs when salts dissolve in water

The Water Molecule is Polar

Polar Water Molecules Interact with the Positive and Negative Ions of a Salt

A solute (substance being dissolved): Dissolves in water or other solvent Changes phase if different from the solvent Present in lesser amount (if the same phase as the solvent) A solvent: Retains its phase (if different from the solute) Present in greater amount (if the same phase as the solute) Rule for predicting solubility: Like dissolves Like

The Nature of Aqueous Solutions: Strong and Weak Electrolytes Strong electrolytes: Conduct current very efficiently. Strong electrolytes are completely ionized when they are dissolved in water (soluble salts, strong acids, strong bases). Weak electrolytes: Conduct only a small current. Weak electrolytes are substances that exhibit a small degree of ionization in water (weak acids, weak bases). Non electrolytes: Permit no current to flow. Non electrolytes are substances that dissolve in water but do not produce any ions. (pure water, sugar solution).

BaCI 2 Dissolving

HCI (aq) is Completely Ionized

Acetic Acid in Water

The Composition of Solutions To perform stoichiometric calculation in solution we need- The nature of the reaction The amounts of chemicals present in the solutions (concentration) The concentration of a solution can be expressed by Molarity (M). Molarity (M) = moles of solute per volume of solution in liters: M = molarity = (moles of solute)/(liters of solution)

Example: If you weigh out 5.84 g of NaCl and dissolved it in 500. mL of water, what is the molarity of the solution? 5.84 g NaCl x (1 mol NaCl)/(58.44 g NaCl) = mol 500. mL x 1 L/1000 mL = L Molarity = (0.100 moles)/(0.500 L) = M

Concentration of Ions Example: Give the concentration of each type of ion in the following solution: a M Co(NO 3 ) 2 b. 1 M Fe(ClO 4 ) 3 c M Al 2 (SO 4 ) 3 a. Solution reaction: Co(NO 3 ) 2(s) H2O Co 2+ (aq) + 2NO 3 - (aq) Each mole of Co(NO 3 ) 2 that is dissolved, the solution contains 1 mol Co 2+ ions and 2 mol NO 3 - ions. Thus a solution of 0.50 M Co(NO 3 ) 2 contains 0.50 M Co 2+ and (2 x 0.50) M or 1.0 M NO 3 -. b. Fe(ClO 4 ) 3(s) H2O Fe 3+ (aq) + 3ClO 4 - (aq) 1 M Fe(ClO 4 ) 3 contains 1 M Fe 3+ ions and 3 M ClO 4 - ions. c. Al 2 (SO 4 ) 3(s) H2O 2Al 3+ (aq) + 3SO 4 2- (aq) 1.75 M Al 2 (SO 4 ) 3 contains (2 x 1.75) M or 3.50 M Al 3+ ions and (3 x 1.75) M or 5.25 M SO 4 2- ions.

Standard Solution A standard solution is a solution whose concentration is accurately known. Finding volume or weight: How many grams of K 2 Cr 2 O 7 are required to prepare 1.00 L of a 0.20 M solution? M = moles/L Moles = M x L = M x 1.00 L = mol K 2 Cr 2 O mol K 2 Cr 2 O 7 x ( g)/(1 mol) = 58.8 g

What volume is needed to prepare a 1.20 M solution from 10.0 g CaCl 2 ? (mm = g) M = moles/L ==> L = moles/M 10.0 g CaCl 2 x (1 mole CaCl 2 )/111.0 g CaCl 2 ) = moles L = (0.090 moles)/(1.20 M) = L = 75.0 mL

Preparation of a Standard Solution

Dilution The desired molarity solutions are often prepared from concentrated stock solutions (routinely used solutions prepared in concentrated form) by adding water. This process is called dilution. Dilution with water does not alter the numbers of moles of solute present. Moles of solute before dilution = moles of solute after dilution M 1 V 1 = M 2 V 2

(a) A Measuring Pipet (b) A Volumetric (transfer) Pipet

Example: What volume of 16 M sulfuric acid must be used to prepare 1.5 L of 0.10 M H 2 SO 4 solution? V 1 = Volume before dilution = ? M 1 = Concentration before dilution = 16 M V 2 = Volume after dilution = 1.5 L M 2 = Concentration after dilution = 0.10 M M 1 V 1 = M 2 V 2 V 1 =M 2 V 2 / M 1 = (0.10 M x 1.5 L)/ 16 M = L = 9.4 mL

Types of Chemical Reactions Solution Reactions are classified as follows: Precipitation reactions Acid – base reactions Oxidation – reduction reactions Other classification: Formation reaction Decomposition reaction Single replacement Double replacement

Precipitation Reactions When two solutions are mixed, an insoluble substance (solid) sometimes forms and separates from the solution. Such a reaction is called a precipitation reaction, and the solid that forms is called a precipitate. Example: K 2 CrO 4(aq) + Ba(NO 3 ) 2(aq)  Products (Soluble) (Soluble) Contains the ions: K +, CrO 4 2-, Ba 2+, NO 3 - Product must contain both anions and cations Most ionic products contain one type of cation and one type of anion

… Precipitation Reactions continued… The possible combinations are- K 2 CrO 4, KNO 3, BaCrO 4, Ba(NO 3 ) 2 (reactant) (reactant) The only real possibility for the solid yellow products are-KNO 3 or BaCrO 4 Both K + and NO 3 - ions are colorless but CrO 4 2- ion is yellow, so the yellow solid is BaCrO 4. K 2 CrO 4(aq) + Ba(NO 3 ) 2(aq)  BaCrO 4(s) + 2KNO 3(aq)

The Reaction of K 2 CrO 4 (aq) and Ba(NO 3 ) 2 (aq)

Describing Reactions in Solutions Regular balanced equation or molecular equation: (gives the overall reaction stoichiometry but not the actual forms of reactants and products) K 2 CrO 4(aq) + Ba(NO 3 ) 2(aq)  BaCrO 4(s) + 2KNO 3(aq) Complete ionic equation: (represents as ions all reactants and products that are strong electrolyte) 2K + (aq) + CrO 4 2- (aq) + Ba 2+ (aq) + 2NO 3 - (aq)  BaCrO 4(s) + 2K + (aq) + 2NO 3 - (aq) Net ionic equation: (Ions participate in reaction) Ba 2+ (aq) + CrO 4 2- (aq)  BaCrO 4(s)

Spectator ions: The ions that do not participate directly in the reaction, present in solution both before and after the reaction. K + and NO 3 - ions. 2K + (aq) + CrO 4 2- (aq) + Ba 2+ (aq) + 2NO 3 - (aq)  BaCrO 4(s) + 2K + (aq) + 2NO 3 - (aq)

Simple Rules for Solubility 1.Most nitrate (NO 3  ) salts are soluble. 2.Most alkali (group 1A) salts and NH 4 + are soluble. 3.Most Cl , Br , and I  salts are soluble (NOT Ag +, Pb 2+, Hg 2 2+ ) 4.Most sulfate salts are soluble (NOT BaSO 4, PbSO 4, HgSO 4, CaSO 4 ) 5.Most OH  salts are only slightly soluble (NaOH, KOH are soluble, Ba(OH) 2, Ca(OH) 2 are marginally soluble) 6.Most S 2 , CO 3 2 , CrO 4 2 , PO 4 3  salts are only slightly soluble.

Stoichiometry of Precipitation Reactions Step 1: Identify the species present in the combined solution, and determine what reaction occurs. Step 2: Write the balanced net ionic equation for the reaction. Step 3: Calculate the moles of the reactants (use volume and molarity). Step 4: Determine which reactant is limiting. Step 5: Calculate the moles of product or products, as required. Step 6: Convert to grams or other units, as required.

Example: Calculate the mass of solid NaCl that must be added to 1.50 L of a M AgNO 3 solution to precipitate all the Ag + ions in the form of AgCl. Ag + NO - 3 Na + Cl - Ag + (aq) + Cl - (aq)  AgCl (s) 1.50 L x (0.100 mol Ag + ) = mol Ag + Ag + and Cl - react in a 1:1 ratio, mol Cl - ions and mol NaCl x (58.45 g NaCl)/(1 mol NaCl) = 8.77 g NaCl

Acid-Base Reactions Arrhenius: Acid is a substance that produces H + ions when dissolved in water Base is substances that produces OH - ions Bronsted-Lowry: An acid is a proton donor A base is a proton acceptor Balanced equation: HCl + NaOH  NaCl + H 2 O Complete ionic equation: H + + Cl - + Na + + OH -  Na + + Cl - + H 2 O Net ionic equation: H + + OH -  H 2 O

Neutralization Reaction Example: What volume of a M HCl solution is needed to neutralize 25.0 mL of M NaOH? The species available for reaction are – H +, Cl -, Na +, OH - NaCl is soluble, so Na +, Cl - are spectator ions, net ionic equation, H + (aq) + OH - (aq)  H 2 O (l) Moles of OH - ions: 25.0 mL x (1L)/(1000 mL) x (0.350 mol OH - )/(L NaOH) = 8.75 x mol OH - H + and OH - ions react in a 1:1 ratio, 8.75 X mol H + ions is required to neutralize OH - ions present. V x (0.100 mol H + )/L = 8.75 x mol H + V = [(8.75 x mol H + )]/[(0.100 mol H + )/L] = 8.75 x L x (1000 mL)/1L = 87.5 mL

Acid-Base Titration Acid-base reaction is called a neutralization reaction (the acid and base are destroyed to leave a “neutral” solution). Volumetric analysis is a technique for determining the amount of a certain substance by doing a titration. A titration involves delivery of a measured volume of a solution of known concentration (the titrant) into a solution containing the substance being analyzed (the analyte). Equivalence point or stoichiometric point: enough titrant added to react exactly with the analyte. Indicator: A substance added at the beginning of the titration that changes color at or near the equivalence point. End point: The point where the indicator actually changes color.

Oxidation-Reduction Reactions In oxidation-reduction (redox) reaction, one species loses electrons while another species gains electrons. Both processed must occur simultaneously. We can’t have a net ionic reaction that contain only oxidation or only reduction. The number of electron lost by the oxidation process must equal the number gained by the reduction process. We cannot create or destroy electrons. Cu  Cu e - (oxidation) 2Ag + + 2e -  2Ag (reduction) Cu + 2Ag +  Cu 2+ +2Ag (net ionic reaction)

Oxidation States Oxidation states or oxidation numbers provides a way to keep track of electrons (accounting for electrons) in redox reactions. Rules for Assigning Oxidation States: 1.The oxidation state of an atom in an element is O (element in its elemental form is O). 2.The oxidation state of a monatomic ion is the same as its charge. 3.In its compound, fluorine is always assigned an oxidation state of –1.

…Oxidation States continued… 4.Oxygen is assigned an oxidation state –2 in covalent compound (except in peroxides where it is –1) 5.Hydrogen is assigned an oxidation state +1 in its covalent compounds with non metals. 6.Sum of the oxidation states must be zero for an electrically neutral compound, for an ion equals the charge of the ion.

Assigning Oxidation States Example: Assign oxidation states to all atoms in the following. a.CO 2  O = -2 (for each oxygen); C = +4 b.SF 6  F = -1 (for each fluorine); S = +6 c.NO 3 -  O = -2 (each oxygen); N = +5 d.MnO 4 -  O = -2 (each oxygen); Mn = +7 e.HSO 3 -  O = -2 (each oxygen); H = +1, S = +4 f.H 2 O  H = +1 (each); O = -2 g.Li 3 N  Li = +1 (each); N = -3

Oxidation-Reduction Oxidation is an increase in oxidation state or a loss of electron. Reduction is a decrease in oxidation state or a gain of electron. Oxidizing Agent is the electron acceptor (reduced). Reducing Agent is the electron donor (oxidized) PbO (s) + CO (g)  Pb (s) + CO 2(g) (each) Pb  reduced, C  oxidized, PbO  oxidizing agent, CO  reducing agent

A Summary of an Oxidation-Reduction Process

Balancing Oxidation-Reduction Equations Redox reactions in acidic solution Step 1: Write separate equation for the oxidation and reduction half reactions. Step 2: For each half reaction- a.Balance all the elements except hydrogen and oxygen b.Balance oxygen using H 2 O c.Balance hydrogen using H + d.Balance the charge using electrons Step 3: If necessary, multiply one or both balanced half- reactions by an integer to equalize the number of electrons transferred Step 4: Add the half reactions and cancel identical species Step 5: Check that the elements and charges are balanced

Balancing Oxidation-Reduction Reactions Example: Balance the equation for the reaction between permanganate and iron (II) ion in acidic solution MnO 4 - (aq) + Fe 2+ (aq) Acid Fe 3+ (aq) + Mn 2+ (aq) Half reactions: MnO 4 -  Mn 2+ (reduction) Fe 2+  Fe 3+ (oxidation) Balance half reactions: MnO 4 -  Mn 2+ a.Manganese is balanced b.Balance oxygen by adding 4H 2 O to the right side of the equation

..continued.. c.Balance hydrogen by adding 8H + to the left side 8H + (aq) + MnO 4 - (aq)  Mn 2+ (aq) + 4H 2 O d.Equalize the charge by adding electrons 5e - + 8H + (aq) + MnO 4 - (aq)  Mn 2+ (aq) + 4H 2 O (l) Fe 2+ (aq)  Fe 3+ (aq) + e - x 5 5 Fe 2+ (aq)  5 Fe 3+ (aq) + 5e - Add half reactions – 5Fe 2+ (aq) + MnO - 4(aq) + 8H + (aq)  5Fe 3+ (aq) + Mn 2+ (aq) + 4H 2 O (l) Check that elements and charges are balanced.

Balancing Oxidation-Reduction Reactions in Basic Solution Step 1: Use the half reaction method as specified for acidic solution. Step 2: Add a number of OH - ions that is equal to the number of H + ions (to both sides). Step 3: Form H 2 O by combining H + and OH -. Step 4: Check that elements and charges are balanced.

Example: ClO - + CrO 2 -  Cl - + CrO 4 2- (basic solution) First half reaction: ClO - +2e -  Cl - ClO - + 2e -  Cl - + H 2 O ClO - + 2e - +2H +  Cl - + H 2 O Second half reaction: CrO 2 -  CrO 4 2- CrO 2 -  CrO e - CrO H 2 O  CrO e - CrO H 2 O  CrO e - + 4H +

…Continued… Multiply first half reaction by 3 and second half reaction by 2 then add two half reactions. 3ClO - + 6e - + 6H +  3Cl - + 3H 2 O 2CrO H 2 O  2CrO e - + 8H + 3ClO - + 2CrO H 2 O  3Cl - + 2CrO H + Add OH - to both sides to remove H + 3ClO - + 2CrO H 2 O + 2OH -  3Cl - + 2CrO H + + 2OH - 3ClO - + 2CrO H 2 O + 2OH -  3Cl - + 2CrO H 2 O 3ClO - + 2CrO OH -  3Cl - + 2CrO H 2 O

Summary Like dissolves Like Strong, weak and non electrolytes Molarity: moles of solute/Liters of solution Standard Solution: concentration accurately known Dilution: M 1 V 1 = M 2 V 2 Precipitation, acid-base and oxidation reduction reactions Spectator ions Acids: produces H + or proton donor Bases: produces OH - and proton acceptor Acid-Base titration: Neutralization reaction Oxidation-Reduction reactions Oxidation States Balancing redox equations