Chapter 11: Acid-Base Titrations Chapter 10: Polyprotic Acid-Base Equilibria 1. Treating Complex Acid-Base Systems Complex systems are defined as solutions made up of: 1) Two acids or bases of different strengths HCl + CH 3 COOH NaOH + CH 3 COO - 2) An acid or base that has two or more acidic protons or basic functional groups H 3 PO 4 Ca(OH) 2 3) An amphiprotic substance that is capable of acting as both acid and base HCO 3 - + H 2 O <=> CO 3 2- + H 3 O + HCO 3 - + H 2 O <=> H 2 CO 3 + OH - N H 3 + CH 2 COO - + H 2 O <=> NH 2 CH 2 COO - + H 3 O + N H 3 + CH 2 COO - + H 2 O <=> NH 3 + CH 2 COOH + OH - 2. Equations for more than one equilibrium are required to describe the characteristics of any of these systems. 3. Applications 1) We encounter these systems in most biological and environmental matrices 2) We need to predict which species will be present (and in what amounts) is important in defining biological acid/base buffering. 3) We can use the titration curve to examine how the species will change as we slowly add titran t (in other words, change the pH). 4. Examples A. Mixtures of Strong and Weak Acids or Strong and Weak Bases B. Polyfunctional Acids and Bases: C. Buffer solutions Involving Polyprotic Acids: D. Calculation of the pH of Solutions of Amphiprotic Salts (NaHA): E. Titration Curves of Polyfunctional Acids: A. Mixtures of Strong and Weak Acids or Strong and Weak Bases 4. Examples A. Mixtures of Strong and Weak Acids or Strong and Weak Bases Example 11-1A: Derive a titration curve for a titration of a 50.00 mL solution containing 0.1000 M strong acid, HCl, and 0.1000 M weak aci d HA (K a = 1 X 10 -4 ) with 0.1000 NaOH. (similar to Example 11-1) The equilibria: HCl + H 2 O => H 3 O + + Cl - [1] HA + H 2 O <=> H 3 O + + A - [2] 2 H 2 O <=> H 3 O + + OH - [3] [H 3 O + ] = C HCl + [A - ] + [OH - ] Since the HCl, which is completely dissociated, will repress the dissociatio n of HA and H 2 O. Assume that [OH - ] and [A - ]<<C HCl so that [H 3 O+] = C HCl . 1. V NaOH =0: The pH before the addition of titrant is determined by the concentration of HCl alone: [H 3 O+] = C HCl 2. 0 <V NaOH <50.00 mL: After titrant has been added, the titration curve will be identical to that for the titration of the HCl alone an d p H will be determined b y the remainin g HCl in the solution. total added3 3 V ]OH[ of Moles]OH[ of Moles ]OH[ −+ + − = 3. V NaOH =50.00 mL: When the HCl has been neutralized (the first equivalence-point), the presence of the weak acid must be considered. At this point, the equilibrium described in eq. [2]: ]AH[ ]HA][OH[ K 2 3 a1 −+ = 100 K C KC]OH[ a HA aHA3 >= + 4. 50.00 mL<V NaOH <100.00mL HA is now reactin g with the titran t and the titration curve will be identical to that of the titration of a weak acid and HA NaA a - a - a C C logpK ]HA[ ][A logpK ]A[ ][HA logpKpH +=+=−= 5. V NaOH =100.00 mL At the point where both HCl and HA are neutralized (the second equivalence-point), the titration solution now contains A - which reacts with water A - + H 2 O <=> HA + OH - and the pH at this equivalence point is determined by concentration o f [A - ] and K b (1.0 x 10 -10 ) 100 K C KC]OH[ b b bb >= − 6. V NaOH >100.00 mL After all of the acid has been neutralized, further addition of titrant results in a mixture of a weak and stron g b ase. p H of the solution is determined b y the concentration of the stron g b ase. total 3added V ]OH[ of Moles]OH[ of Moles ]OH[ +− − − = [...]... more H+ An example of a polyfunctional acid is H3PO4, with the following equilibria: H3PO4 + H2O H3O+ + H2PO4− + [ H 3O ][ H 2 PO 4 ] K a1 = = 7.11x10 −3 [ H 3 PO 4 ] H2PO4- + H2O H3O+ + HPO422− [H3O+ ][HPO4 ] = 6.34x10−8 K a2 = − [H 2 PO4 ] HPO4-2 + H2O H3O+ + PO43+ 3− [H 3O ][PO 4 ] K a3 = = 4.5x10−13 2− [HPO4 ] Equilibria Constants for Overall Reactions H3PO4 + H2O H3O+ + H2PO4-... titration curve includes a titration curve of HCl (a strong acid) with NaOH and a titration curve of HA (weak acid) with NaOH You still can apply the theories and equations of neutralization titration in Chapter 9 into this complex system and (c) 29.00 mL CHCLVHCL =CKOHVKOH V1st Eq.KOH = 25.00x0.1200/0.1000= 30.00mL CHAVHA=CKOHVKOH V2nd Eq.= V1st Eq.KOH+25.00x0.0800/0.1000= 50.00mL ≈ CHCL = 0.1200 M ≈... PO43- Ka3 Kb PO43- + H2O HPO42- + OH- Kb1 HPO42- + H2O H2PO4- + OH- Kb2 H2PO4-+ H2O H3PO4 + OH- Kb3 Ka1 Kb3 = Ka2 Kb2 = Ka3 Kb1= [H3O+][OH-] = Kw Any Questions? C Buffer solutions Involving Polyprotic Acids: (A) Three buffer systems can be described when using a weak diprotic acid, H3A, and its salts, H2A-, HA2- and A3- - H3A/ H2A - 2- H2A /HA pH = pK a1 + log pH = pK a2 + log CH A2 C H 3A... lc u la tio n o f p H a n d s p e c ie s p re s e n t Example 11-2: pH = pK a1 + log CH A2 C H 3A D Calculation of the pH of Solutions of Amphiprotic Salts (NaHA) The pH is determined by the following equilibria: - - HA + H2O H2A + OH HA- + H2O A2- + H3O+ K b2 Kw = K a1 Ka2 At the first equivalence point, the solution will be either acidic or basic based upon the relative magnitude of the equilibrium . Chapter 11: Acid-Base Titrations Chapter 10: Polyprotic Acid-Base Equilibria 1. Treating Complex Acid-Base Systems Complex. Weak Bases B. Polyfunctional Acids and Bases: C. Buffer solutions Involving Polyprotic Acids: D. Calculation of the pH of Solutions of Amphiprotic Salts