Brønsted–Lowry Theory

1. Brønsted–Lowry Theory Brønsted-Lowry Theory of Acids and Bases An acid–base reaction is a proton-transfer reaction in which the proton is transferred from the acid to the base. Acid A compound from which a proton can be removed. Base A compound that can remove a proton.

2. Brønsted–Lowry Theory

3. Brønsted–Lowry Theory

4. Brønsted–Lowry Theory Amphoteric A substance that can behave as an acid in one situation and a base in another. Reversible Reaction The products of the reaction can react with themselves to reproduce the original reactants. Forward Direction: Left to right Reverse Direction: Right to left

5. Brønsted–Lowry Theory

6. Brønsted–Lowry Theory General equation for a Brønsted–Lowry proton-transfer reaction: B + HA HB+ + A– base acid proton proton remover source

7. Lewis Theory Lewis Theory of Acids and Bases Acid Electron-pair acceptor. Base Electron-pair donor.

8. Lewis Theory

9. Acid–Base Theories Summary of Acid–Base Theories

10. Conjugate Acid–Base Pairs B + HA HB+ + A– base acid acid base proton proton proton proton receiver source source receiver Conjugate Acid–Base Pair Two species that transform into each other by gain or loss of a proton, H+. Based on the Brønsted-Lowry theory. HB+ & B and HA & A– are conjugate acid–base pairs

11. Conjugate Acid–Base Pairs

12. Conjugate Acid–Base Pairs

13. Conjugate Acid–Base Pairs Example: What is the conjugate acid of NH3? What are the conjugate acid and conjugate base of HCO3–? Solution: To write the formula of a conjugate acid, add one H+. To write the formula of a conjugate base, subtract one H+. The conjugate acid of NH3 is NH4+. The conjugate acid of HCO3– is H2CO3. The conjugate base of HCO3– is CO32–.

14. Acid/Base Relative Strengths In Chapter 9, we classified acids as strong or weak: Strong acids were considered to be completely ionized: HSt(aq) H+(aq) + St–(aq) Weak acids were considered to be completely un-ionized: HWk(aq)

15. Acid/Base Relative Strengths Nitric acid is a stronger acid; it is a good proton source. Consider the ionization equation for nitric acid: HNO3(aq) H+(aq) + NO3–(aq) The conjugate base of the acid is on the right-hand side of the equation. A base is a proton remover.

16. Acid/Base Relative Strengths Hydrofluoric acid is a weaker acid; it is a poor proton source. Consider the ionization equation for hydrofluoric acid: HF(aq) H+(aq) + F–(aq) The conjugate base of the acid is on the right-hand side of the equation. A base is a proton remover.

17. Acid/Base Relative Strengths

18. Predicting Acid–Base Rxns Predicting the Favored Direction of a Proton-Transfer Reaction The stronger acid will always surrender a proton to the stronger base, yielding the weaker acid and base as favored species at equilibrium.

19. Predicting Acid–Base Rxns Procedure How to Predict the Favored Direction of an Acid–Base Reaction • For a given pair of reactants, write the equation for the transfer of one proton from one species to the other. (Do not transfer two protons.) • Label the acid and base on each side of the equation. • Determine which side of the equation has both the weaker acid and the weaker base (they must both be on the same side). That side identifies the products in the favored direction.

20. Predicting Acid–Base Rxns Example: Write the net ionic equation for the reaction between hydrogen sulfate ion and hydroxide ion. Predict which side will be favored at equilibrium. Solution: Step 1: For a given pair of reactants, write the equation for the transfer of one proton from one species to the other. (Do not transfer two protons.) HSO4–(aq) + OH–(aq) SO42–(aq) + HOH(l)

21. Predicting Acid–Base Rxns Write the net ionic equation for the reaction between hydrogen sulfate ion and hydroxide ion. Predict which side will be favored at equilibrium. Step 2: Label the acid and base on each side of the equation. HSO4–(aq) + OH–(aq) SO42–(aq) + HOH(l) Acid Base Base Acid

22. Predicting Acid–Base Rxns Write the net ionic equation for the reaction between hydrogen sulfate ion and hydroxide ion. Predict which side will be favored at equilibrium. Step 3: Determine which side of the equation has both the weaker acid and the weaker base (they must both be on the same side). That side identifies the products in the favored direction. HSO4–(aq) + OH–(aq) SO42–(aq) + HOH(l) ? ? ? ? Acid Base Base Acid

23. Predicting Acid–Base Rxns

24. Predicting Acid–Base Rxns Write the net ionic equation for the reaction between hydrogen sulfate ion and hydroxide ion. Predict which side will be favored at equilibrium. From Table 17.1, HSO4–(aq) is a stronger acid than HOH(l) SO42–(aq) is a weaker base than OH–(aq): HSO4–(aq) + OH–(aq) SO42–(aq) + HOH(l) Stronger Stronger Weaker Weaker Acid Base Base Acid The reaction is favored in the forward direction.

25. Predicting Acid–Base Rxns

26. The Water Equilibrium Autoionization of water

27. The Water Equilibrium H2O(l) H+(aq) + OH–(aq) Kw = [H+] [OH–] = 1.0  10–14 Kw is the water constant or equilibrium constant for water If [H+] = [OH–] = x Kw = [H+] [OH–] = 1.0  10–14 [H+] = [OH–] = 10–7 moles/liter

28. The Water Equilibrium For water or water solutions: If [H+] = [OH–] = 10–7 M, the solution is neutral. If [H+] > [OH–], the solution is acidic. If [H+] < [OH–], the solution is basic.

29. The Water Equilibrium Example: What is the hydrogen ion concentration in a solution in which the hydroxide ion concentration is 10–4 M? Is the solution acidic or basic? Solution: GIVEN: [OH–] = 10–4 M WANTED: [H+] EQUATION: Kw = [H+] [OH–] = 1.0  10–14 [H+] = 10–10 M Since [H+] = 10–10 < [OH–] = 10–4, the solution is basic

30. pH and pOH (Integer Values) A mathematical function is a rule that describes how to change one quantity to another. The p function: pQ = – log Q Applied to [H+] and [OH–], pH = – log [H+] pOH = – log [OH–]

31. pH and pOH (Integer Values) Inverse Functions pH = – log [H+] –pH = log [H+] antilog (–pH) = antilog (log [H+]) antilog (–pH) = [H+] Similarly, [OH–] = antilog (–pOH)

32. pH and pOH (Integer Values) Example: What is the pH of a solution with [H+] = 10–5 M? What is the [OH–] of a solution with pOH = 6? Solution: pH = – log [H+] = – log 10–5 = 5 [OH–] = antilog (–pOH) = antilog (–6) = 10–6 M

33. pH and pOH (Integer Values) Kw = [H+] [OH–] = 1.0  10–14 [H+] [OH–] = 1.0  10–14 – log ([H+] [OH–]) = – log (1.0  10–14) – log ([H+] [OH–]) = 14 – log [H+] + (– log [OH–]) = 14 pH + pOH = 14

34. pH and pOH (Integer Values) Example: The hydrogen ion concentration of a solution is 10–3 M. What are the pH, pOH, and [OH–] of the solution? Solution: pH = – log [H+] = – log 10–3= 3 pH + pOH = 14 pOH = 14 – pH = 14 – 3 =11 [OH–] = antilog (–pOH) = antilog (–11) = 10–11 M

35. pH and pOH (Integer Values) A solution is neutral if [H+] = 10–7 M A solution is acidic if [H+] > 10–7 M A solution is basic if [H+] < 10–7 M Using pH = – log [H+] and pOH = – log [OH–], A solution is neutral if pH = 7 A solution is acidic if pH < 7 A solution is basic if pH > 7

36. pH and pOH (integer Values)

37. pH and pOH (integer Values)

38. Noninteger pH-[H+]/pOH-[OH–] Significant Figures and Logarithms In a logarithm, the digits to the left of the decimal are not counted as significant figures. Counting significant figures in a logarithm begins at the decimal point.

39. Noninteger pH-[H+]/pOH-[OH–]

40. Noninteger pH-[H+]/pOH-[OH–] Example: The hydrogen ion concentration of a solution is 2.7 10–6 M. What are the pH, pOH, and hydroxide ion concentration? Solution: pH = – log [H+] = – log (2.7  10–6) = 5.57 pH + pOH = 14.00 pOH = 14.00 – pH = 14.00 – 5.57 = 8.43 [OH–] = antilog (–pOH) = antilog (–8.43) = 3.7  10–9 M