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ACIDS AND BASES

ACIDS AND BASES. REVISE. Brønsted-Lowry acids and bases Amphoteric substances Conjugate acid base pairs Neutralisation. Neutral: pH = 7 ([H + ] = [OH - ]) Acidic: pH < 7 ([H + ] > [OH - ]) Basic: pH > 7 ([H + ] < [OH - ]). pH = -log [H + ] pOH = -log [OH - ]

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ACIDS AND BASES

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  1. ACIDS AND BASES

  2. REVISE • Brønsted-Lowry acids and bases • Amphoteric substances • Conjugate acid base pairs • Neutralisation Neutral: pH = 7 ([H+] = [OH-]) Acidic: pH < 7 ([H+] > [OH-]) Basic: pH > 7 ([H+] < [OH-]) pH = -log [H+] pOH = -log [OH-] pH + pOH = 14 at 25oC Kw = [OH-][H+] Kw = 14.0 at 25oC Kw = KaKb

  3. STRONG ACIDS AND BASES Strong acids and bases  react nearly “completely” to produce H+ and OH-  equilibrium constants are large e.g.: HCl H+ + Cl- Complete dissociation: large small Common strong acids: HCl, HBr, HI, H2SO4, HNO3, HClO4 (Why is HF not a strong acid?) Common strong bases: LiOH, NaOH, KOH, RbOH, CsOH, R4NOH

  4. Example: Calculate the pH of 0.1M LiOH. Strong base  Dissociates completely LiOH  Li+ + OH- 0.1M 0 0 Start: 0 0.1M 0.1M Complete rxn: pOH = -log [OH-] = -log (0.1) = 1  pH = 14 - 1 = 13

  5. Problem: What is the pH of 1x10-8 M KOH? As before: pOH = -log (1x10-8) = 8 pH = 14 – 8 = 6 BUT pH 6  acidic conditions and KOH is a strong base IMPOSSIBLE!!!!!!

  6. Since the concentration of KOH is so low (1x10-8 M), we need to take the ionisation of water into account. In pure water [OH-] = 1x10-7 M, which is greater than the concentration of OH- from KOH. We do this by systematic treatment of equilibrium. Charge balance:[K+] + [H+] = [OH-] Mass balance: [K+] = 1x10-8 M Equilibria: [H+][OH-] = Kw = 1x10-14 M 3 equations + 3 unknowns  solve simultaneously Find: pH = 7.02 Hint: You end up with a quadratic equation which you solve using the formula.

  7. Also note that: • Only pure water produces 1x10-7 M H+ and OH-. • If there is say 1x10-4 M HBr in solution, • pH = 4 and [OH-] = 1x10-10 M • But the only source of OH- is from the dissociation of water.  if water produces 1x10-10 M OH-  it can only produce 1x10-10 M H+ due to the dissociation of water. pH in this case is due mainly to the dissociation ofHBr and not the dissociation of water. • It is thus important to look at the concentration • of acid and bases present.

  8. Some guidelines regarding the concentrations of acids and bases: • When conc > 1x10-6 M •  calculate pH as usual • When conc < 1x10-8 M •  pH = 7 • (there is not enough acid or base to affect the pH of water) • When conc  1x10-8 - 1x10-6 M •  Effect of water ionisation and added acid and bases are comparable, thus: • use the systematic treatment of equilibrium approach.

  9. WEAK ACIDS AND BASES Weak acids and bases  react only “partially” to produce H+ and OH-  equilibrium constants are small HA H+ + A- Partial dissociation small large Acid dissociation constant • Common weak acids: • carboxylic acids • (e.g. acetic acid = CH3COOH) • ammonium ions • (e.g. RNH3+, R2NH2+, R3NH+) • Common weak bases: • carboxylate anions • (e.g. acetate = CH3COO-) • amines • (e.g. RNH2, R2NH, R3N)

  10. Base hydrolysis: B + H2O BH+ + OH- • Weak base • partial dissociation • Kb small base hydrolysis constant/ base “dissociation” constant NOTE: pKa = -log Ka pKb = -log Kb As K increases, its p-function decreases and vice versa.

  11. Problem: Find the pH of a solution of formic acid given that the formal concentration is 2 M and Ka = 1.80x10-4. HCOOH H+ + HCOO- H2O H+ + OH- Systematic treatment of equilibria: Charge balance: [H+] = [HCOO-] + [OH-] Mass balance: 2 M = [HCOOH] + [HCOO-] Equilibria: 4 equations  4 unknowns  difficult to solve

  12. Make an assumption: [H+] due to acid dissociation  [H+] due to water dissociation Produces HCOO- Produces OH- [HCOO-] large [OH-] small  [HCOO-] >> [OH-]  Charge balance: [H+]  [HCOO-]

  13. Charge balance: [H+]  [HCOO-] Mass balance: 2 M = [HCOOH] +[H+] Equilibria: Let [H+] = [HCOO-] = x Or x = -0.019  No negative conc’s [H+] = [HCOO-] = 0.019 M  pH = 1.7

  14. OR since [HCOOH] > 1x10-6, we can calculate pH as usual Weak acid equilibrium conditions HCOOH H+ + HCOO- 2M 0 0 Start: 2-x x x Equilibrium: Solve as before

  15. [HCOO-] 0.019 M  = = F 2 M FRACTION OF DISSOCIATION,  Fraction of acid in the form A- For the above problem: = 0.0095  Acid is 0.95% dissociated at 2 M formal concentration Weak electrolytes dissociate more as they are diluted.

  16. B + H2O BH+ + OH- WEAK BASE EQUILIBRIA Charge balance: [BH+] = [OH-] Mass balance: F = [B] + [BH+] Equilibria: Let [BH+] = [OH-] = x FRACTION OF ASSOCIATION

  17. CONJUGATE ACIDS AND BASES Relationship between Ka and Kb for a conjugate acid- base pair: Ka.Kb = Kw = 1x10-14 at 25oC • If Ka is very large (strong acid) Then Kb must be very small (weak conjugate base) And vice versa Base so weak it is not a base at all in water If Ka is very small, say 1x10-6 (weak acid) Then Kb must be small, 1x10-8(weak conjugate base) Greater acid strength, weaker conjugate base strength, and vice versa.

  18. Kw 1x10-14 Kb = = Ka 5 .70x10-10 Problem: Calculate the pH of 0.1 M NH3, given that pKa = 9.244 for ammonia. Kb NH3 + H2O NH4+ + OH- base acid pKa = -log Ka Ka = 5.70x10-10 = 1.75 x10-5

  19. x2 x2 Kb = = F - x 0.1 - x Problem: Calculate the pH of 0.1 M NH3, given that pKa = 9.244 for ammonia. Where x = [OH-] = [NH+] = 1.75 x10-5 Solve for x using the quadratic equation Negative value discarded Find x = 1.31x10-3 M = [OH-] pOH = -log [OH-] = 2.88 pH = 14 – 2.88 = 11.12

  20. BUFFERS Mixture of an acid and its conjugate base. Buffer solution resists change in pH when acids or bases are added or when dilution occurs. Mix: A moles of weak acid + B moles of conjugate base • Find: • moles of acid remains close to A, and • moles of base remains close to B  Very little reaction HA H+ + A- Le Chatelier’s principle

  21. HENDERSON-HASSELBALCH EQUATION For acids: When [A-] = [HA], pH = pKa For bases: pKa applies to this acid Kb  B + H2O BH+ + OH-  Ka base acid acid base

  22. Why does a buffer resist change in pH when small amounts of strong acid or bases is added? ? The acid or base is consumed by A- or HA respectively A buffer has a maximum capacity to resist change to pH. Buffer capacity, :  Measure of how well solution resists change in pH when strong acid/base is added. Larger   more resistance to pH change

  23. A buffer is most effective in resisting changes in pH when: pH = pKa i.e.: [HA] = [A-]  Choose buffer whose pKa is as close as possible to the desired pH. pKa 1 pH unit

  24. NH3 + H2O NH4+ + OH- Ka base acid (0.200) pH = 9.244 + log (0.300) Problem: Calculate the pH of a solution containing 0.200 M NH3 and 0.300 M NH4Cl given that the acid dissociation constant for NH4+ is 5.7x10-10. pKa = 9.244 pKa applies to this acid pH = 9.07

  25. POLYPROTIC ACIDS AND BASES Can donate or accept more than one proton. In general: Diprotic acid: H2L HL- + H+Ka1 K1 HL-L2- + H+Ka2 K2 Diprotic base: L2- + H2O HL- + OH- Kb1 HL-+ H2OH2L + OH- Kb2 Relationships between Ka’s and Kb’s: Ka1. Kb2 = Kw Ka2. Kb1 = Kw

  26. Using pKa values and mass balance equations, the fraction of each species can be determined at a given pH.

  27. ACID-BASE TITRATIONS We will construct graphs to see how pH changes as titrant is added. • Start by: • writing chemical reaction between titrant and analyte • using the reaction to calculate the composition and pH after each addition of titrant

  28. TITRATION OF STRONG BASE WITH STRONG ACID Example: Titrate 50.00 ml of 0.02000 M KOH with 0.1000 M HBr. HBr + KOH KBr + H2O What is of interest to us in an acid-base titration: H+ + OH-H2O Mix strong acid and strong base  reaction goes to completion H+ + OH-H2O

  29. C1V1 C2V2 = n1 n2 Example: Titrate 50.00 ml of 0.02000 M KOH with 0.1000 M HBr. * Calculate volume of HBr needed to reach the equivalence point, Veq: But n1 = n2 = 1  CHBrVeq = CKOHVKOH (0.1000 M)Veq = (0.02000 M)(50.00 ml) Veq = 10.00 ml

  30. There are 3 parts to the titration curve: 1 • Before reaching the equivalence point •  excess OH- present • At the equivalence point •  [H+] = [OH-] 2 3 • After reaching the equivalence point •  excess H+ present

  31. nunreacted COH- = Vtotal • Before reaching the equivalence point •  excess OH- present HBr + KOH KBr + H2O Starting nOH- = (0.02 M)(0.050 L) = 1x10-3 mol Say 2.00 ml HBr has been added. nH+ added = (0.1 M)(0.002 L) = 2x10-4 mol COH- = 0.01538 M Vtotal = 50 + 2 mL = 52 mL = 0.052 L • nOH- unreacted = 8x10-4 mol Kw = [H+][OH-] 1x10-14 = [H+](0.01538 M) [H+] = 6.500x10-13 M  pH = 12.19

  32. At the equivalence point •  nH+ = nOH- pH is determined by dissociation of H2O: H2O H+ + OH- x x Kw = [H+][OH-] 1x10-14 = x2 x = 1x10-7 M  [H+] = 1x10-7 M  pH = 7 pH = 7 at the equivalence point ONLY for strong acid – strong base titrations!!

  33. nexcess CH+ = Vtotal • After reaching the equivalence point •  excess H+ present HBr + KOH KBr + H2O Say 10.10 ml HBr has been added. Starting nOH- = 1x10-3 mol nH+ added = (0.1 M)(0.0101 L) = 1.010x10-3 mol CH+ = 1.664x10-4 M pH = 3.78 • nH+ excess = 1x10-5 mol Vtotal = 50 + 10.1 mL = 60.1 mL = 0.0601 L

  34. Note: A rapid change in pH near the equivalence point occurs. Equivalence point where: • slope is greatest • second derivative is • zero (point of inflection)

  35. Calculate titration curve by calculating pH values after a number of additions of HBr.

  36. TITRATION OF WEAK ACID WITH STRONG BASE Example: Titrate 50.00 ml of 0.02000 M formic acid with 0.1000 M NaOH. HCO2H + NaOH HCO2Na + H2O OR HCO2H + OH- HCO2- + H2O HA A- pKa = 3.745 Equilibrium constant so large  reaction “goes to completion” after each addition of OH- Ka = 1.80x10-4 Kb = 5.56x10-11 Strong and weak react completely

  37. C1V1 C2V2 = n1 n2 Example: Titrate 50.00 ml of 0.02000 M formic acid with 0.1000 M NaOH. HCO2H + OH- HCO2- + H2O * Calculate volume of NaOH needed to reach the equivalence point, Veq: But n1 = n2 = 1  CNaOHVeq = CFAVFA (0.1000 M)Veq = (0.02000 M)(50.00 ml) Veq = 10.00 ml

  38. Before base is added  HA and H2O present. HAweak acid,  pH determined by equilibrium: • HA H+ + A- Ka There are 4 parts to the titration curve: • From first addition of NaOH to immediately before equivalence point  mixture of unreacted HA and A- • HA + OH- A- + H2O • BUFFER!! use Henderson-Hasselbalch eqn for pH 2 1

  39. At the equivalence point •  all HA converted to A-. A- is a weak base whose pH is determined by reaction: • A- + H2O HA + OH- Kb 4 3 4) Beyond the equivalence point  excess OH- added to A-. Good approx: pH determined by strong base (neglect small effect from A-)

  40. HA H+ + A- Ka = 1.80x10-4 F- x x x x2 1.80x10-4 = 0.02 - x • Before base is added •  HA and H2O present. HA = weak acid. x2 + 1.80x10-4x – 3.60x10-6 = 0 x = 1.81x10-3  [H+] = 1.81x10-3  pH = 2.47

  41. - Start 1x10-3 2x10-4 End 8x10-4 - 2x10-4 2x10-4 pH = 3.745 + log 8x10-4 • From first addition of NaOH to immediately before equivalence point  mixture of unreacted HA and A- . BUFFER!! HCO2H + OH- HCO2- + H2O Say 2.00 ml NaOH has been added. Starting nHA = (0.02 M)(0.05 L) = 1x10-3 mol nOH- added = (0.1 M)(0.002 L) = 2x10-4 mol HA + OH- A- + H2O  pH = 3.14

  42. Special condition: When volume of titrant = ½ Veq pH = pKa Since: nHA = nA-

  43. - Start 1x10-3 1x10-3 End - - 1x10-3 • At the equivalence point •  all HA converted to A-. A- = weak base. (nHA = nNaOH) Starting nHA = 1x10-3 mol HA + OH- A- + H2O  nOH- = 1x10-3 mol • Solution contains just A-  a solution of weak base

  44. nA- 1x10-3 mol FA- = = V 0.060 L x2 5.56x10-11 = 0.0167 - x Kb = 5.56x10-11 A- + H2O HA + OH- F- x x x Vtotal = 50 + 10 mL = 60 mL = 0.060 L = 0.0167 M [OH-] = 9.63x10-7 M x2 + 5.56x10-11x – 9.27x10-13 = 0 pOH = 6.02 x = 9.63x10-7 pH = 7.98 pH is slightly basic at equivalence point for strong base-weak acid titrations

  45. CALCULATED TITRATION CURVE

  46. Titration curve depends on Ka of HA. As HA becomes a weaker acid the inflection near the equivalence point decreases until the equivalence point becomes too shallow to detect  not practical to titrate an acid or base that is too weak.

  47. Titration curve depends on extent of dilution of HA. As HA becomes a more dilute the inflection near the equivalence point decreases until the equivalence point becomes too shallow to detect  not practical to titrate a very dilute acid or base.

  48. TITRATION OF WEAK BASE WITH STRONG ACID This is the reverse of the titration of weak base with strong acid. The titration reaction is: B + H+ BH+ Recall: Strong and weak react completely

  49. Before acid is added •  B and H2O present. • B weak base  pH determined by equilibrium: • B + H2O BH+ + OH- Kb F-x x x There are 4 parts to the titration curve:

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