2. End of Section 18-2

3. Section 18-3 Section 18.3 Hydrogen Ions and pH Explain pH and pOH. Le Châtelier’s principle: states that if a stress is applied to a system at equilibrium, the system shifts in the direction that relieves the stress ion product constant for water pH pOH Relate pH and pOH to the ion product constant for water. Calculate the pH and pOH of aqueous solutions. pH and pOH are logarithmic scales that express the concentrations of hydrogen ions and hydroxide ions in aqueous solutions.

4. Acids contain more H + ions than OH - ions Bases contain more OH - ions than H + ions Water is used as the solvent for Acids/Bases because it produces an equal number of OH - and H + ions Remember…

5. Section 18-3 Ion Product Constant for Water Pure water contains equal concentrations of H + and OH – ions. The ion production of water, K w = [H + ][OH – ]. The ion product constant for water is the value of the equilibrium constant expression for the self-ionization of water.ion product constant for water

6. Section 18-3 Ion Product Constant for Water (cont.) With pure water at 298 K, both [H + ] and [OH – ] are equal to 1.0 × 10 –7 M. K w at 298 K w = 1.0 × 10 –14 K w and LeChâtelier’s Principle proves [H + ] × [OH – ] must equal 1.0 × 10 –14 at 298 K, and as [H + ] goes up, [OH – ] must go down. K w = [H + ][OH - ] = 1.0 X 10 -14

7. REMEMBER…

8. Section 18-3 pH and pOH (cont.) For all strong monoprotic acids, the concentration of the acid is the concentration of H + ions. For all strong monoprotic bases, the concentration of the OH – ions available is the concentration of base.

9. 1. What is the [H + ] in a 0.10 M HCl solution? HCl is a strong acid, so it all ionizes. HCl  H + + Cl - 0.10 M 

10. Weak acids and weak bases only partially ionize and K a and K b values must be used to find concentration. I.C.E. Charts

11. Section 18-3 pH and pOH (cont.) What is the pH of a 0.10 M HF solution? (K a = 6.3x10 -4 ) HF ↔ H + + F - K a = [H + ][F - ] [HF] Initial 0.10 0 0 Change -x +x +x Equilibrium 0.10-x x x 6.3x10 -4 = [x][x] [0.10-x]

12. 2. What is the [H + ] in a 0.10 M HF solution? HF is a weak acid, so it partially ionizes. K a = [H + ][F - ] [HF] (K a = 8.6x10 -4 ) 8.6x10 -4 = [x] [0.10-x] x = 9.3x10 -3 HF ↔ H + + F - Initial0.1 00 Change-x +x+x Equilibrium0.1-x x x

13. What is the [H + ] in a 0.20 M H 2 S solution? H 2 S is diprotic, but we only use the first H. K a = [H + ][HS - ] [H 2 S] (K a = 8.9x10 -8 ) 8.9x10 -8 = [x] [0.20-x] x = 1.3x10 -4 H 2 S ↔ H + +HS - Initial0.2000 Change-xxx Equilibrium0.20-xxx

14. A.A B.B C.C D.D Section 18-3 Section 18.3 Assessment In dilute aqueous solution, as [H + ] increases: A.pH decreases B.pOH increases C.[OH – ] decreases D.all of the above

15. A.A B.B C.C D.D Section 18-3 Section 18.3 Assessment What is the pH of a neutral solution such as pure water? A.0 B.7 C.14 D.1.0 × 10 –14

16. End of Section 18-3

17. Section 18-4 Section 18.4 Neutralization Write chemical equations for neutralization reactions. stoichiometry: the study of quantitative relationships between the amounts of reactants used and products formed by a chemical reaction; is based on the law of conservation of mass Explain how neutralization reactions are used in acid-base titrations. Compare the properties of buffered and unbuffered solutions.

18. Section 18-4 Section 18.4 Neutralization (cont.) neutralization reaction salt titration titrant equivalence point In a neutralization reaction, an acid reacts with a base to produce a salt and water. acid-base indicator end point salt hydrolysis buffer buffer capacity

19. Section 18-4 Reactions Between Acids and Bases A neutralization reaction is a reaction in which an acid and a base in an aqueous solution react to produce a salt and water.neutralization reaction A salt is an ionic compound made up of a cation from a base and an anion from an acid.salt Neutralization is a double-replacement reaction.

20. Section 18-4 Reactions Between Acids and Bases (cont.)

21. Hydrochloric Acid and Sodium Hydroxide Neutralization Reactions HCl+ NaOH  NaCl + H 2 O Carbonic Acid and Calcium Hydroxide H 2 CO 3 + Ca(OH) 2  CaCO 3 + H 2 O

22. Section 18-4 Reactions Between Acids and Bases (cont.) Titration is a method for determining the concentration of a solution by reacting a known volume of that solution with a solution of known concentration.Titration In a titration procedure, a measured volume of an acid or base of unknown concentration is placed in a beaker, and initial pH recorded. A buret is filled with the titrating solution of known concentration, called a titrant.titrant

23. Section 18-4 Reactions Between Acids and Bases (cont.) A measured volume of standard solution is SLOWLY added to the solution in the beaker. An abrupt change in pH occurs at the equivalence point. How does titration work in lab? The standard is mixed into the solution in the beaker and the pH is read and recorded after each addition The process continues until the reaction reaches the equivalence point, the point where moles of H + from the acid equals moles of OH – from the base.equivalence point

24. Section 18-4 Reactions Between Acids and Bases (cont.)

25. Section 18-4 Reactions Between Acids and Bases (cont.) Chemical dyes whose color are affected by acidic and basic solutions are called acid- base indicators.acid- base indicators

26. Section 18-4 Reactions Between Acids and Bases (cont.) An end point is the point at which an indicator used in a titration changes color.end point An indicator will change color at the equivalence point. The following equation can be used to determine the unknown molarity M a V a (#of H’s) = M b V b (#of OH’s)

27. A 10.0 mL sample of HCl is neutralized by 40.0 mL of 0.25 M NaOH solution. What was the concentration of the HCl sample? Titration question: HCl + NaOH  NaCl + H 2 O M a V a (#of H’s) = M b V b (#of OH’s) 10.0 mL*M a (1) = 0.25 M * 40.0mL (1) M a = 0.25 M * 40.0 mL 10.0mL M a = 1.00M HCl

28. Section 18-4 Salt Hydrolysis In salt hydrolysis, the anions of the dissociated salt accept the hydrogen ions from water or the cations of the dissociated salt donate hydrogen ions to water.salt hydrolysis

29. Section 18-4 Salt Hydrolysis (cont.) Salts that produce basic solutions −KF is the salt of a strong base (KOH) and a weak acid (HF). KF(s) → K + (aq) + F – (aq)

30. Section 18-4 Salt Hydrolysis (cont.) Salts that produce acidic solutions −NH 4 Cl is the salt of a weak base (NH 3 ) and strong acid (HCl). −When dissolved in water, the salt dissociates into ammonium ions and chloride ions. NH 4 Cl(s) → NH 4 + (aq) + Cl – (aq)

31. Section 18-4 Salt Hydrolysis (cont.) Salts that produce neutral solutions −NaNO 3 is the salt of a strong acid (HNO 3 ) and a strong base (NaOH). −Little or no salt hydrolysis occurs because neither Na + nor NO 3 – react with water.

32. Section 18-4 Buffered Solutions The pH of blood must be kept in within a narrow range. Buffers are solutions that resist changes in pH when limited amounts of acid or base are added.Buffers

33. Section 18-4 Buffered Solutions (cont.) Ions and molecules in a buffer solution resist changes in pH by reacting with any hydrogen ions of hydroxide ions added to the buffered solution. HF(aq) ↔ H + (aq) + F – (aq) When acid is added, the equilibrium shifts to the left.

34. Section 18-4 Buffered Solutions (cont.) Additional H + ions react with F – ions to form undissociated HF molecules but the pH changes little. The amount of acid or base that a buffer solution can absorb without a significant change in pH is called the buffer capacity.buffer capacity

35. Section 18-4 Buffered Solutions (cont.) A buffer is most effective when the concentrations of the conjugate acid-base pair are equal or nearly equal.

36. A.A B.B C.C D.D Section 18-4 Section 18.4 Assessment In a neutralization reaction, an acid and base react to form: A.salt and oxygen gas B.salt and ammonia C.salt and water D.precipitate and water

37. A.A B.B C.C D.D Section 18-4 Section 18.4 Assessment Solutions that resist changes in pH are called ____. A.titrants B.salts C.conjugate pairs D.buffers