Unit 4: Types of Chemical Reactions and Solution Stoichiometry

Unit 4: Types of Chemical Reactions and Solution Stoichiometry
Solutions, Predicting Products, and Balancing Oxidation-Reduction Reactions Copyright © MsRazz ChemClass

Unit 4 Topics 4.1 Water, the Common Solvent 4.2 The Nature of Aqueous Solutions: Strong and Weak Electrolytes 4.3 The Composition of Solutions 4.4 Types of Chemical Reactions 4.5 Precipitation Reactions 4.6 Describing Reactions in Solution 4.7 Stoichiometry of Precipitation Reactions 4.8 Acid–Base Reactions 4.9 Oxidation–Reduction Reactions 4.10 Balancing Oxidation–Reduction Equations

Section 4.1: Water the Common Solvent
One of the most important substances on Earth. Can dissolve many different substances. A polar molecule because of its unequal charge distribution. H2O is “bent” or V-shaped (104.5°C between hydrogens) H2O is a polar molecule because the oxygen atom has a greater attraction for electrons.

Dissolution of a Solid in a Liquid
Animation Link: Dissolving - Ionic and Molecular Compounds

Hydration: as ionic solids dissolve in water (break up into individual anions and cations), the + ends of the water attract the anions and the – ends of the water attract the cations.

The solubility of substances in water depends on two things:
The attraction of the ions to each other within the solid. The attraction of the ions to water molecules.

Though “like dissolves like,” water dissolves many nonionic substances, such as ethanol (C2H5OH), which have polar bonds. Thus ionic and polar substances dissolve well in water, which nonpolar substances do not.

Nature of Aqueous Solutions
Section 4.2: The Nature of Aqueous Solutions A solution is a homogeneous mixture made up of a solute and solvent. Solute – substance being dissolved. Solvent – the substance doing the dissolving. Aqueous solution – solutions in which water is the solvent.

Electrolytes Electrical conductivity is the ability of a solution to conduct electricity. Strong Electrolytes – conduct current very efficiently (bulb shines brightly). Weak Electrolytes – conduct only a small current (bulb glows dimly). Nonelectrolytes – no current flows (bulb remains unlit).

Strong Electrolytes These are completely ionized (completely dissociate into separate ions) when dissolved in water. Three classes: 1. Soluble salts Ex: NaCl  Na+ + Cl- 2. Strong acids – Produce H+ when dissolved in water Ex: HCl  H+ + Cl- 3. Strong bases – contain OH- Ex: NaOH  Na+ + OH-

Weak Electrolytes These dissociate only a little, while the majority of the substance does not. Three classes: 1. Slightly soluble salts Ex: BaF2 2. Weak acids Ex: HC2H3O2, HF, HCN, HNO2 3. Weak bases Ex: NH3

Nonelectrolytes These dissolve in water but do not produce ions.
Ex: C2H5OH (ethanol) – this substance is dispersed in water but does not break into smaller components. All molecular substances will act this way.

Animation Link: Strong vs. Weak Electrolytes
Electrolyte Behavior Animation Link: Strong vs. Weak Electrolytes

Concentration is measured in Molarity
Section 4.3: Composition of Solutions Molarity (M) = moles of solute per volume of solution in liters: Molarity = moles of solute liters of solution

Example A g sample of potassium phosphate is dissolved in enough water to make 1.50 L of solution. What is the molarity of the solution? 1 mol K3(PO4) 212.27gK3(PO4) 500.0g K3(PO4) x = 2.355 mol K3(PO4) mol L 2.355 mol K3(PO4) 1.50L M= = = 1.57M

Realize that the solution concentration is given in terms of the form of the solute before it dissolves. It therefore may not accurately reflect the concentration of ions within the solution.

Concentration of Ions For a 0.25 M CaCl2 solution: CaCl2 → Ca2+ + 2Cl– Ca2+: 1 × 0.25 M = 0.25 M Ca2+ Cl–: 2 × 0.25 M = 0.50 M Cl–

Example Calculate the concentration of Cl- ions if 0.70mol of ZnCl2 are dissolved in 1.75L of solution. When ZnCl2 dissolves: ZnCl2  Zn+2 + 2Cl- 0.70 moles 1.75L soln x 2Cl- = 0.80 M Cl-

Concept Check Which of the following solutions contains the greatest number of ions? 400.0 mL of 0.10 M NaCl. 300.0 mL of 0.10 M CaCl2. 200.0 mL of 0.10 M FeCl3. 800.0 mL of 0.10 M sucrose. a) contains mol of ions (0.400 L × 0.10 M × 2). b) contains mol of ions (0.300 L × 0.10 M × 3). c) contains mol of ions (0.200 L × 0.10 M × 4). d) does not contain any ions because sucrose does not break up into ions. Therefore, letter b) is correct.

Notice The solution with the greatest number of ions is not necessarily the one in which: the volume of the solution is the largest. the formula unit has the greatest number of ions.

Making Solutions: Terminology
Standard solution: Solution whose concentration is accurately known. Stock solution: Solutions in concentrated forms (used for dilutions of different concentrations). Dilution: The process of adding water to stock solutions to attain desired concentration.

Dilution Dilution with water does not alter the numbers of moles of solute present. Moles of solute before dilution = moles of solute after dilution M1V1 = M2V2

Example What volume of 16M sulfuric acid must be used to prepare 1.5L of a 0.10M H2SO4 solution? M1V1=M2V2 (16M)(V1) = (0.10M)(1.5L) V1=0.0094L x 1000mL = 9.4mL 1L

Concept Check A 0.50 M solution of sodium chloride in an open beaker sits on a lab bench. Which of the following would decrease the concentration of the salt solution? Add water to the solution. Pour some of the solution down the sink drain. c) Add more sodium chloride to the solution. d) Let the solution sit out in the open air for a couple of days. e) At least two of the above would decrease the concentration of the salt solution.

Section 4.4: Types of Chemical Reactions
Synthesis (review) Decomposition (review) Single replacement (review) Double replacement (review) Combustion (review) Precipitation Reactions Acid–Base Reactions Oxidation–Reduction Reactions Not in your textbook

Synthesis Elements combine with other elements to form a compound.
sodium + chlorine  magnesium + phosphorus  sulfur + aluminum  sodium chloride magnesium phosphide aluminum sulfide

Decomposition Compounds break down into elements and/or smaller compounds. calcium sulfide potassium chlorate sodium carbonate  magnesium hydroxide  hydrogen peroxide  calcium + sulfur potassium chloride + oxygen sodium oxide + carbon dioxide magnesium oxide + water water + oxygen

Single Replacement A more active element will replace a less active element within a compound. Active metals can replace less active metals lithium + copper(II) chloride  Active metals can replace hydrogen in water sodium + water  Active metals can replace hydrogen in acid potassium + nitric acid  Active nonmetals can replace less active nonmetals fluorine + aluminum chloride  copper + lithium chloride sodium hydroxide + hydrogen potassium nitrate + hydrogen aluminum fluoride + chlorine

Gas-producing Certain combinations of ions will produce gases.
NH4+ + OH- produces NH3 + H2O ammonium chloride + sodium hydroxide ammonia + water + sodium chloride H+ + CO3-2 produces CO2 + H2O hydrochloric acid + calcium carbonate carbon dioxide + water + calcium chloride H+ + SO3-2 produces SO2 + H2O acetic acid + lithium sulfite sulfur dioxide + water + lithium acetate H+ + S-2 produces H2S phosphoric acid + silver sulfide  hydrogen sulfide + silver phosphate

Precipitation Reaction
Section 4.5: Precipitation Reactions A double displacement reaction in which a solid forms and separates from the solution. When ionic compounds dissolve in water, the resulting solution contains the separated ions. Precipitate – the solid that forms.

The Reaction of K2CrO4(aq) and Ba(NO3)2(aq)
Ba2+(aq) + CrO42–(aq) → BaCrO4(s)

Precipitation of Silver Chloride

Precipitates Soluble – solid dissolves in solution; (aq) is used in reaction. Insoluble – solid does not dissolve in solution; (s) is used in reaction. Insoluble and slightly soluble are often used interchangeably.

Simple Rules for Solubility

Concept Check Which of the following ions form compounds with Pb2+ that are generally soluble in water? a) S2– b) Cl– c) NO3– d) SO42– e) Na+ a), b), and d) all form precipitates with Pb2+. A compound cannot form between only Pb2+ and Na+.

Formula Equation (Molecular Equation)
Section 4.6: Describing Reactions in Solution Gives the overall reaction stoichiometry but not necessarily the actual forms of the reactants and products in solution. Reactants and products generally shown as compounds. Use solubility rules to determine which compounds are aqueous and which compounds are solids. Example: AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)

Complete Ionic Equation
Represents as ions all reactants and products that are strong electrolytes. Example: Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq)  AgCl(s) + Na+(aq) + NO3-(aq)

Example: Ag+(aq) + Cl-(aq)  AgCl(s)
Net Ionic Equation Includes only those solution components undergoing a change. Show only components that actually react. Example: Ag+(aq) + Cl-(aq)  AgCl(s) Spectator ions are not included (ions that do not participate directly in the reaction). Na+ and NO3- are spectator ions.

Concept Check Write the correct formula equation, complete ionic equation, and net ionic equation for the reaction between cobalt(II) chloride and sodium hydroxide. Formula Equation: CoCl2(aq) + 2NaOH(aq)  Co(OH)2(s) + 2NaCl(aq) Complete Ionic Equation: Co2+(aq) + 2Cl(aq) + 2Na+(aq) + 2OH(aq)  Co(OH)2(s) + 2Na+(aq) + 2Cl(aq) Net Ionic Equation: Co2+(aq) + 2OH(aq)  Co(OH)2(s)

Section 4.7: Solution Stoichiometry
Solving problems involving precipitates from solution makes use of: MOLARITY SOLUBILITY RULES BALANCING EQUATIONS LIMITING REACTANT CALCULATIONS

Solving Stoichiometry Problems for Reactions in Solution:
Identify the species present in the combined solution, and determine what reaction if any occurs. Write the balanced net ionic equation for the reaction. Calculate the moles of reactants. Determine which reactant is limiting. Calculate the moles of product(s), as required. Convert to grams or other units, as required.

Example When aqueous solutions of NaBr and Pb(NO3)2 are mixed, PbBr2 precipitates. Calculate the mass of PbBr2 formed when 1.25L of M Pb(NO3)2 and 2.00L of M NaBr are mixed. When NaBr(aq) (Na+ and Br- ions) are mixed with Pb(NO3)2(aq) (Pb+2 and NO3- ions) the resulting solution will have the ions Na+, Br-, Pb2+ and NO3-. PbBr2 is insoluble and will precipitate. Write the balanced ionic equation. Pb2+ (aq) + 2Br-(aq)  PbBr2(s) Find moles of reactants M Pb(NO3)2 contains M Pb2+ ions. Convert to moles. 1.25L x ( mol Pb2+) = mol Pb2+ / 1 = L M NaBr contains M Br- ions. Convert to moles. 2.00L x ( mol Br-) = mol Br- / 2 = L

Pb+2(aq) + 2Br-(aq)  PbBr2(s)
Example (cont.) When aqueous solutions of NaBr and Pb(NO3)2 are mixed, PbBr2 precipitates. Calculate the mass of PbBr2 formed when 1.25L of M Pb(NO3)2 and 2.00L of M NaBr are mixed. Find limiting reactant. Br- will be limiting. Calculate moles of product. 0.0500molBr- x 1 mol PbBr2 = molPbBr2 2 mol Br- Pb+2(aq) + 2Br-(aq)  PbBr2(s) Convert to grams molPbBr2 x gPbBr2 = 9.18gPbBr2 1 mol PbBr2

Neutralization of a Strong Acid by a Strong Base
Section 4.8: Acid-Base Reactions Animation courtesy of:

Performing Calculations for Acid–Base Reactions
List the species present in the combined solution before any reaction occurs, and decide what reaction will occur. Write the balanced net ionic equation for this reaction. (*For the reaction of a strong acid with a strong base, the net ionic equation is always: H+ + OH-  H2O(l)) Calculate moles of reactants. (For reactions in solution, use molarity and volume of original solutions.) Determine the limiting reactant, where appropriate. Calculate the moles of the required reactant or product. Convert to grams or volume (of solution), as required.

Example What volume of 0.100M HCl solution is required in order to neutralize 25.0mL of 0.350M NaOH? List species present in combined solution, decide reaction. H+, Cl-, Na+, OH- Write the balanced net ionic equation. H+(aq) + OH-(aq)  H2O(l) Find moles of reactants mL NaOH x L x 0.350mol OH- = 8.75x10-3 mol OH- 1000mL L NaOH Find limiting reactant. There is no need to determine the LR because the problem requires the addition of just enough H+ ions to react exactly with the OH- ions present.

V = 8.75x10-2 L of 0.100MHCl required to
Example (cont.) What volume of 0.100M HCl solution is required in order to neutralize 25.0mL of 0.350M NaOH? Calculate moles of reactant. H+ and OH- react in a 1:1 ratio, so 8.75x10-3 mol H+ ions is required to neutralize OH- ions. Convert to needed volume. V x (0.100molH+/L) = 8.75x10-3molH+ V = 8.75x10-2 L of 0.100MHCl required to neutralize 25.0mL of 0.350M NaOH

An acid-base reaction is also known as a neutralization reaction.
Once the exact amount of base is added to react with an acid in solution, the acid is said to have been neutralized.

Acid–Base Titrations Volumetric analysis – a method to determine the concentration of a particular substance by doing a titration. Titration – 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 – enough titrant added to react exactly with the analyte. Endpoint – the indicator changes color so you can tell the equivalence point has been reached.

For a Titration to be Successful:
The exact reaction with titrant and analyte has to be known. The equivalence point must be marked accurately. The volume of titrant needed to react equivalence point must be known accurately.

Section 4.9: Redox Reactions
Reactions in which one or more electrons are transferred.

Video Link: Reaction of Sodium and Chlorine

Oxidation States Provide a way to keep track of electrons in oxidation-reduction reactions (especially those with covalent substances) Oxidation states of atoms in covalent compounds are obtained arbitrarily by assigning the electrons to particular atoms.

Rules for Assigning Oxidation Numbers
Each atom in a pure element has an oxidation number of zero (0). Example: I2, Cu, Al, Na, O2, P4, S8 all have zero oxidation states.

Monatomic ions have oxidation numbers equal to the charge on the ion. Example: Mg2+ = +2, H1+ = +1

Group 1 elements have an oxidation number of +1, group 2 have +2, and group thirteen have +3. Example: NaCl, Na = +1 CaSO4, Ca = +2 LiOH, Li = +1

Fluorine always has an oxidation number of -1 in compounds with other elements. Example: HF, BF3 -1 -1

The oxidation number of H is +1 and for O is -2 in most compounds. a) Exception #1: When H forms a compound with a metal, it will have an oxidation number of -1. Example: LiH +1 -1

The oxidation number of H is +1 and for O is -2 in most compounds. b) Exception #2: When O forms a peroxide it will have a oxidation number of -1. Example: H2O2 +1 -1

The sum of oxidation numbers must be zero (0) for a neutral compound, and for a polyatomic ion it must be equal to the ion charge. Example: ClO1- +1 -2

Oxidation states do not have to be integers. Example: I31-, each I is = -1/3 Fe3O4, each O is -2, then each Fe must be + 8/3

Examples CO2 Cr2O7-2 H2SO4 NO2+1 Predict the oxidation states for each element in the following compounds. +4 -2 +6 -2 +1 +6 -2 +5 -2

Redox Characteristics
Transfer of electrons Transfer may occur to form ions Oxidation – increase in oxidation state (loss of electrons); reducing agent Reduction – decrease in oxidation state (gain of electrons); oxidizing agent LeO the lion says GeR! Lose electrons Oxidize Gain electrons Reduced

Redox Characteristics
The oxidizing agent (electron acceptor): causes another substance to be oxidized and therefore is itself reduced. The reducing agent (electron donor): causes another substance to be reduced and therefore is itself oxidized.

Concept Check Which of the following are oxidation-reduction reactions? Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g) Cr2O72-(aq) + 2OH-(aq)  2CrO42-(aq) + H2O(l) 2CuCl(aq)  CuCl2(aq) + Cu(s) Zn – reducing agent; HCl – oxidizing agent c) CuCl acts as the reducing and oxidizing agent

Balancing Oxidation–Reduction Reactions by Half Reactions
Section 4.10: Balancing Redox Reactions To balance redox reactions, we use the half-reaction method. Oxidation states are assigned to each element. Elements being oxidized or reduced are identified. We then separate the reaction into two half-reactions: one involving oxidation and the other dealing with reduction. The method for balancing half-reactions varies depending on whether the reactions are taking place in acidic or basic (alkaline) solution.

The Half-Reaction Method for Balancing Redox Reactions in Acidic Solution
Assign oxidation numbers to identify which element is oxidized and which is reduced. Write separate equations each for the reduction and oxidation half reactions. For each half reaction: Balance first for the element changing oxidation states Balance for all other elements except hydrogen and oxygen. Balance oxygen with the use of H2O Balance hydrogen using H+ Add electrons to balance for charge If needed, multiply one or both half reactions by an integer to make the electrons gained equal the electrons lost. Combine half-reactions, cancel identical species that appear on either side. Make sure elements and charges are balanced.

Example: Balance the following reaction in acidic solution: MnO4-1 + Fe+2  Fe+3 + Mn+2
oxd +3 +7 -2 +2 red [ Oxd ½ rxn: Fe+2  Fe+3 + 1e- ] x5 Red ½ rxn: 5e- + 8H+1 + MnO4-1  Mn+2 + 4H2O 5Fe+2  5Fe+3 + 5e- 5e- + 8H+1 + MnO4-1  Mn+2 + 4H2O 8H+1 + MnO4-1 + 5Fe+2  5Fe+3 + Mn+2 + 4H2O

The Half-Reaction Method for Redox Reactions in Basic Solution
Use half-reaction method as specified for solutions that are acidic to obtain balanced half reactions as if H+ were present. For each half reaction: Balance first for the element changing oxidation states Balance for all other elements except hydrogen and oxygen. Balance oxygen with the use of H2O Balance hydrogen using H+ *Add OH- to BOTH sides of the equation for each H+ you added *Combine H+ and OH- on the same side to make H2O *Cancel/Reduce H2O Add electrons to balance for charge If needed multiply one or both half reactions by an integer to make the electrons gained equal the electrons lost. Combine half reactions, cancel identical species that appear on either side. Make sure elements and charges are balanced.

Example: Balance the following redox reaction in basic solution: HS-1 + ClO3-1  S + Cl-1
-2 oxd +1 +5 -2 -1 red H2O Oxd ½ rxn: [ OH-1 + HS-1  S + H+1 + OH-1 + 2e- ] x3 Red ½ rxn: 6H2O 6e- + 6OH-1 + 6H+1 + ClO3-1  Cl-1 + 3H2O + 6OH-1 3OH-1 + 3HS-1  3S + 3H2O + 6e- 6e- + 6H2O + ClO3-1  Cl-1 + 3H2O + 3 6OH-1 3HS-1 + ClO3-1  3S + Cl-1 + 3OH-1

Unit 4: Types of Chemical Reactions and Solution Stoichiometry

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