1. Introductory Chemistry: Concepts & Connections Introductory Chemistry: Concepts & Connections 4 th Edition by Charles H. Corwin Acids and Bases Christopher G. Hamaker, Illinois State University, Normal IL © 2005, Prentice Hall Chapter 15

2. 2 Properties of Acids 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.

3. Chapter 153 Properties of Bases 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.

4. Chapter 154 Acid/Base Neutralization 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)

5. Chapter 155 The pH Scale 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.

6. 6 Acid/Base Classifications of Solutions 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.

7. Chapter 157 Buffers 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.

8. Chapter 158 Arrhenius Acids and Bases 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.

9. Chapter 159 Strengths of Acids 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.

10. Chapter 1510 Strengths of Bases 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.

11. Chapter 1511 Strong & Weak Arrhenius Acids 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%

12. Chapter 1512 Arrhenius Acids in Solution 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.

13. Chapter 1513 Strong & Weak Arrhenius Bases 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

14. Chapter 1514 Arrhenius Bases in Solution 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%)

15. Chapter 1515 Neutralization Reactions 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)

16. Chapter 1516 Predicting Neutralization Reactions 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, and 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)

17. Chapter 1517 Brønsted-Lowry Acids & Bases 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.

18. Chapter 1518 Brønsted-Lowry Acids & Bases Lets 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.

19. Chapter 1519 Amphiprotic Compounds 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.

20. Chapter 1520 Acid-Base Indicators A solution that changes color as the pH changes is an acid-base indicator. Three common indicators are methyl red, bromothymol blue, and phenolphthalein. Each has a different color above and below a certain pH.

21. Chapter 1521 Acid-Base Indicators Shown below are the three indicators at different pH values Methyl Red Bromothymol Blue Phenolphthalein

22. Chapter 1522 Acid-Base Titrations 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.

23. Chapter 1523 Titration Problem Consider the titration of acetic acid with sodium hydroxide. A 10.0 mL sample of acetic acid requires 37.55 mL of 0.223 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

24. Chapter 1524 Titration Problem Continued The molarity of NaOH can be written as the unit factor 0.233 mol NaOH / 1000 mL solution. = 0.00837 mol HC 2 H 3 O 2 37.55 mL solution × 1 mol HC 2 H 3 O 2 1 mol NaOH 0.233 mol NaOH 1000 mL solution × = 0.837 M HC 2 H 3 O 2 1000 mL solution 10.0 mL solution 0.00837 mol HC 2 H 3 O 2 1 L solution ×

25. Chapter 1525 Another Titration Problem A 10.0 mL sample of 0.555 M H 2 SO 4 is titrated with 0.233 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 0.555 mol H 2 SO 4 /1000 mL and 0.233 mol NaOH/1000 mL.

26. Chapter 1526 Problem Continued 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 0.555 mol H 2 SO 4 1000 mL H 2 SO 4 × 0.233 mol NaOH 1000 mL NaOH × 49.8 mL of 0.233 M NaOH is required to neutralize 10.0 mL of 0.555 M H 2 SO 4.

27. Chapter 1527 Acid-Base Standardization 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 25.50 mL are required to neutralize 0.375 g Na 2 CO 3 ? 2 HCl(aq) + Na 2 CO 3 (aq) → 2 NaCl(aq) + H 2 O(l) + CO 2 (g)

28. Chapter 1528 Standardization Continued = 0.277 M HCl× 25.50 mL solution 0.00708 mol HCl 1000 mL solution 1 L solution = 0.00708 mol HCl 0.375 g Na 2 CO 3 × × 1 mol Na 2 CO 3 105.99 g Na 2 CO 3 2 mol HCl 1 mol Na 2 CO 3

29. Chapter 1529 Ionization of Water 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 ions so water is a weak conductor. The concentration of hydrogen ions, [H + ], in pure water is about 1 × 10 -7 mol/L at 25  C.

30. Chapter 1530 Autoionization of Water Since [H + ] is 1 × 10 -7 mol/L at 25  C, the hydroxide ion concentration, [OH - ], must also be 1 × 10 -7 mol/L at 25  C: –H 2 O(l) → H + (aq) + OH - (aq) At 25  C –[H + ][OH - ] = (1 × 10 -7 )(1 × 10 -7 ) = 1.0 × 10 -14 This is the ionization constant of water, K w.

31. Chapter 1531 [H + ] and [OH - ] Relationship At 25  C, [H + ][OH - ] = 1.0 × 10 -14. So, if we know the [H + ], we can calculate [OH - ]. What is the [OH - ] if [H + ] = 0.1 M ? –[H + ][OH - ] = 1.0 × 10 -14 –(0.1)[OH - ] = 1.0 × 10 -14 –[OH - ] = 1.0 × 10 -13

32. Chapter 1532 The pH Concept 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 ten to express the hydrogen ion concentration. Mathematically: pH = –log[H + ] –[H + ] is the molar hydrogen ion concentration

33. Chapter 1533 Calculating pH What is the pH if the hydrogen ion concentration in a vinegar solution is 0.001 M? pH = –log[H + ] pH = –log(0.001) pH = – ( –3) = 3 The pH of the vinegar is 3, so the vinegar is acidic.

34. Chapter 1534 Calculating [H + ] from pH 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 = 0.000001 M [H + ] = 1 × 10 –6 M.

35. Chapter 1535 Advanced pH Calculations What is the pH of blood with [H + ] = 4.8 × 10 –8 M? –pH = –log[H + ] = –log(4.8 × 10 –8 ) = – (–7.32) –pH = 7.32 What is the [H + ] in orange juice with a pH of 2.75? –[H + ] = 10 –pH = 10 –2.75 = 0. 0018 M –[H + ] = 2.75 × 10 –3 M

36. Chapter 1536 Strong & Weak Electrolytes 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.

37. Chapter 1537 Electrolyte Strength Weak acids and bases and insoluble ionic compounds are weak electrolytes. Strong acids and bases and soluble ionic compounds are strong electrolytes.

38. Chapter 1538 Total Ionic Equation 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.

39. Chapter 1539 Net Ionic Equation 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)

40. 40 Writing Net Ionic Equations 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.

41. Chapter 1541 Conclusions 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.

42. Chapter 1542 Conclusions Continued 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

43. Chapter 1543 Conclusions Continued In an aqueous solution, [H + ][OH - ] = 1.0 × 10 -14. 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.