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Chapter 16

Acid-Base Equilibria. Chapter 16 . Revision. Acids and bases change the colours of certain indicators . Acids and bases neutralize each other. Acids and bases react to form salts . Acids react with certain metals to release hydrogen . Bronsted-Lowry Acids and Bases.

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Chapter 16

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  1. Acid-Base Equilibria Chapter 16

  2. Revision Acids and bases change the colours of certain indicators. Acids and bases neutralize each other. Acids and bases react to form salts. Acids react with certain metals to release hydrogen.

  3. Bronsted-Lowry Acids and Bases These two chemists pointed out that acids and bases can be seen as proton transfer reactions. According to the Bronsted-Lowry concept: An acid is the species donating a proton in a proton-transfer reaction A base is the species accepting the proton in a proton-transfer reaction.

  4. If we consider the reaction: HCl (g) + NH3 (g) NH4Cl (s) We see that we can view it as a proton transfer. In any reversible acid-base reaction, both forward and reverse reactions involve proton transfers.

  5. Consider the reaction of NH3 with H2O: NH3 (aq) + H2O (aq) NH4+ (aq) + OH- (aq) A conjugate acid-base pair consists of two species in an acid-base reaction, one acid one base, that differ by the loss or gain of a proton. Note: NH3 and NH4+ are a conjugate acid-base pair. H2O and OH- are also a conjugate acid-base pair.

  6. Example Not in Book! Identify the acid and base species in the following equation: CO32-(aq) + H2O(l) HCO3-(aq) + OH- Acid species : Base species :

  7. An amphiprotic species is a species that can act as either an acid or a base (it can lose or gain a proton), depending on the other reactant. Consider water: H2O + CH3O- OH- + CH3OH acid base H2O + HBr  H3O+ + Br- base acid

  8. Relative strengths of acids and bases The strongest acids have the weakest conjugate bases and the strongest bases have the weakest conjugate acids.

  9. [H3O+] [OH-] Kc = [H2O]2 [H2O]2 Kc = [H3O+] . [OH-] AUTOPROTOLYSIS Water undergoes self-ionisation autoprotolysis, since H2O acts as an acid and a base. H2O + H2O H3O++ OH- The extent of autoprotolysis is very small. The equilibrium constant expression for this reaction is: The concentration of water is essentially constant. Therefore: constant = Kw

  10. We call the equilibrium value of the ion product [H3O+][OH-] the ion-product constant of water. Kw = [H3O+] [OH-] = 1.0 x 10-14 at 25°C Using Kw you can calculate concentrations of H3O+ and OH- in pure water. [H3O+] [OH-] = 1.0 x 10-14 But [H3O+] = [OH-] in pure water  [H3O+] = [OH-] = 1.0 x 10-7 M If you add an acid or a base to water the concentrations of H3O+ and OH- will no longer be equal. But Kw will still hold.

  11. The pH Scale Because concentration values may be very small, it is often more convenient to express acidity in terms of pH. Definition of pH pH is defined as the negative logarithm of the molar hydronium-ion concentration. pH = -log [H3O+] often written as : pH = -log [H+]

  12. For a solution with a hydronium-ion concentration of 1.0 x 10-3 M, the pH is: Note that the number of places after the decimal point in the pH equals the number of significant figures reported in the hydronium-ion concentration.

  13. Example Not in Book! Calculating the pH from the hydronium-ion concentration Calculate the pH of typical adult blood, which has a hydronium-ion concentration of 4.0 x 10-8M.

  14. Example Not in Book! Calculating the hydronium-ion concentration from the pH. The pH of natural rain is 5.60. Calculate its hydronium-ion concentration.

  15. Other “p” scales – pOH In the same manner that we defined pH we can also define pOH: pOH = -log [OH-] also remember that: Kw = [H3O+] . [OH-] = 1.0 x 10-14 therefore we can show that: pH + pOH = 14.00 -log ([H+] [OH-]) = -log (1x10-14) -log [H+] + -log [OH-] = -log (1x10-14) pH + pOH = 14

  16. Example Not in Book! Calculating concentrations of H3O+ and OH- in solutions of a strong acid or base. Calculate the concentrations of hydronium ion and hydroxide ion at 25°C in 0.10 M HCl.

  17. Eg. Acetic Acid: CH3COOH + H2O H3O+ + CH3COO- Weak Acids An acid reacts with water to produce hydronium ion and the conjugate base ion. This process is called acid ionization or acid dissociation.

  18. [H3O+] . [A-] Kc = [HA] . [H2O] [H3O+] . [A-] Ka = [H2O] Kc = [HA] In general for an acid, HA, we can write: HA(aq) + H2O H3O+ + A-(aq) and the corresponding equilibrium constant expression would be: For a dilute solution, [H2O] would be nearly constant, hence:

  19. Example Not in Book! Determining Ka from the solution pH. Sore-throat medications sometimes contain the weak acid phenol, HC6H5O. A 0.10 M solution of phenol has a pH of 5.43 at 25°C. What is the acid-dissociation constant, Ka, for this acid at 25°C? What is its degree of ionization?

  20. Solution HA + H2O H3O++ A- Initial  Eqlbm pH = 5.43

  21. Solution Degree of ionisation (fraction of ions that were ionised): HA + H2O H3O++ A- x of the original 0.10 M was ionised. Degree of ionisation =

  22. Not in Book! Example(See discussion p609-) Calculating Concentrations of Species in a Weak Acid Solution Using Ka. (Approximation Method) Para-hydroxybenzoic acid is used to make certain dyes. What are the concentrations of this acid (i.e.hydrogen ion) and para-hydroxybenzoate anion in a 0.200 M aqueous solution at 25°C? What is the pH of the solution and the degree of ionisation of the acid? The Ka of this acid is 2.6 x 10-5.

  23. Solution HA + H2O H3O++ A- Initial  Eqlbm Now we must solve this quadratic. One possible way of simplifying things is via an assumption!

  24. Because Ka is small (2.6 x 10-5), Hence: [H3O+] = [A-] = Degree of ionisation [HA] = =

  25. Polyprotic Acids So far we have only dealt with acids releasing only one H3O+ ion. Some acids, have two or more such protons; these acids are called polyprotic acids. Remember the strong acid H2SO4

  26. H2CO3 + H2O H3O+ + HCO3- HCO3- + H2O H3O+ + CO32- [H3O+] [HCO3-] = 4.3 x 10-7 Ka1= [H2CO3] [H3O+] [CO32-] Ka2= = 4.8 x 10-11 [HCO3-] For a weak diprotic acid like carbonic acid, H2CO3, there are two simultaneous equilibria to consider. Each equilibrium has an associated acid dissociation constant. For the loss of the first proton: and for the loss of the second proton:

  27. In general, the second ionisation constant, Ka2, of a polyprotic acid is much smaller then the first ionisation constant, Ka1. In the case of a triprotic acid, the third ionisation constant, Ka3, is much smaller that the second one, Ka2.

  28. Weak Bases Equilibria involving weak bases are treated similarly to those for weak acids. In general, a weak base B with the base ionization: B(aq) + H2O HB+ + OH- has a base-ionization constant, Kb, equal to: [HB+] [OH-] Kb = [B]

  29. Example Not in Book! Calculating concentrations of species in a weak base solution using Kb. Aniline, C6H5NH2, is used in the manufacturing of some perfumes. What is the pH of a 0.035 M solution of aniline at 25°C? The Kb = 4.2 x 10-10 at 25°C.

  30. B + H2O HB+ + OH- Initial Solution  Eqlbm Now

  31. HA(aq) + H2O H3O+ + A-(aq) [H3O+] [A-] Ka = [HA] A-(aq) + H2O HA(aq) + OH- [HA] [OH-] Kb = [A-] [H3O+] [A-] [HA] [OH-] Ka .Kb = x [HA] [A-] Relationship between Ka and Kb Remember that for we have: and for we can write : Multiplying: = [H3O+] [OH-]  Ka .Kb= Kw

  32. Example Not in Book! Obtaining Ka from Kb or Kb from Ka Obtain the Kb for the F- ion, the ion added to public water supplies to protect teeth. For HF, Ka = 6.8 x 10-4. Kb.Ka = Kw

  33. Buffers A buffer is a solution characterised by the ability to resist changes in pH when limited amounts of acid or base are added to it. Buffers contain either a weak acid and its conjugate base or a weak base and its conjugate acid. Thus, a buffer solution contains both an acid species and a base species in equilibrium.

  34. Example Not in Book! Calculate the pH of a buffer from given volumes. What is the pH of a buffer made by mixing 1.00 L of 0.020 M benzoic acid, HC7H5O2, with 3.00 L of 0.060 M sodium benzoate, NaC7H5O2? The Ka for benzoic acid is 6.3 x 10-5.

  35. Example Not in Book! Calculating the pH of a buffer when a strong acid or strong base is added. Calculate the pH change that will result from the addition of 5.0 mL of 0.10 M HCl to 50.0 mL of a buffer containing 0.10 M NH3 and 0.10 M NH4+. How much would the pH of 50.0 mL of water change if the same amount of acid were added.?

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