1A5PPT - Chemistry 1A Chapter 5 Water, H2O Water...

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Unformatted text preview: Chemistry 1A Chapter 5 Water, H2O Water Attractions Liquid Water Solutions • A solution, also called a homogeneous mixture, is a mixture whose particles are so evenly distributed that the relative concentrations of the components are the same throughout. • Water solutions are called aqueous solutions. Solution (Homogeneous Mixture) Solute and Solvent • In solutions of solids dissolved in liquids, we call the solid the solute and the liquid the solvent. • In solutions of gases in liquids, we call the gas the solute and the liquid the solvent. • In other solutions, we call the minor component the solute and the major component the solvent. Solution of an Ionic Compound Solution of an Ionic Compound (cont.) Liquid-Liquid Solution Precipitation Reactions • In a precipitation reaction, one product is insoluble in water. • As that product forms, it emerges, or precipitates, from the solution as a solid. • The solid is called a precipitate. • For example, Ca(NO3)2(aq) + Na2CO3(aq) → CaCO3(s) + 2NaNO3(aq) Precipitation Questions • • • • Describe the solution formed at the instant water solutions of two ionic compounds are mixed (before the reaction takes place). Describe the reaction that takes place in this mixture. Describe the final mixture. Write the complete equation for the reaction. Solution of Ca(NO3)2 Solution of Ca(NO3)2 and Na2CO3 at the time of mixing, before the reaction Product Mixture for the reaction of Ca(NO3)2 and Na2CO3 Complete Ionic Equation Spectator Ions • Ions that are important for delivering other ions into solution but that are not actively involved in the reaction are called spectator ions. • Spectator ions can be recognized because they are separate and surrounded by water molecules both before and after the reaction. Net Ionic Equations • An equation written without spectator ions is called a net ionic equation. Ca2+(aq) + CO32−(aq) → CaCO3(s) Writing Precipitation Equations • Step 1: Determine the formulas for the possible products using the general double-displacement equation. AB + CD → AD + CB • Step 2: Predict whether either of the possible products is water insoluble. If either possible product is insoluble, a precipitation reaction takes place, and you may continue with step 3. If neither is insoluble, write “No reaction”. Water Solubility • Ionic compounds with the following ions are soluble. – NH4+, group 1 metal ions, NO3−, and C2H3O2− • Ionic compounds with the following ions are usually soluble. – Cl−, Br−, I− except with Ag+ and Pb2+ – SO42− except with Ba2+ and Pb2+ • Ionic compounds with the following ions are insoluble. – CO32−, PO43−, and OH− except with NH4+ and group 1 metal cations – S2− except with NH4+ and group 1 and 2 metal cations Writing Precipitation Equations (cont) • Step 3: Follow these steps to write the complete equation. – Write the formulas for the reactants separated by a “+”. – Separate the formulas for the reactants and products with a single arrow. – Write the formulas for the products separated by a “+”. – Write the physical state for each formula. • The insoluble product will be followed by (s). • Water-soluble ionic compounds will be followed by (aq). – Balance the equation. Writing Precipitation Equations (cont) • Write the complete ionic equation. – Describe aqueous ionic compounds as ions. – Describe the solid with a complete formula. – Make sure that you have correctly done the following • Balanced the equation • Included charges on ions • Included states Writing Precipitation Equations (cont) • Write the net ionic equation. – Eliminate ions that are in the same form on each side of the complete ionic equation. – Rewrite what’s left and balance. Skills to Master (1) • Convert between names and symbols for the common elements. • Identify whether an element is a metal or a nonmetal. • Determine the charges on many of the monatomic ions. • Convert between the name and formula for polyatomic ions. Skills to Master (2) • Convert between the name and formula for ionic compounds. • Balance chemical equations. • Predict the products of double displacement reactions. • Predict ionic solubility. Arrhenius Acid Definition • An acid is a substance that generates hydronium ions, H3O+ (often described as H+), when added to water. • An acidic solution is a solution with a significant concentration of H3O+ ions. Characteristics of Acids • Acids have a sour taste. • Acids turn litmus from blue to red. • Acids react with bases. Strong and Weak Acids • Strong Acid = due to a completion reaction with water, generates close to one H3O+ for each acid molecule added to water. • Weak Acid = due to a reversible reaction with water, generates significantly less than one H3O+ for each molecule of acid added to water. Strong Acid and Water When HCl dissolves in water, hydronium ions, H3O+, and chloride ions, Cl−, ions form. Solution of a Strong Acid Acetic Acid Weak Acid and Water Acetic acid reacts with water in a reversible reaction, which forms hydronium and acetate ions. Solution of Weak Acid Strong and Weak Acids Recognizing Acids • From name – (something) acid – (metal) hydrogen sulfate – (metal) dihydrogen phosphate • From formula – HX(aq), HaXbOc, HC2H3O2, RCO2H (RCOOH) – (symbol for metal)HSO4 – (symbol for metal)H2PO4 Strong and Weak Acids • Strong Acids – Monoprotic – HCl(aq), HBr(aq), HI(aq), HNO3, HClO4 – H2SO4 • Weak Acid – The rest Arrhenius Base Definitions • A base is a substance that generates OH− when added to water. • A basic solution is a solution with a significant concentration of OH− ions. Characteristics of Bases • Bases have a bitter taste. • Bases feel slippery on your fingers. • Bases turn litmus from red to blue. • Bases react with acids. Strong Bases • Anions in ionic compounds except – Neutral anions - Cl−, Br−, I−, NO3−, ClO4− – Acidic anions – HSO4−, H2PO4− • Uncharged, molecular bases – NH3 Strong Bases • Strong Base = due to a completion reaction with water, generates close to one (or more) OH− for each formula unit of base added to water. – Metal hydroxides (e.g. NaOH) NaOH(s) → Na+(aq) + OH −(aq) – Metal hydrides (e.g. LiH) LiH(s) → Li+(aq) + H−(aq) → H−(aq) + H2O(l) H2(aq) + OH−(aq) – Metal amides (e.g. NaNH2) NaNH2(s) → Na+(aq) + NH2−(aq) NH2−(aq) + H2O(l) → NH3(aq) + OH−(aq) Weak Base • Weak Base = due to a reversible reaction with water, generates significantly less than one OH− for each formula unit of base added to water. – All other bases Ammonia and Water Ammonia reacts with water in a reversible reaction, which forms ammonium and hydroxide ions. Ammonia Solution pH • Acidic solutions have pH values less than 7, and the more acidic the solution is, the lower its pH. • Basic solutions have pH values greater than 7, and the more basic the solution is, the higher its pH. pH Range Neutralization Reactions • Reactions between Arrhenius acids and Arrhenius bases are called neutralization reactions. HNO3(aq) + NaOH(aq) → H2O(l) + NaNO3(aq) Aqueous Nitric Acid Mixture of HNO3 and NaOH Before Reaction. Strong Acid and Strong Base Reaction The hydronium ion, H3O+, from the strong acid reacts with the hydroxide ion, OH−, from the strong base to form water, H2O. Mixture of HNO3 and NaOH After the Reaction Electrolytes • Strong electrolytes ionize (strong acids) or dissociate (water-soluble ionic compounds) completely when added to water, causing the water to conduct electric currents strongly. • Weak electrolytes ionize (weak acids and ammonia) incompletely when added to water, causing the water to conduct electric currents weakly. • Nonelectrolytes (such as alcohols and sugars) do not form ions in solution, and therefore, do not cause water to conduct electric currents. Steps to Neutralization Equations • Do you have an acid and a base? • If you have a strong acid or a strong base or if both are strong, write a single arrow. If both are weak, write a double arrow. • Write the formulas and states for the products. – If the base is ammonia, NH3(aq) + HX(aq) → NH4X(aq) – Otherwise, AB + CD → AD + CB • Balance the complete equation. Steps to Neutralization Equations (2) • Write the complete ionic equation. – Describe strong electrolytes as ions. – Describe everything else with a complete formula. – Make sure that you have correctly done the following • Balanced the equation • Included charges on ions • Included states Steps to Neutralization Equations (3) • Write the net ionic equation. – Eliminate ions that are in the same form on each side of the complete ionic equation. – Rewrite what’s left and balance. Steps to Neutralization Equations (4) • Check to be sure the following is true. – Strong acids described as H+. – Weak acids described with a complete formula. – H2SO4 described as H+ and HSO4−. – Pure (s), (l), or (g) described with complete formula. – Ions in strong electrolytes on both sides of the equation are eliminated. Reaction between an Acid and a Hydroxide Base. • The reaction has the double displacement form. AB + CD → AD + CB – The positive part of the acid is H+. • The hydroxide base can be soluble or insoluble. • The products are water and a water-soluble ionic compound. Reaction between an Acid and a Carbonate Base. • The reaction has the double displacement form. AB + CD → AD + CB – The positive part of the acid is H+. • The products are water, carbon dioxide, and a water-soluble ionic compound. The H2O and the CO2 come from the decomposition of the initial product H2CO3. Arrhenius Acid-Base Reactions? NH3(aq) + HF(aq) base acid H2O(l) + HF(aq) neutral acid NH3(aq) + H2O(l) base neutral H2PO4−(aq) + HF(aq) acid acid NH4+(aq) + F−(aq) H3O+(aq) + F−(aq) NH4+(aq) + OH−(aq) H3PO4(aq) + F−(aq) Acid and Base Definitions • Acid – Arrhenius: a substance that generates H3O+ in water – Brønsted-Lowry: a proton, H+, donor • Base – Arrhenius: a substance that generates OHin water – Brønsted-Lowry: a proton, H+, acceptor • Acid-Base Reaction – Arrhenius: between an Arrhenius acid and base – Brønsted-Lowry: a proton (H+) transfer Brønsted-Lowry Acids and Bases NH3(aq) + HF(aq) base acid H2O(l) + HF(aq) base acid NH3(aq) + H2O(l) base acid H2PO4−(aq) + HF(aq) base acid NH4+(aq) + F−(aq) H3O+(aq) + F−(aq) NH4+(aq) + OH−(aq) H3PO4(aq) + F−(aq) Why Two Definitions for Acids and Bases? (1) • Positive Aspects of Arrhenius Definitions – All isolated substances can be classified as acids (generate H3O+ in water), bases (generate OH- in water), or neither. – Allows predictions, including (1) whether substances will react with a base or acid, (2) whether the pH of a solution of the substance will be less than 7 or greater than 7, and (3) whether a solution of the substance will be sour. • Negative Aspects of Arrhenius Definitions – Does not include similar reactions (H+ transfer reactions) as acid-base reactions. Why Two Definitions for Acids and Bases? (2) • Positive Aspects of Brønsted-Lowry Definitions – Includes similar reactions (H+ transfer reactions) as acid-base reactions. • Negative Aspects of Brønsted-Lowry Definitions – Cannot classify isolated substances as acids (generate H3O+ in water), bases (generate OH− in water), or neither. The same substance can sometimes be an acid and sometimes a base. – Does not allow predictions of (1) whether substances will react with a base or acid, (2) whether the pH of a solution of the substance will be less than 7 or greater than 7, and (3) whether a solution of the substance will be sour. Conjugate Acid-Base Pairs Brønsted-Lowry Acids and Bases NH3(aq) + HF(aq) base acid H2O(l) + HF(aq) base acid NH3(aq) + H2O(l) base acid H2PO4−(aq) + HF(aq) base acid NH4+(aq) + F−(aq) acid base H3O+(aq) + F−(aq) acid base NH4+(aq) + OH−(aq) acid base H3PO4(aq) + F−(aq) acid base Amphoteric Substances Can be a Brønsted-Lowry acid in one reaction and a Brønsted-Lowry base in another? HCO3−(aq) + HF(aq) CO2(g) + H2O(l) + F−(aq) base acid HCO3−(aq) + OH−(aq) CO32−(aq) + H2O(l) acid base H2PO4−(aq) + HF(aq) H3PO4(aq) + F−(aq) base acid H2PO4−(aq) + 2OH−(aq) → PO43−(aq) + 2H2O(l) acid base Oxidation • Historically, oxidation meant reacting with oxygen. 2Zn(s) + O2(g) → 2ZnO(s) Zn → Zn2+ + 2e− or 2Zn → 2Zn2+ + 4e− O + 2e− → O2− or O2 + 4e− → 2O2− Oxidation Redefined (1) • Many reactions that are similar to the reaction between zinc and oxygen were not considered oxidation. • For example, both the zinc-oxygen reaction and the reaction between sodium metal and chlorine gas (described on the next slide) involve the transfer of electrons. Oxidation and Formation of Binary Ionic Compounds Similar to Oxidation of Zinc 2Na(s) + Cl2(g) Na → 2NaCl(s) → Na+ + e− or 2Na → 2Na+ + 2e− Cl + e− → Cl− or Cl2 + 2e− → 2Cl− Oxidation = Loss of Electrons Oxidation Redefined (2) • To include the similar reactions in the same category, oxidation was redefined as any chemical change in which at least one element loses electrons. Zinc Oxide Reduction • The following equation describes one of the steps in the production of metallic zinc. ZnO(s) + C(g) → Zn(s) + CO(g) • Because zinc is reducing the number of bonds to oxygen atoms, historically, zinc was said to be reduced. • When we analyze the changes taking place, we see that zinc ions are gaining two electrons to form zinc atoms. Zn2+ + 2e− → Zn • The definition of reduction was broadened to coincide with the definition of oxidation. According to the modern definition, when something gains electrons, it is reduced. Reduction • The loss of electrons (oxidation) by one substance is accompanied by the gain of electrons by another (reduction). Reduction is any chemical change in which at least one element gains electrons. Memory Aid Oxidizing and Reducing Agents • A reducing agent is a substance that loses electrons, making it possible for another substance to gain electrons and be reduced. The oxidized substance is always the reducing agent. • An oxidizing agent is a substance that gains electrons, making it possible for another substance to lose electrons and be oxidized. The reduced substance is always the oxidizing agent. Identifying Oxidizing and Reducing Agents 2Zn(s) + O2(g) → 2ZnO(s) Zn → Zn2+ + 2e− O + 2e− → O2− • Zinc atoms lose electrons, making it possible for oxygen atoms to gain electrons and be reduced, so zinc is the reducing agent. • Oxygen atoms gain electrons, making it possible for zinc atoms to lose electrons and be oxidized, so O2 is the oxidizing agent. Partial Loss and Gain of Electrons N 2 (g) + O2(g) → 2NO(g) • The N-O bond is a polar covalent bond in which the oxygen atom attracts electrons more than the nitrogen atom. • Thus the oxygen atoms gain electrons partially and are reduced. • The nitrogen atoms lose electrons partially and are oxidized. • N2 is the reducing agent. • O2 is the oxidizing agent. Redox Terms (1) Redox Terms (2) • Oxidation-Reduction Reaction – an electron transfer reaction • Oxidation – complete or partial loss of electrons • Reduction – complete or partial gain of electrons • Oxidizing Agent – the substance reduced; gains electrons, making it possible for something to lose them. • Reducing Agent – the substance oxidized; loses electrons, making it possible for something to gain them. Questions Answered by Oxidation Numbers Is the reaction redox? If any atoms change their oxidation number, yes. What’s oxidized? The element that increases its oxidation number What’s reduced? The element that decreases its oxidation number What’s the reducing agent? The substance with the element oxidized The substance with the element reduced What’s the oxidizing agent? Steps for Determination of Oxidation Numbers • Step 1: Assign oxidation numbers to as many atoms as you can using the guidelines described on the next slide. • Step 2: To determine oxidation numbers for atoms not described on the pervious slide, use the following guideline. – The sum of the oxidation numbers for each atom in the formula is equal to the overall charge on the formula. (This includes uncharged formulas where the sum of the oxidation numbers is zero.) Oxidation Numbers uncharged element 0 no exceptions monatomic ions no exceptions combined fluorine charge on ion -1 no exceptions combined oxygen -2 -1 in peroxides covalently bonded hydrogen +1 no exceptions Single Displacement Single Displacement Reaction Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s) Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) oxidation: Zn(s) → Zn2+(aq) + 2e− reduction: Cu2+(aq) + 2e− → Cu(s) Single Displacement Reaction Example Voltaic Cell • The system in which two halfreactions for a redox reaction are separated allowing the electrons transferred in the reaction to be passed between them through a wire is called voltaic cell. Voltaic Cell Electrodes • The electrical conductors placed in the half-cells are called electrodes. • They can be active electrodes, which participate in the reaction, or passive electrodes, which transfer the electrons into or out of a halfcell but do not participate in the reaction. Anode • The anode is the site of oxidation. • Because oxidation involves loss of electrons, the anode is the source of electrons. For this reason, it is described as the negative electrode. • Because electrons are lost forming more positive (or less negative) species at the anode, the surroundings tend to become more positive. Thus anions are attracted to the anode. Cathode • The cathode is the site of reduction. • By convention, the cathode is the positive electrode. • Because electrons come to the cathode and substances gain these electrons to become more negative (or less positive), the surroundings tend to become more negative. Thus cations are attracted to the cathode. Other Cell Components • A device called a salt bridge can be used to keep the charges balanced. • The portion of the electrochemical cell that allows ions to flow is called the electrolyte. Leclanché Cell or Dry Cell Anode oxidation: Zn(s) → Zn2+(aq) + 2e− Cathode reduction: 2MnO2(s) + 2NH4+(aq) + 2e− → Mn2O3(s) + 2NH3(aq) + H2O(l) Overall reaction: Zn(s) + 2MnO2(s) + 2NH4+(aq) → Zn2+(aq) + Mn2O3(s) + 2NH3(aq) + H2O(l) Dry Cell Image Alkaline Batteries Anode oxidation: Zn(s) + 2OH−(aq) → ZnO(s) + H2O(l) + 2e− Cathode reduction: 2MnO2(s) + H2O(l) + 2e− → Mn2O3(s) + 2OH−(aq) Overall reaction: Zn(s) + 2MnO2(s) → ZnO(s) + Mn2O3(s) Electrolysis • Voltage, a measure of the strength of an electric current, represents the force that moves electrons from the anode to the cathode in a voltaic cell. • When a greater force (voltage) is applied in the opposite direction, electrons can be pushed from what would normally be the cathode toward the voltaic cell’s anode. This process is called electrolysis. • In a broader sense, electrolysis is the process by which a redox reaction is made to occur in the nonspontaneous direction. 2NaCl(l) → 2Na(l) + Cl2(g) Primary and Secondary Batteries • Batteries that are not rechargeable are called primary batteries. • A rechargeable battery is often called a secondary battery or a storage battery. Nickel-Cadmium Battery Anode reaction: Cd(s) + 2OH−(aq) Cd(OH)2(s) + 2e− Cathode reaction: NiO(OH)(s) + H2O(l) + e− Ni(OH)2(s) + OH−(aq) Net Reaction: Cd(s) + 2NiO(OH)(s) + 2H2O(l) Cd(OH)2(s) + 2Ni(OH)2(s) Lead Acid Battery Pb(s) + HSO4−(aq) + H2O(l) PbSO4(s) + H3O+(aq) + 2e− Cathode reaction: PbO2(s) + HSO4−(aq) + 3H3O+(aq) + 2e− PbSO4(s) + 5H2O(l) Net Reaction: Pb(s) + PbO2(s) + 2HSO4−(aq) + 2H3O+(aq) 2PbSO4(s) + 4H2O(l) Conversions to Moles Molarity • Converts between moles of solute and volume of solution Equation Stoichiometry (1) Equation Stoichiometry (1) • Tip-off - The calculation calls for you to convert from amount of one substance to amount of another, both of which are involved in a chemical reaction. • General Steps 1. If you are not given it, write and balance the chemical equation for the reaction. 2. Start your dimensional analysis in the usual way. Equation Stoichiometry (2) 3. Convert from the units that you are given for substance 1 to moles of substance 1. – For pure solids and liquids, this means converting mass to moles using the molar mass of the substance. – Molarity can be used to convert from volume of solution to moles of solute. Equation Stoichiometry (3) 4. Convert from grams of substance 1 to moles of substance 1. 5. Convert from moles of substance 2 to the desired units for substance 2. – – For pure solids and liquids, this means converting moles to mass using the molar mass of substance 2. Molarity can be used to convert from moles of solute to volume of solution. 6. Calculate your answer and report it with the correct significant figures (in scientific notation, if necessary) and unit. Titration • Titration involves the addition of one solution (solution 1) to another solution (solution 2) until a chemical reaction between the components in the solutions is complete. Solution 1 is called the titrant, and we say that it is used to titrate solution 2. The completeness of the reaction is usually shown by a change of color caused by a substance called an indicator. Titration Apparatus Steps for Titration (1) • A specific volume of the solution to be titrated (solution 2) is added to an Erlenmeyer flask. For example, 25.00 mL of a phosphoric acid solution of unknown concentration might be added to a 250-mL Erlenmeyer flask. Steps for Titration (2) • A solution of a substance that reacts with the solute in the solution in the Erlenmeyer flask is added to a buret. This solution in the buret, which has a known concentration, is the titrant. The buret is set up over the Erlenmeyer flask so the titrant can be added to the solution to be titrated. For example, a 1.02 M NaOH solution might be added to a buret, which is set up over the Erlenmeyer flask containing the phosphoric acid solution. Steps for Titration (3) • An indicator is added to the solution being titrated. The indicator is a substance that changes color when the reaction is complete. In our example, phenolphthalein, which is a common an acid-base indicator, is added to the phosphoric acid solution in the Erlenmeyer flask. Phenolphthalein • Phenolphthalein is a substance that has two forms. In acidic conditions, it is in the acid form, which is colorless. In basic conditions, an H+ ion is removed from each phenolphthalein molecule, converting it to its base form, which is red. Steps for Titration (4) • The titrant is slowly added to the solution being titrated until the indicator changes color, showing that the reaction is complete. This stage in the procedure is called the endpoint. In our example, the NaOH solution is slowly added from the buret until the mixture in the Erlenmeyer flask changes from colorless to red. At this point, 34.2 mL of 1.02 M NaOH has been added. Making Solutions • From pure solids • From a pure or almost pure acid (e.g. 17 M HC2H3O2 or 18 M H2SO4) • From any other more concentrated solution Making a Solution from Pure Solid Making a Solution From Concentrated Acid Add Concentrated Acid to Water Dilution Problems mol solute concentrated = mol solute dilute ⎛ --- mol solute conc ⎞ --- L conc soln ⎜ ⎟ = --- L dil soln L conc soln ⎝ ⎠ Vconc M conc = Vdilute M dilute M C VC = M D VD ⎛ --- mol solute dil ⎞ ⎜ ⎟ L dil soln ⎝ ⎠ Making a Solution from a More Concentrated Solution (not concentrated acid) ...
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