2. pH = - log [H + ] or pH = - log [H 3 O + ]

4. Example: If [H + ] = 1 X 10 -10 pH = - log 1 X 10 -10 pH = - (- 10) pH = 10  What would be the pH of a 0.000018 M HNO 3 solution? Example: If [H + ] = 1.8 X 10 -5 pH = - log 1.8 X 10 -5 pH = - (- 4.74) pH = 4.74 pH Problems

5. If the pH of Coke is 3.12, [H + ] = ??? [H + ] = 10 -3.12 = 7.6 x 10 -4 M More pH Problems

6. Identify as SA, SB, WA, WB Big or Small K value? H 2 O can function as both an ACID and a BASE. In pure water there can be AUTOIONIZATION Pure Water – Acid or Base? H 2 O + H 2 O  H 3 O + + OH - WB SA WA SB

7. Equilibrium constant for water = Kw Kw = [H 3 O + ] [OH - ] = 1.00 x 10 -14 at 25 oC K w = [H 3 O + ] [OH - ] In pure water, [H 3 O + ] = [OH - ] so K w = [x][x] = [x] 2 and so, [H 3 O + ] = [OH - ] = 1.00 x 10 -7 THEREFORE, the pH is… pH of Pure Water

8. H + vs OH - What would be the pH of a 1.0 M HCl solution? Of a 0.01 M HCl solution? [H + ] and [OH - ] have an inverse relationship in aqueous solution

9. pOH is from the base perspective pOH = - log [OH - ] Since we are dealing with aqueous solution… [H + ] [OH - ] = 1.00 x 10 -14 pH + pOH always equals14 pOH

10. pHpOH[H + ][OH - ]Acid or Base 2.45 4.75 3.5 x 10 -10 0.00084

11.  Why is the pH of a 0.1 or 10 -1 M acetic acid not 1?

12. A strong acid ionize 100% but a weak acid does not! Weak acid has K a < 1 Leads to small [H 3 O + ] Strong vs Weak Acids

13. Weak base has K b < 1 Leads to small [OH - ] Strong vs Weak Bases

15. Strong Acid vs Strong Base Complete ionization! Virtually no reactants left No equilibrium The equivalence point is the point where the number of moles of base equal the number of moles of acid. http://www. youtube.co m/watch?v= ILn79QpYw Pc http://www.you tube.com/watc h?v=HnGy8Um 6ibM http://www.y outube.com/ watch?v=H63 dHo-T1TM

16. Weak Acid and Strong Base At equivalence point, the pH > 7

17. Weak Base and Strong Acid

18. Ka of polyprotic acids  A polyprotic acid has two or more hydrogen which can ionize in multiple steps. H 2 CO 3 (aq)  H + + HCO 3 - K 1 = 4.5 x10 -7 HCO 3 (aq)  H + + CO 3 -2 K 2 = 4.7 x10 -11 H 2 CO 3 (aq)  2 H + + CO 3 -2 Ka = ?  (Overall Ka = K 1 x K 2 )

19. Practice Problem  What is the Ka of the equation below? H 3 PO 4 (aq)  3 H + + PO 4 -3 H 3 PO 4  H + + H 2 PO 4 - K 1 = 7.1 x 10 -3 H 2 PO 4 -  H + + HPO 4 -2 K 2 = 6.3 x 10 -8 HPO 4 -2  H + + PO 4 -3 K 3 = 4.5 x 10 -13

20. Titration Curve of a Weak Diprotic acid with a Strong Base

21. Acidic Salts  When dissolved, the salt created from a strong acid and a weak base will be acidic. NH 4 Cl (s)  NH 4 + + Cl - Why? NH 4 + + H 2 O  NH 4 OH + H 3 O + The ammonium ion acts as an acid. It will have a K a value. conjugate of NH 3 conjugate of HCl

22. Basic Salts  When dissolved, the salt created from a strong base and a weak acid will be basic. NaC 2 H 3 O 2 (s)  Na + + C 2 H 3 O 2 - So? C 2 H 3 O 2 - + H 2 O  HC 2 H 3 O 2 + OH - The acetate ion acts as a base. It will have a K b value. How will the products react with H + and OH - ?

23. Ka and Kb  For an acid-base conjugate pair:  The conjugate of a strong is weak, and the conjugate of a weak is strong. Why? (K a )(K b ) = K w = 1.4 x 10 -14 Ka is the acid ionization constant Kb is the base ionization constant Kw is the ionization constant of water

24. Practice Problem  Acetic acid has a Ka value of 1.7 x 10 -5. What is the conjugate base? What is the Kb value of the conjugate base?  Carboxyl or organic acid group (-COOH)

25. Practice Problem  Methylamine has a K b value of 4.38 x 10 -4. What is the conjugate acid? What is the Ka value of the conjugate base? Amine group (-NH 2 )   Methyl group (-CH 3 )

26. Calculating the pH of a Salt Information Given  Concentration of salt  K a or K b value Additional Information Needed  Identify as Acidic, Basic, or Neutral  Equation of ions with water  K eq expression

27. Practice Problem  Find the pH of a 0.500 M solution of KCN. The Ka value of HCN is 5.8 x 10 -10.

28. Steps  Dissociation of Salt KCN + H 2 O  K + + CN -  Identify base and acid KOH (strong base) + HCN (weak acid) The salt must be basic. CN - is a strong conjugate base

29. Steps (continued)  Reaction between CN - and water CN - + H 2 O  OH - + HCN  Write Kb expression Kb = [HCN][OH - ] [CN - ]

30. Steps (continued 2)  Use Ka of HCN to find Kb of CN - KaKb = Kw (5.8 x 10 -10 )(Kb) = (1.0 x 10 -14 ) Kb = 1.7 x 10 -5

31. Steps (continued 3)  Use Kb and Kb expression to solve for [OH - ] (set up ice table first) CN - + H 2 O  OH - + HCN I.500M n/a 0 0 C -x +x +x E.500 - x x x 1.7 x 10 -5 = [x][x] [.500 - x] X = 2.9 x 10 -3

32. Steps (continued 4)  Using [OH - ], find pOH and pH pOH = -log[2.9 x 10 -3 ] pOH = 2.54 pOH + pH = 14 2.54 + pH = 14 pH = 11.46

33. Practice Problem  What would be the pH of a 0.200 M ammonium chloride solution? Kb of ammonia is 1.7 x 10 -5. Write dissociation of ammonium chloride Write reaction of ammonium and water Write Ka expression for ammonium Calculate Ka value using Kb and Kw Solve for [H + ] Find pH

34. Practice Problem  What would be the pH of a 0.200 M ammonium chloride solution? Kb of ammonia is 1.7 x 10 -5. Ka of NH 4 + = [H 3 O + ] [NH 3 ] [NH4 + ] 5.88 x 10 -10 = __x 2 __ (.200) x = 1.08 x 10 -5 pH = 4.96

35. Buffer  A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base.  A buffer contains: A weak acid or weak base, AND The salt of the weak acid or base NH 3 and NH 4 Cl (Weak Base and Acidic Salt) HC 2 H 3 O 2 and NaC 2 H 3 O 2 (Weak Acid and Basic Salt)

36. HOW DO BUFFERS WORK? BUFFER: NH 3 and NH 4 Cl  The NH 3 (base) will neutralize any acid i.e. HCl (extra H + ions) by combining with the extra H + ions to form NH 4 + ions.  The NH 4 + (conjugate acid) will neutralize any base i.e. KOH (extra OH - ions) by donating its H + ion to form HOH.

37. HCl + CH 3 COO - CH 3 COOH + Cl - BUFFER: some H +,some C 2 H 3 O 2 -, HC 2 H 3 O 2 and Na +,C 2 H 3 O 2 - ADD: HCl ADD: KOH KOH + CH 3 COOH CH 3 COOK + HOH

38. Common Ion Consider mixture of HC 2 H 3 O 2 and NaC 2 H 3 O 2 The common ion (C 2 H 3 O 2 - ) suppresses the ionization of the weak acid. This is called the common ion effect. NaC 2 H 3 O 2 (aq)  Na + + C 2 H 3 O 2 - HC 2 H 3 O 2 (aq)  H + + C 2 H 3 O 2 -

39. Practice Problem  Which of the following are buffer systems? (a) KCl/HF (b) NH 4 NO 3 /NH 3 (c) KCl/HCl (d) NaHCO 3 /H 2 CO 3 (e) Ca(OH) 2 /CaSO 4 (f) NH 3 /HNO 2  Which buffers have a pH above 7? --------------No Common Ion ---------------Strong Acid (not weak) ----------------------------- Strong Base (not weak) -------------------- Weak Base and Acid (no salt)

40. Finding the pH of the buffer  What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? Ka for HCOOH = 1.8 x 10 -4 Mixture of weak acid and conjugate base! Initial (M) Change (M) Equilibrium (M) HCOOH (aq) H + (aq) + HCOO - (aq) -x-x+x+x 0.30 - x +x+x x0.52 + x 0.300.000.52 x = 1.04 X 10 -4 pH = 4.0

41. Finding the pH of a buffer EASIER METHOD! Henderson-Hasselbach equation (on reference sheet)  For weak acid and its salt  For weak base and its salt pH = pK a + log [A - ] [HA] pOH = pK b + log [HB + ] [B]

42. Henderson-Hasselbach  What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? Ka for HCOOH = 1.8 x 10 -4 pH = 3.77 + log [0.52] [0.30] = 4.0

43. Practice Problem  Calculate the pH of the 0.30 M NH 3 /0.36 M NH 4 Cl buffer system. The Kb of NH 3 is 1.8 x 10 -5 NH 3 + HOH  NH 4 + + OH -

44. pH after NaOH is added  What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution? NH 3 + HOH  NH 4 + + OH - (.30)(.08) n/a (.36)(.08) (.050)(.020) I.024 moles.0288.0010 moles C +.001 -.0010 -.0010 E.025.0278 0 E. 25 M.278 M

45. Weak Acid and Strong Base At equivalence point, the pH > 7

46. Helpful Guides  Halfway to equivalence, [H + ] = Ka this means that the pH = pKa  Prior to equivalence, [H + ] > Ka this means that the pH < pKa  Between halfway to equivalence, [H + ] < Ka this means that the pH >pKa

47. Weak Base and Strong Acid

48. Choosing the Correct Indicator Which indicator would you use for a titration of HNO 3 with NH 3 ?

49. Summary  Strong Acid vs. Strong Base 100 % ionized! pH = 7 No equilibrium!  Weak Acid vs. Strong Base Acid is neutralized; Need K b for conjugate base equilibrium  Strong Acid vs. Weak Base Base is neutralized; Need K a for conjugate acid equilibrium  Weak Acid vs. Weak Base Depends on the strength of each; could be conjugate acid, conjugate base, or pH 7

50.  These are usually formed from a transition metal surrounded by ligands (polar molecules or negative ions).  As a "rule of thumb" you place twice the number of ligands around an ion as the charge on the ion... example: the dark blue Cu(NH 3 ) 4 2+ (ammonia is used as a test for Cu 2+ ions), and Ag(NH 3 ) 2 +.  Memorize the common ligands.

51. LigandsNames used in the ion H2OH2Oaqua NH 3 ammine OH-hydroxy Cl-chloro Br-bromo CN-cyano SCN-thiocyanato (bonded through sulphur) isothiocyanato (bonded through nitrogen)

52.  Names: ligand first, then cation Examples: tetraamminecopper(II) ion: Cu(NH 3 ) 4 2+ diamminesilver(I) ion: Ag(NH 3 ) 2 +. tetrahydroxyzinc(II) ion: Zn(OH) 4 2-  The charge is the sum of the parts (2+) + 4(-1)= -2.

53.  Aluminum also forms complex ions as do some post transitions metals. Ex: Al(H 2 O) 6 3+  Transitional metals, such as Iron, Zinc and Chromium, can form complex ions.  The odd complex ion, FeSCN 2+, shows up once in a while  Acid-base reactions may change NH 3 into NH 4 + (or vice versa) which will alter its ability to act as a ligand.  Visually, a precipitate may go back into solution as a complex ion is formed. For example, Cu 2+ + a little NH 4 OH will form the light blue precipitate, Cu(OH) 2. With excess ammonia, the complex, Cu(NH 3 ) 4 2+, forms.  Keywords such as "excess" and "concentrated" of any solution may indicate complex ions. AgNO 3 + HCl forms the white precipitate, AgCl. With excess, concentrated HCl, the complex ion, AgCl 2 -, forms and the solution clears.

54.  Total number of bonds from the ligands to the metal atom.  Coordination numbers generally range between 2 and 12, with 4 (tetracoordinate) and 6 (hexacoordinate) being the most common.

55. molecular formula Lewis base/ligand Lewis aciddonor atom coordination number Ag(NH 3 ) 2 + NH 3 Ag + N2 [Zn(CN) 4 ] 2- CN-Zn 2+ C4 [Ni(CN) 4 ] 2- CN-Ni 2+ C4 [PtCl 6 ] 2- Cl-Pt 4+ Cl6 [Ni(NH 3 ) 6 ] 2+ NH 3 Ni 2+ N6