Christopher G. Hamaker, Illinois State University, Normal IL © 2008, Prentice Hall Chapter 15 Acids and Bases INTRODUCTORY CHEMISTRY INTRODUCTORY CHEMISTRY Concepts & Connections Fifth Edition by Charles H. Corwin

Chapter 15 2 An acid is any substance that releases hydrogen ions, H +, into water. Blue litmus paper turns red in the presence of hydrogen ions. Blue litmus is used to test for acids. Acids have a sour taste; lemons, limes, and vinegar are acidic. Properties of Acids

Chapter 15 3 A base is a substance that releases hydroxide ions, OH –, into water. Red litmus paper turns blue in the presence of hydroxide ions. Red litmus is used to test for bases. Bases have a slippery, soapy feel. Bases also have a bitter taste; milk of magnesia is a base. Properties of Bases

Chapter 15 4 Recall that an acid and a base react with each other in a neutralization reaction. When an acid and a base react, water and a salt are produced. For example, nitric acid reacts with sodium hydroxide to produce sodium nitrate and water: HNO 3 (aq) + NaOH(aq) → NaNO 3 (aq) + H 2 O(l) Acid/Base Neutralization

Chapter 15 5 A pH value expresses the acidity or basicity of a solution. Most solutions have a pH between 0 and 14. Acidic solutions have a pH less than 7. –As a solution becomes more acidic, the pH decreases. Basic solutions have a pH greater than 7. –As a solution becomes more basic, the pH increases. The pH Scale

Chapter 15 6 A solution can be classified according to its pH. Strongly acidic solutions have a pH less than 2. Weakly acidic solutions have a pH between 2 and 7. Weakly basic solutions have a pH between 7 and 12. Strongly basic solutions have a pH greater than 12. Neutral solutions have a pH of 7. Acid/Base Classifications of Solutions

Chapter 15 7 A buffer is a solution that resists changes in pH when an acid or a base is added. A buffer is a solution of a weak acid and one of its salts: –Citric acid and sodium citrate make a buffer solution. When acid is added to the buffer, the citrate reacts with the acid to neutralize it. When base is added to the buffer, the citric acid reacts with the base to neutralize it. Buffers

Chapter 15 8 Svante Arrhenius proposed the following definitions for acids and bases in 1884: –An Arrhenius acid is a substance that ionizes in water to produce hydrogen ions. –An Arrhenius base is a substance that ionizes in water to release hydroxide ions. For example, HCl is an Arrhenius acid and NaOH is an Arrhenius base. Arrhenius Acids and Bases

Chapter 15 9 Acids have varying strengths. The strength of an Arrhenius acid is measured by the degree of ionization in solution. Ionization is the process where polar compounds separate into cations and anions in solution. The acid HCl ionizes into H + and Cl – ions in solution. Strengths of Acids

Chapter Bases also have varying strengths. The strength of an Arrhenius base is measured by the degree of dissociation in solution. Dissociation is the process where cations and anions in an ionic compound separate in solution. A formula unit of NaOH dissociates into Na + and OH – ions in solution. Strengths of Bases

Chapter Strong acids ionize extensively to release hydrogen ions into solution. –HCl is a strong acid and ionizes nearly 100%. Weak acids only ionize slightly in solution. –HF is a weak acid and ionizes only about 1%. Strong and Weak Arrhenius Acids

Chapter All Arrhenius acids have a hydrogen atom bonded to the rest of the molecule by a polar bond. This bond is broken when the acid ionizes. Polar water molecules help ionize the acid by pulling the hydrogen atom away: HCl(aq) + H 2 O(l) → H 3 O + (aq) + Cl – (aq) (~100%) HC 2 H 3 O 2 (aq) + H 2 O(l) → H 3 O + (aq) + C 2 H 3 O 2 – (aq) (~1%) The hydronium ion, H 3 O +, is formed when the aqueous hydrogen ion attaches to a water molecule. Arrhenius Acids in Solution

Chapter Strong bases dissociate extensively to release hydroxide ions into solution. –NaOH is a strong base and dissociates nearly 100%. Weak bases only ionize slightly in solution. –NH 4 OH is a weak base and only partially dissociates Strong and Weak Arrhenius Bases

Chapter When we dissolve Arrhenius bases in solution, they dissociate, giving a cation and a hydroxide anion. Strong bases dissociate almost fully, and weak bases dissociate very little: NaOH(aq) → Na + (aq) + OH – (aq) (~100%) NH 4 OH(aq) → NH 4 + (aq) + OH – (aq) (~1%) Arrhenius Bases in Solution

Chapter Chemistry Connection: Svante Arrhenius Svante Arrhenius noted that NaCl solutions conductedelectricity; while sugar solutions did not. He also noticed that the freezing point of NaCl solutions were lowered twice as much as sugar solutions at the same concentration. He proposed that NaCl produces ions when dissolved, while sugar was in solution as molecules. It was nearly 20 years before his ideas were accepted by the scientific community.

Chapter Recall, an acid neutralizes a base to produce a salt and water. –HCl(aq) + NaOH(aq) → NaCl(aq) + H 2 O(l) The reaction produces the aqueous salt NaCl. If we have an acid with two hydrogens (sulfuric acid, H 2 SO 4 ), we need two hydroxide ions to neutralize it. –H 2 SO 4 (aq) + 2 NaOH(aq) → Na 2 SO 4 (aq) + 2 H 2 O(l) Neutralization Reactions

Chapter We can identify the Arrhenius acid and base that react in a neutralization reaction to produce a given salt such as calcium sulfate, CaSO 4. The calcium must be from calcium hydroxide, Ca(OH) 2 ; the sulfate must be from sulfuric acid, H 2 SO 4. –H 2 SO 4 (aq) + Ca(OH) 2 (aq) → CaSO 4 (aq) + 2 H 2 O(l) Predicting Neutralization Reactions

Chapter The Brønsted-Lowry definitions of acids and bases are broader than the Arrhenius definitions. A Brønsted-Lowry acid is a substance that donates a hydrogen ion to any other substance. It is a proton donor. A Brønsted-Lowry base is a substance that accepts a hydrogen ion. It is a proton acceptor. Brønsted-Lowry Acids and Bases

Chapter Let’s look at two acid-base reactions: –HCl(aq) + NaOH(aq) → NaCl(aq) + H 2 O(l) –HCl(aq) + NH 3 (aq) → NH 4 Cl(aq) HCl donates a proton in both reactions and is a Brønsted-Lowry acid. In the first reaction, the NaOH accepts a proton and is the Brønsted-Lowry base. In the second reaction, NH 3 accepts a proton and is the Brønsted-Lowry base. Brønsted-Lowry Acids and Bases

Chapter A substance that is capable of both donating and accepting a proton is an amphiprotic compound. NaHCO 3 is an example: –HCl(aq) + NaHCO 3 (aq) → NaCl(aq) + H 2 CO 3 (aq) –NaOH(aq) + NaHCO 3 (aq) → Na 2 CO 3 (aq) + H 2 O(l) NaHCO 3 accepts a proton from HCl in the first reaction and donates a proton to NaOH in the second reaction. Amphiprotic Compounds

Chapter A solution that changes color as the pH changes is an acid-base indicator. Shown below are the three indicators at different pH values. Methyl Red Bromothymol Blue Phenolphthalein Acid-Base Indicators

Chapter A titration is used to analyze an acid solution using a solution of a base. A measured volume of base is added to the acid solution. When all of the acid has been neutralized, the pH is 7. One extra drop of base solution after the endpoint increases the pH dramatically. When the pH increases above 7, phenolphthalein changes from colorless to pink, indicating the endpoint of the titration. Acid-Base Titrations

Chapter Consider the titration of acetic acid with sodium hydroxide. A 10.0 mL sample of acetic acid requires mL of M NaOH. What is the concentration of the acetic acid? HC 2 H 3 O 2 (aq) + NaOH(aq) → NaC 2 H 3 O 2 (aq) + H 2 O(l) We want concentration acetic acid; we have concentration sodium hydroxide. conc NaOH  mol NaOH  mol HC 2 H 3 O 2  conc HC 2 H 3 O 2 Titration Problem

Chapter The molarity of NaOH can be written as the unit factor mol NaOH / 1000 mL solution. = mol HC 2 H 3 O mL solution × 1 mol HC 2 H 3 O 2 1 mol NaOH mol NaOH 1000 mL solution × = M HC 2 H 3 O mL solution 10.0 mL solution mol HC 2 H 3 O 2 1 L solution × Titration Problem, continued

Chapter A 10.0 mL sample of M H 2 SO 4 is titrated with M NaOH. What volume of NaOH is required for the titration? We want mL of NaOH; we have 10.0 mL of H 2 SO 4. Use mol H 2 SO 4 /1000 mL and mol NaOH/1000 mL. Another Titration Problem

Chapter H 2 SO 4 (aq) + 2 NaOH(aq) → Na 2 SO 4 (aq) + H 2 O(l) = 49.8 mL NaOH 10.0 mL H 2 SO 4 × 1 mol H 2 SO 4 2 mol NaOH mol H 2 SO mL H 2 SO 4 × mol NaOH 1000 mL NaOH × 49.8 mL of M NaOH is required to neutralize 10.0 mL of M H 2 SO 4. Titration Problem, continued

Chapter A standard solution is a solution where the concentration is precisely known. Acid solutions are standardized by neutralizing a weighed quantity of a solid base. What is the molarity of a hydrochloric acid solution if mL are required to neutralize g Na 2 CO 3 ? 2 HCl(aq) + Na 2 CO 3 (aq) → 2 NaCl(aq) + H 2 O(l) + CO 2 (g) Acid-Base Standardization

Chapter = M HCl× mL solution mol HCl 1000 mL solution 1 L solution = mol HCl g Na 2 CO 3 × × 1 mol Na 2 CO g Na 2 CO 3 2 mol HCl 1 mol Na 2 CO 3 Standardization, continued

Chapter Water undergoes an autoionization reaction. Two water molecules react to produce a hydronium ion and a hydroxide ion: –H 2 O(l) + H 2 O(l) → H 3 O + (aq) + OH - (aq) or –H 2 O(l) → H + (aq) + OH - (aq) Only about 1 in 5 million water molecules is present as an ion, so water is a weak conductor. The concentration of hydrogen ions, [H + ], in pure water is about 1 × mol/L at 25  C. Ionization of Water

Chapter Since [H + ] is 1 × mol/L at 25  C, the hydroxide ion concentration [OH - ] must also be 1 × mol/L at 25  C: –H 2 O(l) → H + (aq) + OH - (aq) At 25  C –[H + ][OH - ] = (1 × )(1 × ) = 1.0 × This is the ionization constant of water, K w. Autoionization of Water

Chapter At 25  C, [H + ][OH - ] = 1.0 × So if we know the [H + ], we can calculate [OH - ]. What is the [OH - ] if [H + ] = 0.1 M ? –[H + ][OH - ] = 1.0 × –(0.1)[OH - ] = 1.0 × –[OH - ] = 1.0 × [H + ] and [OH - ] Relationship

Chapter Recall that pH is a measure of the acidity of a solution. A neutral solution has a pH of 7, an acidic solution has a pH less than 7, and a basic solution has a pH greater than 7. The pH scale uses powers of 10 to express the hydrogen ion concentration. Mathematically: pH = –log[H + ] –[H + ] is the molar hydrogen ion concentration The pH Concept

Chapter What is the pH if the hydrogen ion concentration in a vinegar solution is M? pH = –log[H + ] pH = –log(0.001) pH = – ( –3) = 3 The pH of the vinegar is 3, so the vinegar is acidic. Calculating pH

Chapter If we rearrange the pH equation for [H + ], we get: [H + ] = 10 –pH Milk has a pH of 6. What is the concentration of hydrogen ion in milk? [H + ] = 10 –pH = 10 –6 = M [H + ] = 1 × 10 –6 M. Calculating [H + ] from pH

Chapter What is the pH of blood with [H + ] = 4.8 × 10 –8 M? –pH = –log[H + ] = –log(4.8 × 10 –8 ) = – (–3.82) –pH = 3.82 What is the [H + ] in orange juice with a pH of 2.75? –[H + ] = 10 –pH = 10 –2.75 = M –[H + ] = 2.75 × 10 –3 M Advanced pH Calculations

Chapter Critical Thinking: Acid Rain Nitrogen oxides and sulfur oxides, produced from the combustion of fossil fuels, react with rainwater to produce nitric and sulfuric acids. These strong acids reduce the pH of rainwater. Acid rain refers to rain with a pH below 5. Acid rain can lower the pH of lakes and cause corrosion of metal and degradation of limestone and marble statues.

Chapter An aqueous solution that is a good conductor of electricity is a strong electrolyte. An aqueous solution that is a poor conductor of electricity is a weak electrolyte. The greater the degree of ionization or dissociation, the greater the conductivity of the solution. Strong and Weak Electrolytes

Chapter Weak acids and bases are weak electrolytes. Strong acids and bases are strong electrolytes. Insoluble ionic compounds are weak electrolytes. Soluble ionic compounds are strong electrolytes. Electrolyte Strength

Chapter Strengths of Electrolytes

Chapter The concept of ionization allows us to portray ionic solutions more accurately by showing strong electrolytes in their ionized form. –HCl(aq) + NaOH(aq) → NaCl(aq) + H 2 O(l) Write strong acids and bases and soluble ionic compounds as ions: H + (aq) + Cl - (aq) + Na + (aq) + OH - (aq) → Na + (aq) + Cl - (aq) + H 2 O(l) This is the total ionic equation. Each species is written as it predominantly exists in solution. Total Ionic Equations

Chapter H + (aq) + Cl - (aq) + Na + (aq) + OH - (aq) → Na + (aq) + Cl - (aq) + H 2 O(l) Notice that Na + and Cl - appear on both sides of the equation. They are spectator ions. Spectator ions are in the solution, but do not participate in the overall reaction. We can cancel out the spectator ions to give the net ionic equation. The net ionic equation is: H + (aq) + OH - (aq) → H 2 O(l) Net Ionic Equations

Chapter Complete and balance the non-ionized chemical equation. Convert the non-ionized equation into the total ionic equation –Write strong electrolytes in the ionized form. –Write weak electrolytes, water, and gases in the non- ionized form. Cancel all the spectator ions to obtain the net ionic equation. –If all species are eliminated, there is no reaction. Writing Net Ionic Equations

Chapter pH is a measure of the acidity of a solution. The typical range for pH is 0 to 14. Neutral solutions have a pH of 7. Below are some properties of acids and bases: Chapter Summary

Chapter An Arrhenius acid is a substance that ionizes in water to produce hydrogen ions. An Arrhenius base is a substance that ionizes in water to release hydroxide ions. A Brønsted-Lowry acid is a substance that donates a hydrogen ion to any other substance. It is a proton donor. A Brønsted-Lowry base is a substance that accepts a hydrogen ion. It is a proton acceptor. Chapter Summary, continued

Chapter In an aqueous solution, [H + ][OH - ] = 1.0 × This is the ionization constant of water, K w. pH = –log[H + ] [H + ] = 10 –pH Strong electrolytes are mostly dissociated in solution. Weak electrolytes are slightly dissociated in solution. Chapter Summary, continued