1. Some Definitions • Arrhenius • Acid: Substance that, when dissolved in water, increases the concentration of hydrogen ions. • Base: Substance that, when dissolved in water, increases the concentration of hydroxide ions. • Limitation: Restricted to aqueous system!

2. Some Definitions • Brønsted–Lowry • Broader definition! • Acid: Proton (H+ ) donor • Base: Proton(H+ ) acceptor

3. A Brønsted–Lowry acid… …must have a removable (acidic) proton. A Brønsted–Lowry base… …must have a pair of nonbonding electrons.

4. If it can be either… ...it is amphiprotic. HCO3− HSO4− H2O

5. What Happens When an Acid Dissolves in Water? • Water acts as a Brønsted–Lowry base and abstracts a proton (H+) from the acid. • As a result, the conjugate base of the acid and a hydronium ion are formed.

6. Conjugate Acids and Bases: • Reactions between acids and bases always yield their conjugate bases and acids.

7. What are products? • a) HNO3 + OH-

8. What are products? • CH3NH2 + H2O 

9. What are products? • c) OH- + HPO4-2

10. NH3 + H2O 

11. Acid and Base Strength • Strong acids are completely dissociated in water. • Have large K values!! • Their conjugate bases are quite weak. • Weak acids only dissociate partially in water. • Their conjugate bases are weak bases. • Small K values

12. What are the six strong Acids?? Do you remember??

13. What are the six strong Acids?? Do you remember?? • HCl • HBr • HI • (not HF) • HNO3 • H2SO4 • HClO4

14. Acid and Base Strength • Substances with negligible acidity do not dissociate in water. • Their conjugate bases are exceedingly strong.

15. Acid and Base Strength In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base. HCl(aq) + H2O(l) H3O+(aq) + Cl−(aq) H2O is a much stronger base than Cl−, so the equilibrium lies so far to the right K is not measured (K>>1).

16. HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2−(aq) Acid and Base Strength Acetate is a stronger base than H2O, so the equilibrium favors the left side (K<1).

17. H2O(l) + H2O(l) H3O+(aq) + OH−(aq) Autoionization of Water • As we have seen, water is amphoteric. • In pure water, a few molecules act as bases and a few act as acids. • This is referred to as autoionization.

18. Ion-Product Constant • The equilibrium expression for this process is Kc = [H3O+] [OH−] • This special equilibrium constant is referred to as the ion-product constant for water, Kw. • At 25°C, Kw = 1.0  10−14

19. pH pH is defined as the negative base-10 logarithm of the hydronium ion concentration. pH = −log [H3O+]

20. pH • In pure water, Kw = [H3O+] [OH−] = 1.0  10−14 • Because in pure water [H3O+] = [OH−], [H3O+] = (1.0  10−14)1/2 = 1.0  10−7

21. pH • Therefore, in pure water, pH = −log (1.0  10−7) = 7.00 • An acid has a higher [H3O+] than pure water, so its pH is <7 • A base has a lower [H3O+] than pure water, so its pH is >7.

22. pH These are the pH values for several common substances.

23. Other “p” Scales • The “p” in pH tells us to take the negative log of the quantity (in this case, hydrogen ions). • Some similar examples are • pOH −log [OH−] • pKw−log Kw

24. Watch This! Because [H3O+] [OH−] = Kw = 1.0  10−14, we know that −log [H3O+] + −log [OH−] = −log Kw = 14.00 or, in other words, pH + pOH = pKw = 14.00

25. How Do We Measure pH? • For less accurate measurements, one can use • Litmus paper • “Red” paper turns blue above ~pH = 8 • “Blue” paper turns red below ~pH = 5 • An indicator How Do We Measure pH?

26. How Do We Measure pH? For more accurate measurements, one uses a pH meter, which measures the voltage in the solution.

27. Strong Acids • You will recall that the six strong acids are HCl, HBr, HI, HNO3, H2SO4, and HClO4. • These are, by definition, strong electrolytes and exist totally as ions in aqueous solution. • For the monoprotic strong acids, [H+] = [acid].

28. What is the Hydrogen ion concentrations of .013 M HCl? • What is the Hydrogen ion concentrations of .0125 M HNO3? • HClO4 has a [H+] of 2.33 x 10-3 M.. What is the concentration of the acid?

29. Strong Bases • Strong bases are the soluble hydroxides, which are the alkali metal (LiOH, NaOH, KOH, RbOH, CsOH)and heavier alkaline earth metal hydroxides – • Ca(OH )2, Sr(OH )2 Ba(OH)2 • Again, these substances dissociate completely in aqueous solution.

30. Kc = [H3O+] [A−] [HA] HA(aq) + H2O(l) A−(aq) + H3O+(aq) Weak acids/bases • They do not ionize much • Equilibium lies far to the left! the equilibrium expression would be • This equilibrium constant is called the acid-dissociation constant, Ka.

31. Dissociation Constants The greater the value of Ka, the stronger the acid.

32. Write the Ka expression for acetic acid:

33. Write the Ka expression for acetic acid: • H C2H3O2(aq)+H2O(l) DC2H3O2−(aq)+H3O+(aq)

34. Weak bases • Weak bases without OH react with water to produce hydroxide ion. • Examples: NH3 CH3NH2 (methylamine) • lone pair of electrons forms a bond with H+

35. Equilibrium expression for bases is Kb • B + H2O(l) D HB+ OH- Kb=[HB] [OH−] [B]

36. Weak Bases Kb can be used to find [OH−] and, through it, pH.

37. Calculating pH from Ka Calculate the pH of a 0.30 M solution of acetic acid, HC2H3O2, at 25°C. HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2−(aq) Ka for acetic acid at 25°C is 1.8  10−5.

38. [H3O+] [C2H3O2−] [HC2H3O2] Ka = Calculating pH from Ka The equilibrium constant expression is

39. Calculating pH from Ka We next set up a table… We are assuming that x will be very small compared to 0.30 and can, therefore, be ignored.

40. When can we neglect x? • When the initial concnetration is very large compared to 100 Ka . If initial concentration is large by comparison, neglect x. • Quantitatively, if the quantity x is more than about 5% of the initial value, it is better to use the quadratic formula

41. 1.8  10−5 = (x)2 (0.30) Calculating pH from Ka Now, (1.8  10−5) (0.30) = x2 5.4  10−6 = x2 2.3  10−3 = x

42. Calculating pH from Ka pH = −log [H3O+] pH = −log (2.3  10−3) pH = 2.64

43. Calculating pH from Ka Calculate the pH of a 1.00 x 10-4 Msolution of acetic acid, HC2H3O2, at 25°C. HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2−(aq) Ka for acetic acid at 25°C is 1.8  10−5.

44. Calculating pH from Ka We next set up a table… x will is not very small compared to 1.00 x 10-4 and cannot be ignored.

45. Use the quadratic equation (see solver ) • X2 = 1.8 x 10-9 - 1.8 x 10-5 x Or • X2 + 1.8 x 10-5 x - 1.8 x 10-9 = 0 • X = 3.44 x 10-4 = [H+]

46. pH = - log (3.44 x 10-4 ) = 3.46

47. Calculating the pH of weak acid mixtures • Only the acid in the mixture with the larger Ka will donate an appreciable [H+ ]. • Determine pH Based on this acid, and ignore all others!

48. [H3O+] [COO−] [HCOOH] Ka = Calculating Ka from the pH • The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate Ka for formic acid at this temperature. • We know that

49. [H3O+] [COO−] [HCOOH] Ka = Calculating Ka from the pH • The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate Ka for formic acid at this temperature. • We know that

50. Calculating Ka from the pH • The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate Ka for formic acid at this temperature. • To calculate Ka, we need the equilibrium concentrations of all three things. • We can find [H3O+], which is the same as [HCOO−], from the pH.