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ACIDS & BASES-

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  1. ACIDS & BASES-

  2. ACIDS & BASES- Electrolytes: Liquid that allows electricity to flow. Dissociation: Electrolytes break down into ions that conduct electricity.

  3. ACIDS & BASES REVIEW Acids turn Litmus Red. Acids have a pH below 7. Acids are clear in Phenolphthalein. Bases turn Litmus Blue. Bases have a pH above 7. Bases are pink in Phenolphthalein.

  4. Steven Arrhenius (1887) • Arrhenius Acids dissociate to form Hydrogen ions (H+) (aq) • Arrhenius Bases dissociate to form Hydroxide ions (OH-) HCl  H+ + Cl- NaOH OH- + Na+

  5. Arrhenius • Strong acids ionize completely, weak ones do not • High concentration of H+ • the H3O+ (hydronium ion) is equivalent to the Hydrogen Ion.

  6. BRONSTED-LOWRY THEORY

  7. BRONSTED-LOWRY THEORY • Johannes Bronsted & T. M. Lowry • Is based on the transfer of a Proton. • Bronsted-Lowry Acid: Anything that can donate a proton (H+ ion). • Bronsted-Lowry Base: A proton acceptor.

  8. Conjugate Pairs- • This is a pair that differs by one Proton. • Conjugate Acid = Conjugate Base plus a H+

  9. Conjugate Pairs- They have an inverse relationship. The stronger the conjugate acid, the weaker the base.

  10. Conjugate Pairs- Strong Acids: Completely transfers its Protons Weak Acids: Only partially transfers.

  11. Every acid-base reaction contains two acids + two bases HF + H2O  H3O+ + F- HNO3 + H2O  H3O+ + NO2-

  12. Amphoteric Substances- • Amphoteric: When a substance can act as a base or acid. +1 -1 H2O + HCl  H3O + Cl -1 -1 H2O + Br  HBr + OH

  13. Amphoteric Substances-

  14. AUTOIONIZATION When water is by itself and it acts like both an acid and a base. H2O + H2O  OH- + H3O+

  15. Ion-Product Constant +1 -1 -14 Kw= [H3O] [OH] = 1.0 X 10 or +1 -1 -14 Kw= [H] [OH] = 1.0 X 10 Kw = Ion-Product constant

  16. +1Finding H +1 -1 -14 Kw= [H] [OH] = 1.0 X 10 +1 -1 -9 Find [H] when [OH] is 8.8 x 10 +1 -6 [H] = 1.1 x 10

  17. +1Finding H +1 -1 Find [H] when [OH] is .00045M -1 -4 [OH] = 4.5 x 10 +1 -11 [H] = 2.2 x 10

  18. INDICATORS

  19. pH- Soren Sorensen (1909) • an indicator

  20. pH • exponent of the [H3O+] concentration. pH = -log [H3O+] or pH = -log [H+]

  21. Finding pH -10 the pH of 1 x 10 = 10 +1 -6 Find the pH when [H] = 5.6 x 10 pH = -log [H+] pH = 5.25

  22. Finding pH +1 Find the pH when [H] = .0006M -4 pH = -log [6.0 x 10] pH = 3.2

  23. +1Finding [H] +1 Find the [H3O] whenthe pH = 5.2 +1 pH = -log [H3O] +1 -5.2 = log [H3O] +1 -6 [H] = 6 x 10

  24. pOH -1 pOH = -log [OH] pH + pOH = 14

  25. -1Finding [OH] -1 Find the [OH] whenthe pH = 9.18 pH + pOH = 14 pOH = 4.82 -1 4.82 = - log [OH] -1 -5 [OH] = 1.5 x 10

  26. -1Finding [OH] -1 +1 -3 Find the [OH] whenthe [H] = 7.5 x 10 -3 pH = -log [7.5 x 10] pH = 2.12 2.12 + pOH = 14 pOH = 11.88 -1 -11.88 = log [OH] -1 -12 [OH] = 1.3 x 10

  27. STRONG ACIDS • These Dissociate fully. • They exist entirely as Ions. • They include the following: HCl HNO3 H2SO4 HBr HClO3 HI HClO4 These completely Ionize- +1 -1 HClO4  [H](aq) + ClO4 (aq)

  28. a STRONG ACID Problem • What is the pH of a .05M solution of HCl that is completely ionized? +1 -1 • [H] = [Cl] = .05M pH = -Log [.05] pH = 1.3

  29. a STRONG ACID Problem • What is the Hydrogen Ion concentration, when HCl has a pH of 2.34? +1 pH = -Log [H] 2.34 = - Log [X] The Hydrogen Ion concentration is .0046

  30. STRONG BASES • As with the Acids, these fully dissociate. • They include the Alkali Metal Hydroxides. Such as: NaOH LiOH • They also include some of the heavier Alkaline Earth Metal Hydroxides. Such as: Sr(OH)2

  31. PLEASE NOTE! +1 -1 -14 [H][OH] = 1 x 10

  32. a STRONG BASE Problem • What is the pH of a .028M solution of KOH? +1 -1 -14 [H] [OH] = 1 x 10 -14 [x] [.028] = 1 x 10 +1 -13 [H] = 3.57 x 10M-13 pH = -Log 3.57 x 10M pH = 12.45

  33. -OR- • What is the pH of a .028M solution of KOH? pOH = -Log [.028] = 1.55 pH + pOH = 14 pH + 1.55 = 14 pH = 12.45

  34. a STRONG BASE Problem • What is the pH of a .0011M solution of Ca(OH)2 ****Ca(OH)2 contains two Hydroxides* [OH] = .0011 x 2 = .0022M pOH = -Log [.0022] = 2.66 pH = 14 - 2.66 = 11.34

  35. a STRONG BASE Problem • What is the Hydroxide concentration of a solution of Ca(OH)2 that has a pH of 11.68? pH + pOH = 14 11.68 + X = 14 pOH = 2.32 2.32 = -Log [X] X = .00479 ** since there are two Hydroxides: .00479 / 2 = .00239 is the concentration

  36. HELLO

  37. WEAK ACIDS • Most Acids are Weak acids. • They only Ionize partially. • HA stands for a weak acid. +1 -1 HA (aq) ↔H(aq) + A(aq)

  38. ACID DISSOCIATION CONSTANT • The Acid Dissociation Constant is represented by Ka. • This measures the extent of the Ionization. • The larger the Ka , the stronger the Acid. • Stronger acids have a larger Percent Ionization.

  39. ACID DISSOCIATION CONSTANT +1 -1 Ka = [H] [A] / HA Appendix D (page 1062) has a list of the Ka ‘s.

  40. POLYPROTIC ACIDS Polyprotic Acids have more than 1 Ionizable Hydrogen Ion. Ka1 = The Acid Dissociation Constant Ka2 = The Acid Dissociation Constant for the removal of the 2nd proton. This is always much smaller than the Ka1.

  41. POLYPROTIC ACIDS

  42. POLYPROTIC ACIDS • It is always easier to remove the first Proton than it is to remove the second one. • The Ka1 is so much higher than the Ka2. +1 • Please note that the H(aq) usually comes from the first Ionization, so treat it as Monoprotic.

  43. FINDING THE Ka Find the Ka of a 0.10M Formic Acid (HCOOH) solution that has a pH of 2.38. +1 -1 HCOOH(aq) ↔ H(aq) + HCOO(aq) +1 -1 Ka = [H] [HCOO] / HCOOH

  44. FINDING THE Ka ICE +1 +1 -3 pH = -log[H] = 2.38[H] = 4.2 x10M +1 -1 HCOOH ↔ H HCOO I .10M 0 0 -3 -3 -3 C - 4.2 x10M 4.2 x10M 4.2 x10M -3 -3 -3 E .10M - 4.2 x10M 4.2 x10M 4.2 x10M **** .10M – .0042M ≈ .10M

  45. FINDING THE Ka -3 -3 Ka = [4.2 x10M] [4.2 x10M] / .10M -4 Ka = 1.8 x 10

  46. ANOTHER Ka PROBLEM Find the Ka of a 0.25M Lactic Acid solution that has a pH of 2.44 Lactic Acid = CH3CHOHCOOH +1 CH3CHOHCOOH(aq) ↔ H(aq) + -1 CH3CHOHCOO(aq) +1 +1 -3 pH = -log[H] = 2.44 H = 3.6 x 10

  47. ANOTHER Ka ICE +1 -1 CH3CHOHCOOH(aq) ↔ H(aq) + CH3CHOHCOO(aq) I .25M 0 0 -3 -3 -3 C - 3.6 x10M 3.6 x10M 3.6 x10M -3 -3 -3 E .25M – 3.6 x10M 3.6 x10M 3.6 x10M **** .25M - .0036M ≈ .25M

  48. THE Ka is: -3 -3 Ka = [3.6 x10M] [3.6 x10M] / .25M -5 Ka = 5.2 x 10

  49. PERCENT IONIZATION % IONIZATION = Concentration ion (100) /Original % IONIZATION = +1 [H]equilib (100) / [HA]init

  50. PERCENT IONIZATION PROBLEM Find the % Ionization of a .035M HNO2 Solution that has dissociated into -3 7.4 x 10M of Hydrogen ions. % IONIZATION = +1 [H]equilib (100) / [HA]init