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ACID-BASE EQUILIBRIA

ACID-BASE EQUILIBRIA. pH Scale. pH = - log [H + ] or pH = - log [H 3 O + ]. Why a logarithm scale?. pH Problems. 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?

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ACID-BASE EQUILIBRIA

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  1. ACID-BASE EQUILIBRIA

  2. pH Scale pH = - log [H+] or pH = - log [H3O+]

  3. Why a logarithm scale?

  4. pH Problems Example: If [H+] = 1 X 10-10pH = - log 1 X 10-10 pH = - (- 10) pH = 10 • What would be the pH of a 0.000018 M HNO3 solution? Example: If [H+] = 1.8 X 10-5pH = - log 1.8 X 10-5 pH = - (- 4.74) pH = 4.74

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

  6. Pure Water – Acid or Base? Identify as SA, SB, WA, WB Big or Small K value? H2O can function as both an ACID and a BASE. In pure water there can be AUTOIONIZATION SA SB WA WB H2O + H2O  H3O+ + OH-

  7. pH of Pure Water Equilibrium constant for water = Kw Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 oC Kw = [H3O+] [OH-] In pure water, [H3O+] = [OH-] so Kw = [x][x] = [x]2 and so, [H3O+] = [OH-] = 1.00 x 10-7 THEREFORE, the pH is…

  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 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

  10. PRACTICE PROBLEM

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

  12. Strong vs Weak Acids A strong acid ionize 100% but a weak acid does not! Weak acid has Ka < 1 Leads to small [H3O+]

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

  14. Ka and Kb

  15. Strong Acid vs Strong Base http://www.youtube.com/watch?v=H63dHo-T1TM Complete ionization! Virtually no reactants left No equilibrium http://www.youtube.com/watch?v=HnGy8Um6ibM http://www.youtube.com/watch?v=ILn79QpYwPc The equivalence point is the point where the number of moles of base equal the number of moles of acid.

  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. H2CO3 (aq)  H+ + HCO3- K1 = 4.5 x10-7 HCO3 (aq)  H+ + CO3-2 K2 = 4.7 x10-11 H2CO3 (aq)  2 H+ + CO3-2 Ka = ? • (Overall Ka = K1 x K2)

  19. Practice Problem What is the Ka of the equation below? H3PO4 (aq)  3 H++ PO4-3 H3PO4  H++ H2PO4- K1= 7.1 x 10-3 H2PO4- H++ HPO4-2K2= 6.3 x 10-8 HPO4-2 H++ PO4-3 K3= 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. NH4Cl (s)  NH4+ + Cl- Why? NH4+ + H2O  NH4OH + H3O+ The ammonium ion acts as an acid. It will have a Ka value. conjugate of HCl conjugate of NH3

  22. Basic Salts When dissolved, the salt created from a strong base and a weak acid will be basic. NaC2H3O2 (s)  Na+ + C2H3O2- So? C2H3O2- + H2O  HC2H3O2+ OH- The acetate ion acts as a base. It will have a Kb 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? (Ka)(Kb) = Kw = 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 Kb value of 4.38 x 10-4. What is the conjugate acid? What is the Ka value of the conjugate base? Amine group (-NH2)  Methyl group (-CH3) 

  26. Calculating the pH of a Salt Information Given Concentration of salt Ka or Kb value Additional Information Needed Identify as Acidic, Basic , or Neutral Equation of ions with water Keq 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+ H2O  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-+ H2O  OH-+ HCN Write Kb expression Kb = [HCN][OH-] [CN-]

  30. Steps (continued 2) Use Ka of HCN to find Kb of CN- Ka•Kb = 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-+ H2O  OH-+ HCN I .500M n/a 0 0 C -x +x +x E .500 - x xx 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. • [H3O+] [NH3] Ka of NH4+ = • [NH4+] __x2__ (.200) 5.88 x 10-10 = 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 NH3 and NH4Cl (Weak Base and Acidic Salt) HC2H3O2 and NaC2H3O2(Weak Acid and Basic Salt)

  36. HOW DO BUFFERS WORK?BUFFER: NH3 and NH4Cl The NH3 (base) will neutralize any acid i.e. HCl (extra H+ ions) by combining with the extra H+ ions to form NH4 + ions. The NH4+ (conjugate acid) will neutralize any base i.e. KOH (extra OH- ions) by donating its H+ ion to form HOH.

  37. HCl+ CH3COO- CH3COOH + Cl- • BUFFER:some H+,some C2H3O2-, HC2H3O2 and Na+,C2H3O2- ADD: HCl ADD: KOH KOH + CH3COOH CH3COOK + HOH

  38. Common Ion Consider mixture of HC2H3O2 and NaC2H3O2 The common ion (C2H3O2-) suppressesthe ionization of the weak acid. This is called the common ion effect. NaC2H3O2 (aq)  Na+ + C2H3O2- HC2H3O2 (aq)  H+ + C2H3O2-

  39. Practice Problem Which of the following are buffer systems? (a) KCl/HF (b) NH4NO3/NH3 (c) KCl/HCl (d) NaHCO3/H2CO3 (e) Ca(OH)2/CaSO4 (f) NH3/HNO2 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! HCOOH (aq) H+(aq) + HCOO-(aq) Initial (M) Change (M) Equilibrium (M) 0.30 0.00 0.52 -x +x +x 0.30 - x x 0.52 + x x = 1.04 X 10 -4 pH = 4.0

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

  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 [0.52] pH = 3.77 + log [0.30] = 4.0

  43. Practice Problem Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. The Kb of NH3 is 1.8 x 10-5 NH3+ HOH NH4++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? NH3+ HOH NH4++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 HNO3 with NH3 ?

  49. Summary Strong Acid vs. Strong Base 100 % ionized! pH = 7 No equilibrium! Weak Acid vs. Strong Base Acid is neutralized; Need Kb for conjugate base equilibrium Strong Acid vs. Weak Base Base is neutralized; Need Ka 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. Complex Ion Formation 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(NH3)42+ (ammonia is used as a test for Cu2+ ions), and Ag(NH3)2+. Memorize the common ligands.

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