Chapter 15. Titration Curves for Complex Acid/Base Systems

Transcription

1 Chapter 15 Titration Curves for Complex Acid/Base Systems Polyfunctional acids and bases Carbonic acid/bicarbonate buffer system Buffers for human blood ph = CO 2(g) + H 2 O H 2 CO 3(aq) H 2 CO 3 + H 2 O H 3 O + + HCO 3 - HCO 3- + H 2 O H 3 O + + CO 3 2-

2 Complex Acid/Base Systems Complex systems may be described as solutions made up of two acids or two bases of different strengths, an acid or a base that has two or more acidic or basic functional groups, oran amphiprotic substance, acting as both an acid and a base. HCl + CH 3 COOH or H 2 A HA - A - It is possible to determine each of the components in a mixture containing a strong acid and a weak acid (or a strong base and a weak base) provided that the concentrations of the two are of the same order of magnitude and that the dissociation constant for the weak acid or base is somewhat less than about 10 4.

3 Titration Curves for Complex Acid/Base Systems Two acids or two bases of different strengths An acid or a base that has two or more acidic or basic functional groups An amphiprotic substance (act both as an acid and a base)

4 [H 3 O + ] in Solutions of Weak Acids a [H 3 O [HA] 2 ] C HA [H 3 O 2 ] [H 3 O ] Because the acid is a weak acid, [H 3 O + ] <<C HA a [H O ] 2 3 [H 3 O ] a C HA C HA If [H 3 O + ] is not <<C HA, error introduced from above equation will become significant. C HA / a =10 1 Error % = 17 % C HA / a =10 2 Error % = 5.3 % C HA / a =10 3 Error % = 1.6 % C HA / a =10 4 Error % = 0.5 % C HA / a =10 5 Error % = 0 %

5 25 ml Assume [A ] << C HCl And [H 3 O + ] = C HCl Early stages of titration identical to titrating strong acid by strong base

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8 After the first end point, identical to the titration of weak acid by strong base

9 A Mixtures of Strong/Weak Acids or Strong/Weak Bases It is possible to determine each of the components in a mixture of strong acid and weak provided that (i) [A] = [a] (ii) A / a > 10 4 The composition of a mixture of a strong acid and a weak acid can be determined by titration with suitable indicators if the weak acid has a dissociation constant that lies between 10 4 and 10 8 and the concentrations of the two acids are of the same order of magnitude.

10 Polyfunctional Acids and Bases With a polyfunctional acid such as phosphoric acid (H 3 PO 4 ), the protonated species (H 3 PO 4,H 2 PO 4,HPO 4 2 )differenoughintheirdissociationconstantsthat they exhibit multiple end points in a neutralization titration. H 2 PO 4 +H 2 O H 2 PO 4 +H 3 O + H 2 PO 4 +H 2 O HPO 4 2 +H 3 O + HPO 4 2 +H 2 O PO 4 3 +H 3 O + a1 > a2 often by a factor of 10 4 to 10 5 because of electrostatic forces and locations. H 3 PO 4 +2H 2 O HPO H 3 O + H 3 PO 4 +3H 2 O PO H 3 O + a1 a HPO4 [ HO ] [ ] [ HPO] PO x10 [ HO ] [ ] a1a2a3 2.0x10 [ HPO] 10 22

11 The Carbon Dioxide/Carbonic Acid System When carbon dioxide is dissolved in water, a dibasic acid system is formed. CO 2(aq) +H 2 O H 2 CO 3 hyd [ HCO 2 3] [ CO ] 2( aq) 2.8x10 3 H 2 CO 3 +H 2 O H 3 O + +HCO HCO3 [ HO ][ ] [ HCO] x10 4 HCO 3 +H 2 O H 3 O + +CO CO3 [ HCO3 ] [ HO ][ ] 4.69x10 11 Combining the first two, CO 2(aq) +2H 2 O H 3 O + +HCO 3 a1 3 HCO3 [ HO ][ ] [ CO ] 2( aq) 4.2x10 7

12 Triprotic Acid Base Systems Triprotic acid H 3 A + H 2 O H 2 A + H 3 O + H 2 A + H 2 O HA 2 + H 3 O + HA 2 + H 2 O A 3 + H 3 O + Triprotic base A 3 + H 2 O HA 2 + OH HA 2 + H 2 O H 2 A + OH H 2 A + H 2 O H 3 A + OH a1 a2 a3 b1 b2 b3 Conjugate triprotic acid base systems; H 3 A=H 3 PO 4, A 3 =PO 3 4 a1 x b3 = w a2 x b2 = w a3 x b1 = w

13 Calculation of the ph of Buffer Solutions HA H A a b [ H H 2 2 O 3 [ HA] O O H ][ A HO O [ HO ][ HA] [ A ] 3 ] HA W A a (1) (2) Determinant factors: a / b and [HA]/[A - ] (1) lies farther to the right than (2) Acidic (2) lies farther to the right than (1) Basic

14 a1 3 HCO3 [ HO ][ ] [ CO ] 2( aq) 4.2x10 7

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17 Calculation of the ph of Solutions of NaHA

18 Calculation of the ph of Solutions of NaHA

19 Na 2 HA

20 NaH 2 A NaHA

21 Titration Curves for Polyfunctional Acids Compounds with two or more acidic functional groups yield multiple end points in a titration if the functional groups differ sufficiently in strength as acids. Titration of ml of M H 2 Awith M NaOH. a1 =1.00 x 10 3 and a2 =1.00 x 10 7 If a1 / a2 >10 3,the theoretical titration curves can be calculated. a1 / a2 >>10 3 less titration error a1 / a2 smaller, bigger titration error

22 For Diprotic Acid (H 2 A) Titrating of 25.0 ml of M maleic acid (HOOC CH=CH COOH) with M NaOH. For maleic acid, a1 =1.3x10 2 and a2 =5.9x10 7. Step 1: Calculating titrant volume at first equivalence point (V eq1 ) and second equivalence point (V eq2 ) Step 2: Calculating ph at various titrant volume (V b ) Region A: At beginning (V b = 0) Region B: At first buffer region (0 < V b < V eq1 ) Region C: At first equivalence point (V b = V eq1 ) Region D: At second buffer region (V eq1 < V b < V eq2 ) Region E: At second equivalence point (V b = V eq2 ) Region F: After second equivalence point (V b > V eq2 )

23 Equivalence point V e1 x ( M) = (25.0 ml) x ( M) V e1 = 25.0 ml V e2 = 2 x V e1 = 50.0 ml Region A: At Beginning (V b = 0) H 2 A + H 2 O HA + H 3 O + C H2A x x x C H x 2 2 A x C H [H 2 A 3 O [H ] 3 2 O ] [H 3 O [H ] 3 2 O ] a1 = [H 3 O + ] x10 2 [H 3 O + ] 1.3x10 3 = 0 [H 3 O + ] = 3.01x10 2, ph = 1.52

24 Region B: First buffer region (0 < V b < V eq1 ) H 2 A + H 2 O HA + H 3 O + Example: For addition 5.00 ml of M NaOH: C C HA H 2 A 5.00 x x10 M x x [H3O ](1.67x10 +[H3O ]) -2 + a1 1.3x10 (6.67x10 -[H O ]) 3 [H O + ](C 6.67x10 [H 3 O + ] 2 + (2.97x10 2 ) [H 3 O + ] 8.67x10 4 = 0 [H 3 O + ] = 1.81x10 2, ph = [H O + + [H O ][HA ] 3 3 HA 3 a1 + [H2A] (CH A [H3O 2-2 ]) + ]) 2 M

25 Region C: At first equivalence point (V b = V eq1 ) Example: For addition ml of M NaOH: C HA 25.00x a2 5.00x10 Using the equation of NaHA dissolved in water: [H O ] C HA HA 2 a1 w 1+(C / ) M x10 x5.00x x (5.00x10 /1.3x10 ) 7.80x10 5 ph = 4.11

26 Region D: At second buffer region (V eq1 < V b < V eq2 ) HA + H 2 O A 2 + H 3 O + Example: For addition ml of M NaOH: C HA C 2 A a2 [H 3 (25.00 x ) ( )(0.1000) ( )x O C + HA ]C A + [H3O ][A [HA ] 2 2 [H ] [H [H 3 O + ] = 2.89x10 5, ph = O (C + ](C HA A 2 [H O ](0.050/50.50) (2.45/50.50) +[H 3 O 3 + O ]) + ]) 5.9x Assume: C HA >> [H 3 O + ] and C A 2 >> [H 3 O + ]

27 Region E: At second equivalence point (V b = V eq2 ) Example: For addition ml of M NaOH: Assuming all becoming A 2 firstly, the calculating by: A 2 + H 2 O HA + OH C A 2 x x x C 2 A ( ) x b1 w a2 1.00x10 5.9x x10 8 x C A [OH ] [OH ] = 2.38x10 5, poh = 4.62, ph = 9.38

28 Point F: After second equivalence point (V b > V eq2 ) Example: For addition ml of M NaOH: Calculating net excess moles of OH, the [OH ], then ph: ( )x0.100 [OH ] poh = 2.88, ph = x V b (NaOH), ml ph ph Volume of M NaOH, ml

29 Present Equation F = 0 (V b =0) H 2 A [H ] C a1 H 2A 0 < F < 1 (0<V b <V eq1 ) H 2 A/HA ph p a1 C log C HA H 2 A F = 1 (1st eq.; V b= V eq1 ) HA [H ] a1 a2 1 < F < 2 (V eq1 <V b <V eq2 ) HA /A 2 ph p a2 C log C A 2 HA F = 2 (2nd eq.; V b= V eq2 ) A 2 [OH ] w a2 C A 2 F > 2 (V b >V eq2 ) OH /A 2 [OH ] [excess titrant]

30 Titration Curves for the Polyprotic Acids Curve A= triprotic phosphoric acid a1 / a2 =10 5 a2 / a3 =10 5 Curve B = diprotic oxalic acid a1 / a2 =10 3 Curve C = diprotic sulfuric acid a2 =

31 For Triprotic Acid (H 3 A) Step 1: Calculating titrant volume at first equivalence point (V eq1 ), second equivalence point (V eq2 ), and third equivalence point (V eq3 ) Step 2: Calculating ph at various titrant volume (V b ) Region A: At beginning (V b = 0): Region B: At first buffer region (0 < V b < V eq1 ) Region C: At first equivalence point (V b = V eq1 ) Region D: At second buffer region (V eq1 < V b < V eq2 ) Region E: At second equivalence point (V b = V eq2 ) Region F: At third buffer region (V eq2 < V b < V eq3 ) Region G: At third equivalence point (V b = V eq3 ) C D E F G H Region H: After third equivalence point (V b > V eq3 ) A B

32 Titration Curves for Polyfunctional Bases Titration curve for the titration of ml of M Na 2 CO 3 with M HCl. Two end points appear in the titration. The important equilibrium constants are CO 2 3 +H 2 O OH +HCO 3 14 w 1.00x10 b1 11 a2 4.69x x10 4 HCO 3 +H 2 O OH +CO 2(aq) 14 w 1.00x10 b2 7 a1 4.2x10 2.4x10 8

33 Titration Curves for Amphiprotic Species An amphiprotic substance when dissolved in a suitable solvent behaves both as a weak acid and as a weak base. If either of its acidic or basic characters predominates, titration of the substance with a strong base or a strong acid may be feasible.

34 Composition of Polyprotic Acid Solutions Monoprotic acid (acetic acid) HM + H 2 O M + H 3 O + + [HM] [HM] [H ] 0= HM= = = - + C [HM]+[M ] [H ]+ [ M] [M ] a 1= = = = M - + C [HM]+[M ] [H ] + - * C [HM] +[M ] T 0 1 T T 1 a a H n A

35 Composition of Polyprotic Acid Solutions Diprotic acid (carbonic acid) H 2 M + H 2 O HM + H 3 O + HM + H 2 O M 2 + H 3 O + 2 [H2M] [H ] 0= H2M= = 2 C [H ] [H ] [HM ] a1[h ] 1= = = HM 2 C [H ] [H ] 2 [M ] a1a2 2= 2-= = M 2 C [H ] [H ] 2 * C T [H2M] [HM ]+[M ] * T a1 a1 a2 T a1 a1 a2 H M a1 a1 a2

36 Triprotic acid (phosphoric acid) 3 [H3M] [H ] 0= = 3 2 C [H ] [H ] [H ]+ T a1 a1 a2 a1 a2 a3-2 [H2M ] a1[h ] 1= = 3 2 C [H ] [H ] [H ]+ T a1 a1 a2 a1 a2 a3 2 [HM ] a1 a2[h ] 2= = 3 2 C [H ] [H ] [H ]+ T a1 a1 a2 a1 a2 a3 3 [M ] 3= = 3 2 C [H ] [H ] a1 a2 a3 T a1 a1 a2[h ]+a1a2a3 2 3 * C HA [H3M] [H2M ]+[HM ]+[M ] 3 *