Water – “the universal solvent”

Water – “the universal solvent”
One of the most important substances on Earth. Can dissolve many different substances. A polar molecule because of its unequal charge distribution. Copyright © Cengage Learning. All rights reserved

Dissolution of a solid in a liquid
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A solution is a homogenous mixture of 2 or more substances
The solute is(are) the substance(s) present in the smaller amount(s) The solvent is the substance present in the larger amount Solution Solvent Solute Soft drink (l) H2O Sugar, CO2 Air (g) N2 O2, Ar, CH4 Soft Solder (s) Pb Sn

Ions in Aqueous Solution
Ionic Theory of Solutions Many ionic compounds dissociate into independent ions when dissolved in water These compounds that “freely” dissociate into independent ions in aqueous solution are called electrolytes. Their aqueous solutions are capable of conducting an electric current.

Ionic Theory of Solutions Not all electrolytes are ionic compounds. Some molecular compounds dissociate into ions. The resulting solution is electrically conducting, and so we say that the molecular substance is an electrolyte.

Ionic Theory of Solutions Some molecular compounds dissolve but do not dissociate into ions. These compounds are referred to as nonelectrolytes. They dissolve in water to give a nonconducting solution.

Ionic Theory of Solutions Electrolytes are substances that dissolve in water to give an electrically conducting solution. Thus, in general, ionic solids that dissolve in water are electrolytes. Some molecular compounds, such as acids, also dissociate in aqueous solution and are considered electrolytes..

An electrolyte is a substance that, when dissolved in water, results in a solution that can conduct electricity. A nonelectrolyte is a substance that, when dissolved, results in a solution that does not conduct electricity. nonelectrolyte weak electrolyte strong electrolyte

Conduct electricity in solution?
Cations (+) and Anions (-) Strong Electrolyte – 100% dissociation NaCl (s) Na+ (aq) + Cl- (aq) H2O Weak Electrolyte – not completely dissociated CH3COOH CH3COO- (aq) + H+ (aq)

Ionization of acetic acid
CH3COOH CH3COO- (aq) + H+ (aq) A reversible reaction. The reaction can occur in both directions. Acetic acid is a weak electrolyte because its ionization in water is incomplete.

Ionic Theory of Solutions Strong and weak electrolytes. A strong electrolyte is an electrolyte that exists in solution almost entirely as ions. Most ionic solids that dissolve in water do so almost completely as ions, so they are strong electrolytes.

Ionic Theory of Solutions Strong and weak electrolytes. A weak electrolyte is an electrolyte that dissolves in water to give a relatively small percentage of ions. Most soluble molecular compounds are either nonelectrolytes or weak electrolytes.

Ionic Theory of Solutions Solutions of weak electrolytes contain only a small percentage of ions. We denote this situation by writing the equation with a double arrow. Strong and weak electrolytes.

Hydration is the process in which an ion is surrounded by water molecules arranged in a specific manner. d+ d- H2O

Nonelectrolyte does not conduct electricity?
No cations (+) and anions (-) in solution C6H12O6 (s) C6H12O6 (aq) H2O

Chemical Reactions of Solutions
We must know: The nature of the reaction. The amounts of chemicals present in the solutions. Copyright © Cengage Learning. All rights reserved

Working with Solutions
The majority of chemical reactions discussed here occur in aqueous solution. When you run reactions in liquid solutions, it is convenient to dispense the amounts of reactants by measuring out volumes of reactant solutions.

Molar Concentration When we dissolve a substance in a liquid, we call the substance the solute and the liquid the solvent. The general term concentration refers to the quantity of solute in a standard quantity of solution.

Molar Concentration Molar concentration, or molarity [M], is defined as the moles of solute dissolved in one liter (cubic decimeter) of solution.

Solution Stoichiometry
The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. M = molarity = moles of solute liters of solution What mass of KI is required to make 500. mL of a 2.80 M KI solution? M KI M KI volume of KI solution moles KI grams KI 1 L 1000 mL x 2.80 mol KI 1 L soln x 166 g KI 1 mol KI x 500. mL = 232 g KI

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EXERCISE! 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? 1.57 M 500.0 g is equivalent to mol K3PO4 (500.0 g / g/mol). The molarity is therefore 1.57 M (2.355 mol/1.50 L). Copyright © Cengage Learning. All rights reserved

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. Copyright © Cengage Learning. All rights reserved

Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution. Dilution Add Solvent Moles of solute before dilution (i) after dilution (f) = MiVi MfVf =

Diluting Solutions The molarity of a solution and its volume are inversely proportional. Therefore, adding water makes the solution less concentrated. This inverse relationship takes the form of: So, as water is added, increasing the final volume, Vf, the final molarity, Mf, decreases.

How would you prepare 60.0 mL of 0.200 M
HNO3 from a stock solution of 4.00 M HNO3? MiVi = MfVf Mi = 4.00 Mf = 0.200 Vf = 0.06 L Vi = ? L Vi = MfVf Mi = 0.200 x 0.06 4.00 = L = 3 mL 3 mL of acid + 57 mL of water = 60 mL of solution

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. For letter a), adding water to the solution will increase the total volume of solution and therefore decrease the concentration. For letter b), pouring some of the solution down the drain will not change the concentration of the salt solution remaining. For letter c), adding more sodium chloride to the solution will increase the number of moles of salt ions and therefore increase the concentration. For letter d), water will evaporate from the solution and decrease the total volume of solution and therefore increase the concentration. Therefore, since only letter a) would decrease the concentration, letter e) cannot be correct. Copyright © Cengage Learning. All rights reserved

EXERCISE! What is the minimum volume of a 2.00 M NaOH solution needed to make mL of a M NaOH solution? 60.0 mL The minimum volume needed is 60.0 mL. M1V1 = M2V2 (2.00 M)(V1) = (0.800 M)(150.0 mL) Copyright © Cengage Learning. All rights reserved

Types of Chemical Reactions
Most of the reactions we will study fall into one of the following categories Precipitation Reactions Acid-Base Reactions Oxidation-Reduction Reactions

Precipitation Reaction
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. Copyright © Cengage Learning. All rights reserved

Precipitation Reactions A precipitation reaction occurs in aqueous solution because one product is insoluble. A precipitate is an insoluble solid compound formed during a chemical reaction in solution. For example, the reaction of sodium chloride with silver nitrate forms AgCl(s), an insoluble precipitate.

Precipitation of Silver Chloride

Reaction of magnesium chloride and silver nitrate.

Figure 4.4

Precipitation of Lead Iodide
Pb2+ + 2I PbI2 (s) PbI2

The Reaction of K2CrO4(aq) and Ba(NO3)2(aq)
Ba2+(aq) + CrO42–(aq) → BaCrO4(s) Copyright © Cengage Learning. All rights reserved

Precipitation Reactions Solubility rules Substances vary widely in their solubility, or ability to dissolve, in water. For example, NaCl is very soluble in water whereas calcium carbonate, CaCO3, is insoluble in water.

Precipitation Reactions Predicting Precipitation Reactions. To predict whether a precipitate will form, we need to look at potential insoluble products.

Precipitates Soluble – solid dissolves in solution; (aq) is used in reaction equation. Insoluble – solid does not dissolve in solution; (s) is used in reaction equation. Insoluble and slightly soluble are often used interchangeably. Copyright © Cengage Learning. All rights reserved

Simple Rules for Solubility
Most nitrate (NO3-) salts are soluble. Most alkali metal (group 1A) salts and NH4+ are soluble. Most Cl-, Br-, and I- salts are soluble (except Ag+, Pb2+, Hg22+). Most sulfate salts are soluble (except BaSO4, PbSO4, Hg2SO4, CaSO4). Most OH- are only slightly soluble (NaOH, KOH are soluble, Ba(OH)2, Ca(OH)2 are marginally soluble). Most S2-, CO32-, CrO42-, PO43- salts are only slightly soluble, except for those containing the cations in Rule 2. Copyright © Cengage Learning. All rights reserved

Solubility is the maximum amount of solute that will dissolve in a given quantity of solvent at a specific temperature.

Chemistry In Action: - An Undesirable Precipitation Reaction
Ca2+ (aq) + 2HCO3 (aq) CaCO3 (s) + CO2 (aq) + H2O (l) - CO2 (aq) CO2 (g)

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+. Copyright © Cengage Learning. All rights reserved

Formula Equation (Molecular Equation)
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. AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq) Copyright © Cengage Learning. All rights reserved

Complete Ionic Equation
All substances that are strong electrolytes are represented as ions. Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq) AgCl(s) + Na+(aq) + NO3-(aq) Copyright © Cengage Learning. All rights reserved

Net Ionic Equation Includes only those solution components undergoing a change. Show only components that actually react. 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. Copyright © Cengage Learning. All rights reserved

Precipitation Reactions
Precipitate – insoluble solid that separates from solution PbI2 precipitate Pb(NO3)2 (aq) + 2NaI (aq) PbI2 (s) + 2NaNO3 (aq) molecular equation Pb2+ + 2NO3- + 2Na+ + 2I PbI2 (s) + 2Na+ + 2NO3- ionic equation Pb2+ + 2I PbI2 (s) net ionic equation Na+ and NO3- are spectator ions

Writing Net Ionic Equations
Write the balanced molecular equation. Write the ionic equation showing the strong electrolytes completely dissociated into cations and anions. Cancel the spectator ions on both sides of the ionic equation Check that charges and number of atoms are balanced in the net ionic equation Write the net ionic equation for the reaction of silver nitrate with sodium chloride. AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq) Ag+ + NO3- + Na+ + Cl AgCl (s) + Na+ + NO3- Ag+ + Cl AgCl (s)

Molecular and Ionic Equations A molecular equation is one in which the reactants and products are written as if they were molecules, even though they may actually exist in solution as ions. Note that Ca(OH)2, Na2CO3, and NaOH are all soluble compounds but CaCO3 is not.

Molecular and Ionic Equations An ionic equation, however, represents strong electrolytes as separate independent ions. This is a more accurate representation of the way electrolytes behave in solution.

Molecular and Ionic Equations Complete and net ionic equations A complete ionic equation is a chemical equation in which strong electrolytes (such as soluble ionic compounds) are written as separate ions in solution. (strong) (strong) (insoluble)

Molecular and Ionic Equations Complete and net ionic equations. A net ionic equation is a chemical equation from which the spectator ions have been removed. A spectator ion is an ion in an ionic equation that does not take part in the reaction.

Molecular and Ionic Equations Complete and net ionic equations Let’s try an example. First, we start with a molecular equation. Nitric acid, HNO3, and magnesium nitrate, Mg(NO3)2, are both strong electrolytes.

Molecular and Ionic Equations Complete and net ionic equations Separating the strong electrolytes into separate ions, we obtain the complete ionic equation. Note that the nitrate ions did not participate in the reaction. These are spectator ions.

Molecular and Ionic Equations Complete and net ionic equations Eliminating the spectator ions results in the net ionic equation. This equation represents the “essential” reaction.

Precipitation Reactions Predicting Precipitation Reactions. Suppose you mix together solutions of nickel(II) chloride, NiCl2, and sodium phosphate, Na3PO4. How can you tell if a reaction will occur, and if it does, what products to expect?

Precipitation Reactions Predicting Precipitation Reactions. Precipitation reactions have the form of an “exchange reaction.” An exchange (or metathesis) reaction is a reaction between compounds that, when written as a molecular equation, appears to involve an exchange of cations and anions.

Precipitation Reactions Predicting Precipitation Reactions. Now that we have predicted potential products, we must balance the equation. We must verify that NiCl2 and Na3PO4 are soluble and then check the solubilities of the products.

Precipitation Reactions Predicting Precipitation Reactions. The solubility table indicates that our reactants, nickel(II) chloride and sodium phosphate are both soluble. (aq) (aq) (s) (aq) Looking at the potential products we find that nickel(II) phosphate is not soluble although sodium chloride is.

Precipitation Reactions Predicting Precipitation Reactions. We predict that a reaction occurs because nickel(II) phosphate is insoluble and precipitates from the reaction mixture. To see the reaction that occurs on the ionic level, we must rewrite the molecular equation as an ionic equation.

Precipitation Reactions Predicting Precipitation Reactions. First write strong electrolytes (the soluble ionic compounds) in the form of ions to obtain the complete ionic equation

Precipitation Reactions Predicting Precipitation Reactions. After canceling the spectator ions, you obtain the net ionic equation. This equation represents the “essential” reaction.

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) + 2Cl-(aq) Co(OH)2(s) Copyright © Cengage Learning. All rights reserved

Solving Stoichiometry Problems for Reactions in Solution
Identify the species present in the combined solution, and determine what reaction 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. Copyright © Cengage Learning. All rights reserved

Quantitative Analysis
Gravimetric Analysis Gravimetric analysis is a type of quantitative analysis in which the amount of a species in a material is determined by converting the species into a product that can be isolated and weighed. Precipitation reactions are often used in gravimetric analysis. The precipitate from these reactions is then filtered, dried, and weighed.

Gravimetric Analysis Dissolve unknown substance in water
React unknown with known substance to form a precipitate Filter and dry precipitate Weigh precipitate Use chemical formula and mass of precipitate to determine amount of unknown ion

Gravimetric Analysis Consider the problem of determining the amount of lead in a sample of drinking water. Adding sodium sulfate (Na2SO4) to the sample will precipitate lead(II) sulfate. The PbSO4 can then be filtered, dried, and weighed.

Gravimetric Analysis Suppose a 1.00 L sample of polluted water was analyzed for lead(II) ion, Pb2+, by adding an excess of sodium sulfate to it. The mass of lead(II) sulfate that precipitated was mg. What is the mass of lead in a liter of the water? Express the answer as mg of lead per liter of solution.

Gravimetric Analysis First we must obtain the mass percentage of lead in lead(II) sulfate, by dividing the molar mass of lead by the molar mass of PbSO4, then multiplying by 100. Then, calculate the amount of lead in the PbSO4 precipitated.

CONCEPT CHECK! (Part I) 10.0 mL of a 0.30 M sodium phosphate solution reacts with 20.0 mL of a 0.20 M lead(II) nitrate solution (assume no volume change). What precipitate will form? lead(II) phosphate, Pb3(PO4)2 What mass of precipitate will form? 1.1 g Pb3(PO4)2 The balanced molecular equation is: 2Na3PO4(aq) + 3Pb(NO3)2(aq) → 6NaNO3(aq) + Pb3(PO4)2(s). mol Na3PO4 present to start and mol Pb(NO3)2 present to start. Pb(NO3)2 is the limiting reactant, therefore mol of Pb3(PO4)2 is produced. Since the molar mass of Pb3(PO4)2 is g/mol, 1.1 g of Pb3(PO4)2 will form. Copyright © Cengage Learning. All rights reserved

Let’s Think About It Where are we going? To find the mass of solid Pb3(PO4)2 formed. How do we get there? What are the ions present in the combined solution? What is the balanced net ionic equation for the reaction? What are the moles of reactants present in the solution? Which reactant is limiting? What moles of Pb3(PO4)2 will be formed? What mass of Pb3(PO4)2 will be formed? Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! (Part II) 10.0 mL of a 0.30 M sodium phosphate solution reacts with 20.0 mL of a 0.20 M lead(II) nitrate solution (assume no volume change). What is the concentration of nitrate ions left in solution after the reaction is complete? 0.27 M The concentration of nitrate ions left in solution after the reaction is complete is 0.27 M. Nitrate ions are spectator ions and do not participate directly in the chemical reaction. Since there were mol of Pb(NO3)2 present to start, then mol of nitrate ions are present. The total volume in solution is 30.0 mL. Therefore the concentration of nitrate ions = mol / L = 0.27 M. Copyright © Cengage Learning. All rights reserved

Let’s Think About It Where are we going? To find the concentration of nitrate ions left in solution after the reaction is complete. How do we get there? What are the moles of nitrate ions present in the combined solution? What is the total volume of the combined solution? Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! (Part III) 10.0 mL of a 0.30 M sodium phosphate solution reacts with 20.0 mL of a 0.20 M lead(II) nitrate solution (assume no volume change). What is the concentration of phosphate ions left in solution after the reaction is complete? 0.011 M The concentration of phosphate ions left in solution after the reaction is complete is M. Phosphate ions directly participate in the chemical reaction to make the precipitate. There were mol of Na3PO4 present to start, therefore there was mol of phosphate ions present to start mol of phosphate ions were used up in the chemical reaction, therefore mol of phosphate ions is leftover ( – mol). The total volume in solution is 30.0 mL. Therefore the concentration of phosphate ions = mol / L = M. Copyright © Cengage Learning. All rights reserved

Let’s Think About It Where are we going? To find the concentration of phosphate ions left in solution after the reaction is complete. How do we get there? What are the moles of phosphate ions present in the solution at the start of the reaction? How many moles of phosphate ions were used up in the reaction to make the solid Pb3(PO4)2? How many moles of phosphate ions are left over after the reaction is complete? What is the total volume of the combined solution? Copyright © Cengage Learning. All rights reserved

Acids Have a sour taste. Vinegar owes its taste to acetic acid. Citrus
fruits contain citric acid. Cause color changes in plant dyes. React with certain metals to produce hydrogen gas. 2HCl (aq) + Mg (s) MgCl2 (aq) + H2 (g) React with carbonates and bicarbonates to produce carbon dioxide gas 2HCl (aq) + CaCO3 (s) CaCl2 (aq) + CO2 (g) + H2O (l) Aqueous acid solutions conduct electricity.

Bases Have a bitter taste. Feel slippery. Many soaps contain bases.
Cause color changes in plant dyes. Aqueous base solutions conduct electricity.

Pg 129

Household acids and bases. Photo courtesy of American Color.

Acid-Base Reactions Acids and bases are some of the most important electrolytes. They can cause color changes in certain dyes called acid-base indicators. Household acids and bases. Red cabbage juice as an acid-base indicator.

Acid-Base Reactions The Arrhenius Concept The Arrhenius concept defines acids as substances that produce hydrogen ions, H+, when dissolved in water. An example is nitric acid, HNO3, a molecular substance that dissolves in water to give H+ and NO3-.

Acid-Base Reactions The Arrhenius Concept The Arrhenius concept defines bases as substances that produce hydroxide ions, OH-, when dissolved in water. An example is sodium hydroxide, NaOH, an ionic substance that dissolves in water to give sodium ions and hydroxide ions.

Arrhenius acid is a substance that produces H+ (H3O+) in water
Arrhenius base is a substance that produces OH- in water

Hydronium ion, hydrated proton, H3O+

Acid-Base Reactions The Arrhenius Concept The molecular substance ammonia, NH3, is a base in the Arrhenius view, because it yields hydroxide ions when it reacts with water.

Acid-Base Reactions The Brønsted-Lowry Concept The Brønsted-Lowry concept of acids and bases involves the transfer of a proton (H+) from the acid to the base. In this view, acid-base reactions are proton-transfer reactions.

A Brønsted acid is a proton donor A Brønsted base is a proton acceptor
A Brønsted acid must contain at least one ionizable proton!

Acid-Base Reactions The Brønsted-Lowry Concept The Brønsted-Lowry concept defines an acid as the species (molecule or ion) that donates a proton (H+) to another species in a proton-transfer reaction. A base is defined as the species (molecule or ion) that accepts the proton (H+) in a proton-transfer reaction.

Acid-Base Reactions The Brønsted-Lowry Concept In the reaction of ammonia with water, H+ the H2O molecule is the acid because it donates a proton. The NH3 molecule is a base, because it accepts a proton.

Acid-Base Reactions The Brønsted-Lowry Concept The H+(aq) ion associates itself with water to form H3O+(aq). This “mode of transportation” for the H+ ion is called the hydronium ion.

Acid-Base Reactions The Brønsted-Lowry Concept The dissolution of nitric acid, HNO3, in water is therefore a proton-transfer reaction where HNO3 is an acid (proton donor) and H2O is a base (proton acceptor). H+

Acid-Base Reactions In summary, the Arrhenius concept and the Brønsted-Lowry concept are essentially the same in aqueous solution. The Arrhenius concept acid: proton (H+) donor base: hydroxide ion (OH-) donor

Acid-Base Reactions In summary, the Arrhenius concept and the Brønsted-Lowry concept are essentially the same in aqueous solution. The Brønsted-Lowry concept acid: proton (H+) donor base: proton (H+) acceptor

CH3COO- (aq) + H+ (aq) CH3COOH (aq) Brønsted base
Identify each of the following species as a Brønsted acid, base, or both. (a) HI, (b) CH3COO-, (c) H2PO4- HI (aq) H+ (aq) + I- (aq) Brønsted acid CH3COO- (aq) + H+ (aq) CH3COOH (aq) Brønsted base H2PO4- (aq) H+ (aq) + HPO42- (aq) Brønsted acid H2PO4- (aq) + H+ (aq) H3PO4 (aq) Brønsted base

Neutralization of a Strong Acid by a Strong Base

Acid-Base Reactions Strong and Weak Acids and Bases A strong acid is an acid that ionizes completely in water; it is a strong electrolyte.

Acid-Base Reactions Strong and Weak Acids and Bases 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.

Acid-Base Reactions Strong and Weak Acids and Bases A strong base is a base that is present entirely as ions, one of which is OH-; it is a strong electrolyte. The hydroxides of Group IA and IIA elements, except for beryllium hydroxide, are strong bases.

Acid-Base Reactions Strong and Weak Acids and Bases A weak base is a base that is only partially ionized in water; it is a weak electrolyte. Ammonia, NH3, is an example.

Acid-Base Reactions Strong and Weak Acids and Bases You will find it important to be able to identify an acid or base as strong or weak. When you write an ionic equation, strong acids and bases are represented as separate ions. Weak acids and bases are represented as undissociated “molecules” in ionic equations.

Table 4.3

Monoprotic acids Diprotic acids Triprotic acids HCl H+ + Cl-
Strong electrolyte, strong acid HNO H+ + NO3- Strong electrolyte, strong acid CH3COOH H+ + CH3COO- Weak electrolyte, weak acid Diprotic acids H2SO H+ + HSO4- Strong electrolyte, strong acid HSO H+ + SO42- Weak electrolyte, weak acid Triprotic acids H3PO H+ + H2PO4- Weak electrolyte, weak acid H2PO H+ + HPO42- Weak electrolyte, weak acid HPO H+ + PO43- Weak electrolyte, weak acid

Neutralization Reaction
acid + base salt + water HCl (aq) + NaOH (aq) NaCl (aq) + H2O H+ + Cl- + Na+ + OH Na+ + Cl- + H2O H+ + OH H2O

Acid-Base Reactions Neutralization Reactions One of the chemical properties of acids and bases is that they neutralize one another. A neutralization reaction is a reaction of an acid and a base that results in an ionic compound and water. The ionic compound that is the product of a neutralization reaction is called a salt. acid base salt

Acid-Base Reactions Neutralization Reactions The net ionic equation for each acid-base neutralization reaction involves a transfer of a proton. Consider the reaction of the strong acid , HCl(aq) and a strong base, LiOH(aq).

Acid-Base Reactions Neutralization Reactions Writing the strong electrolytes in the form of ions gives the following complete ionic equation.

Acid-Base Reactions Neutralization Reactions Canceling the spectator ions results in the net ionic equation. Note the proton transfer. H+

Acid-Base Reactions Neutralization Reactions In a reaction involving HCN(aq), a weak acid, and KOH(aq), a strong base, the product is KCN, a strong electrolyte Referring to Tables 4.1, 4.2 and 4.3, we obtain this net ionic equation: H+ Note the proton transfer.

Acid-Base Reactions Acid-Base Reactions with Gas Formation Carbonates react with acids to form CO2, carbon dioxide gas. Sulfites react with acids to form SO2, sulfur dioxide gas.

Acid-Base Reactions Acid-Base Reactions with Gas Formation Sulfides react with acids to form H2S, hydrogen sulfide gas.

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. Calculate moles of reactants. Determine the limiting reactant, where appropriate. Calculate the moles of the required reactant or product. Convert to grams or volume (of solution), as required. Copyright © Cengage Learning. All rights reserved

Volumetric Analysis An important method for determining the amount of a particular substance is based on measuring the volume of the reactant solution. Titration is a procedure for determining the amount of substance A by adding a carefully measured volume of a solution with known concentration of B until the reaction of A and B is just complete. Volumetric analysis is a method of analysis based on titration.

Equivalence point – the point at which the reaction is complete
Titrations In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point – the point at which the reaction is complete Indicator – substance that changes color at (or near) the equivalence point Slowly add base to unknown acid UNTIL the indicator changes color

The pH of a Solution The pH of a solution can accurately be measured using a pH meter 2

What volume of a 1.420 M NaOH solution is
Required to titrate mL of a 4.50 M H2SO4 solution? WRITE THE CHEMICAL EQUATION! H2SO4 + 2NaOH H2O + Na2SO4 M acid rx coef. M base volume acid moles acid moles base volume base 4.50 mol H2SO4 1000 mL soln x 2 mol NaOH 1 mol H2SO4 x 1000 ml soln 1.420 mol NaOH x 25.00 mL = 158 mL

Volumetric Analysis Consider the reaction of sulfuric acid, H2SO4, with sodium hydroxide, NaOH: Suppose a beaker contains 35.0 mL of M H2SO4. How many milliliters of M NaOH must be added to completely react with the sulfuric acid?

Volumetric Analysis First we must convert the L (35.0 mL) to moles of H2SO4 (using the molarity of the H2SO4). Then, convert to moles of NaOH (from the balanced chemical equation). Finally, convert to volume of NaOH solution (using the molarity of NaOH).

CONCEPT CHECK! For the titration of sulfuric acid (H2SO4) with sodium hydroxide (NaOH), how many moles of sodium hydroxide would be required to react with 1.00 L of M sulfuric acid to reach the endpoint? 1.00 mol NaOH The balanced equation is: H2SO4 + 2NaOH → 2H2O + Na2SO moles of sulfuric acid is present to start. Due to the 1:2 ratio in the equation, 1.00 mol of NaOH would be required to exactly react with the sulfuric acid. 1.00 mol of sodium hydroxide would be required. Copyright © Cengage Learning. All rights reserved

Let’s Think About It Where are we going? To find the moles of NaOH required for the reaction. How do we get there? What are the ions present in the combined solution? What is the reaction? What is the balanced net ionic equation for the reaction? What are the moles of H+ present in the solution? How much OH– is required to react with all of the H+ present? Copyright © Cengage Learning. All rights reserved

Oxidation-Reduction Reactions Oxidation-reduction reactions involve the transfer of electrons from one species to another. Oxidation is defined as the loss of electrons. Reduction is defined as the gain of electrons. Oxidation and reduction always occur simultaneously.

Zn (s) + CuSO4 (aq) ZnSO4 (aq) + Cu (s)
Zn Zn2+ + 2e- Zn is oxidized Zn is the reducing agent Cu2+ + 2e Cu Cu2+ is reduced Cu2+ is the oxidizing agent Copper wire reacts with silver nitrate to form silver metal. What is the oxidizing agent in the reaction? Cu (s) + 2AgNO3 (aq) Cu(NO3)2 (aq) + 2Ag (s) Cu Cu2+ + 2e- Ag+ + 1e Ag Ag+ is reduced Ag+ is the oxidizing agent

Oxidation-Reduction Reactions The reaction of an iron nail with a solution of copper(II) sulfate, CuSO4, is an oxidation- reduction reaction The molecular equation for this reaction is:

Oxidation-Reduction Reactions The net ionic equation shows the reaction of iron metal with Cu2+(aq) to produce iron(II) ion and copper metal. Loss of 2 e-1 oxidation Gain of 2 e-1 reduction

Oxidation-Reduction Reactions Oxidation Numbers The concept of oxidation numbers is a simple way of keeping track of electrons in a reaction. The oxidation number (or oxidation state) of an atom in a substance is the actual charge of the atom if it exists as a monatomic ion. Alternatively, it is hypothetical charge assigned to the atom in the substance by simple rules.

Rules for Assigning Oxidation States
Oxidation state of an atom in an element = 0 Oxidation state of monatomic ion = charge of the ion Oxygen = -2 in covalent compounds (except in peroxides where it = -1) Hydrogen = +1 in covalent compounds. When it is bonded to metals in binary compounds In these cases, its oxidation number is –1. Fluorine = -1 in compounds Sum of oxidation states = 0 in compounds Sum of oxidation states = charge of the ion in ions Copyright © Cengage Learning. All rights reserved

The oxidation numbers of elements in their compounds

IF7 F = -1 7x(-1) + ? = 0 I = +7 K2Cr2O7 NaIO3 Na = +1 O = -2 O = -2
Oxidation numbers of all the elements in the following ? F = -1 7x(-1) + ? = 0 I = +7 K2Cr2O7 NaIO3 Na = +1 O = -2 O = -2 K = +1 3x(-2) ? = 0 7x(-2) + 2x(+1) + 2x(?) = 0 I = +5 Cr = +6

EXERCISE! Find the oxidation states for each of the elements in each of the following compounds: K2Cr2O7 CO32- MnO2 PCl5 SF4 K = +1; Cr = +6; O = –2 C = +4; O = –2 Mn = +4; O = –2 P = +5; Cl = –1 S = +4; F = –1 K2Cr2O7; K = +1; Cr = +6; O = -2 CO32-; C = +4; O = -2 MnO2; Mn = +4; O = -2 PCl5; P = +5; Cl = -1 SF4; S = +4; F = -1 Copyright © Cengage Learning. All rights reserved

Review: 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 Copyright © Cengage Learning. All rights reserved

Balancing Oxidation–Reduction Reactions by Oxidation States
Write the unbalanced equation. Determine the oxidation states of all atoms in the reactants and products. Show electrons gained and lost using “tie lines.” Use coefficients to equalize the electrons gained and lost. Balance the rest of the equation by inspection. Add appropriate states. Copyright © Cengage Learning. All rights reserved

3. How are electrons gained and lost?
1 e– gained (each atom) Zn(s) + HCl(aq) Zn2+(aq) + Cl–(aq) + H2(g) – – 2 e– lost The oxidation state of chlorine remains unchanged. Copyright © Cengage Learning. All rights reserved

4. What coefficients are needed to equalize the electrons gained and lost? 1 e– gained (each atom) × 2 Zn(s) + HCl(aq) Zn2+(aq) + Cl–(aq) + H2(g) – – 2 e– lost Zn(s) + 2HCl(aq) Zn2+(aq) + Cl–(aq) + H2(g) Copyright © Cengage Learning. All rights reserved

Oxidation-Reduction Reactions Describing Oxidation-Reduction Reactions Look again at the reaction of iron with copper(II) sulfate. We can write this reaction in terms of two half-reactions.

Oxidation-Reduction Reactions Describing Oxidation-Reduction Reactions A half-reaction is one of the two parts of an oxidation-reduction reaction. One involves the loss of electrons (oxidation) and the other involves the gain of electrons (reduction). oxidation half-reaction reduction half-reaction

Oxidation-Reduction Reactions
(electron transfer reactions) 2Mg Mg2+ + 4e- Oxidation half-reaction (lose e-) O2 + 4e O2- Reduction half-reaction (gain e-) 2Mg + O2 + 4e Mg2+ + 2O2- + 4e- 2Mg + O MgO

Oxidation-Reduction Reactions Some Common Oxidation-Reduction Reactions Most of the oxidation-reduction reactions fall into one of the following simple categories: Combination Reactions Decomposition Reactions Combustion Reactions Displacement Reactions

Oxidation-Reduction Reactions Combination Reactions A combination reaction is a reaction in which two substances combine to form a third substance. Antimony and chlorine combine in a fiery reaction.

Oxidation-Reduction Reactions Combination Reactions Other combination reactions involve compounds as reactants.

Types of Oxidation-Reduction Reactions
Combination Reaction A + B C +3 -1 2Al + 3Br AlBr3

Oxidation-Reduction Reactions Decomposition Reactions A decomposition reaction is a reaction in which a single compound reacts to give two or more substances. Ammonium dichromate volcano

Decomposition Reaction C A + B +1 +5 -2 +1 -1 2KClO KCl + 3O2

Oxidation-Reduction Reactions Combustion Reactions A combustion reaction is a reaction in which a substance reacts with oxygen, usually with the rapid release of heat to produce a flame.

Combustion Reaction A + O B +4 -2 S + O SO2 +2 -2 2Mg + O MgO

The burning of calcium metal in oxygen.

The burning of calcium metal in chlorine.

Oxidation-Reduction Reactions Displacement Reactions A displacement reaction (also called a single- replacement reaction) is a reaction in which an element reacts with a compound, displacing an element from it.

Displacement Reaction A + BC AC + B +1 +2 Sr + 2H2O Sr(OH)2 + H2 Hydrogen Displacement +4 +2 TiCl4 + 2Mg Ti + 2MgCl2 Metal Displacement -1 -1 Cl2 + 2KBr KCl + Br2 Halogen Displacement

The Activity Series for Metals
Hydrogen Displacement Reaction M + BC AC + B M is metal BC is acid or H2O B is H2 Ca + 2H2O Ca(OH)2 + H2 Pb + 2H2O Pb(OH)2 + H2

The Activity Series for Halogens
F2 > Cl2 > Br2 > I2 Halogen Displacement Reaction -1 -1 Cl2 + 2KBr KCl + Br2 I2 + 2KBr KI + Br2

Disproportionation Reaction Element is simultaneously oxidized and reduced. +1 -1 Cl2 + 2OH ClO- + Cl- + H2O Chlorine Chemistry

Classify the following reactions.
Ca2+ + CO CaCO3 Precipitation NH3 + H NH4+ Acid-Base Zn + 2HCl ZnCl2 + H2 Redox (H2 Displacement) Ca + F CaF2 Redox (Combination)

Oxidation-Reduction Reactions Balancing Simple Oxidation-Reduction Reactions At first glance, the equation representing the reaction of zinc metal with silver(I) ions might appear to be balanced. However, a balanced equation must have a charge balance as well as a mass balance.

Oxidation-Reduction Reactions Balancing Simple Oxidation-Reduction Reactions Since the number of electrons lost in the oxidation half-reaction must equal the number gained in the reduction half-reaction, oxidation half-reaction reduction half-reaction we must double the reaction involving the reduction of the silver.

Oxidation-Reduction Reactions Balancing Simple Oxidation-Reduction Reactions Adding the two half-reactions together, the electrons cancel, oxidation half-reaction reduction half-reaction which yields the balanced oxidation-reduction reaction.

Worked Example 4.1

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Worked Example 4.12

Water – “the universal solvent”

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