15 Titration Curves for Complex Acid/Base Systems

Transcription

1 15 Titration Curves for Complex Acid/Base Systems (1) To acids or to bases of different strength. () An acid or a base that has to or more acids or bases of different strengths () An amphiprotic substance 15A Mixtures of Strong and Weak Acids or Strong and eak Bases Ex. 151, Calculate the ph of a mixture (5 ml) that is M in HCl and M in HA ( a 1.00 ) during its titration ith 0.00 M H at (a) 0.00 ml (b) 5.00 ml and (c) 9.00 ml. (a) 0.00 ml H added c HCl [A [A M, ph 0.9 (assume[a << 0.100M) Check this assumption: [ A a 1.00 [A 8., A A c HA A [A M A 8. [ (b) After adding 5.00 ml of base [A (1.0 )[A M, [A M c HCl 0.08 M [A 0.08 M, ph 1.08 Check this assumption: [A 1.00 A M, [A A, c HA A [A [A A [A (8. )[A M [A M ( << 0.08 M ) (c) After adding 9.00 ml of base c HCl M 9

2 cha.70 M c HCl [A 1.85 [A, c HA A [A.70 M [A H [A A, [A [ [ A, H (1.00 ) (1.85 ) (1.75 ) M, ph.5 [ ( from HA) 1.85 ( from HCl) Fig 151 Curves for the titration of strong acid/eak acid mixtures ith 0.00 M NaH. Each titration is on 5.00 ml of a solution that is M in HCl and M in HA. 15B Polyfunctional Acids and Bases 15B1 The phosphoric acid system H P H H P H P 7.11 P H P H HP H P 8 a 6. P HP H P H [P 1 a.5 P > a > a 9

3 H P H P H a a [P P B The carbon dioxide carbonic acid system C (aq) H H C hyd.8 [C (aq) H C H H C HC C HC H H [C 11 C.69 C C (aq) H H HC C 9 C [C (aq) HC H H [C 11 C.69 C Ex. 15 Calculate the ph of a solution that is M C. c C [C ( aq) C C [C [ C ( aq ) >> C C [C, [C ( aq) cc From chargebalance C [C [H, [C [H << C C. [C (aq) 7, ph log(1.0 ) C Buffer Solutions Involving Polyprotic Acids M 1.0 Ex. 15 Calculate the for a buffer solution that is.00 M in H P and 1.50M in H P. H P H H P H P 7.11 P M

4 H P H HP H P 8 a 6. P Assume: P and [P << P and P [ H P c H.00 M, P P ch P 1.50 M Check the assumption P 9.8 P 8 6. P 1.50 P M and [P < P M the assumption is valid *For a buffer prepared from NaHA and Na A HA H H A H is disregarded A << A and [A Ex. 155 Calculate the for a buffer solution that is M in potassium hydrogen phthalate (HP) and0.150m in potassium phthalate ( P). HP H H P [P 6 a.91 P Assume: P is negligible [ HP c HP M, [P c P Check the assumption (1.0 )(0.0500) 61.1 P M M P 6 5 M and P << P and [P, the assumption is valid 15D Calculation of the ph of Solution of NaHA HA H A H [A a A HA H H A H A[H b A For mass balance: c NaHA A A [A

5 For charge balance: [Na [H A [A c NaHA A A [A [H A [A [A [H A A A, aa [A aa A A ( ) A aa 1 aa a A A 1 ac NaHA C NaHA 1 if CNaHA/ >> 1 and acnaha >> a 1a Ex. 156 Calculate the of a 1.00 M Na HP solution. a 6. 8 and a.5 1 CNa HP / a (1.00 )/(6. 8 ) >> 1 and a CNa HP < 1 1 acna HP CNa HP 1 a 8 6. Ex. 157 Find the of a 0.00 M NaH P solution and a 6. 8 CNaH P / 0.00/7.11 and acnah P >> 8 acnah P CNaH P

6 Ex. 158 Calculate the of a 0.0 M NaHC solution. C H H HC C [C (aq) HC H H [C 11 C.69 C CNaHA/ >> 1 and acnaha >> M 15E Titration Curves for Polyfunctional Acids Fig. 15 Titration of 0.00 ml of 0.00 M H A ith 0.00 M NaH. For H A, 1.00 and a The method of ph calculation is shon for several points and regions on the titration curve. Volume of 0.0 M NaH, ml Ex Construct a curve for the titration of 5.00 ml of 0.00 M maleic acid, HCCHCHCH, ith 0.00 M NaH. H M H H HM 1. HM H H M a / a is large ( ) Initial ph M, M M 0.00 M 0.00 M a , ph log

7 First Buffer Region: addition of 5.00 ml of base C NaHM M / M CH M M ( ) 0.00/ M << CH M or C HM is not valid M 1.67 [H M 6.67 [H a [ H (1.67 ) ph log Just Prior to First Equivalence Point C NaHM M / M CH M M ( ) 0.00/ M Mass balance: CH M C NaHM M M [M Charge balance : [Na M [M [H C NaHM M [M CH M [M M chm am M M 1 ch M M 0 a M.99 M, CH M.00 M 1. a ph.99 at adding.99 ml of titrant, ph. First Equivalence Point M C NaHM / M 7 ac NaHM C / 1 (5.00 ) /(1. ) NaHM a ph log ( ).11 Just After the First Equivalence Point C HM [5.00( ) 0.00/ M C M ( ) 0.00/ M 98

8 Mass balance: C HM CM MM [M M Charge balance : [Na M [M [H [Na / M [M M ( ) am M ± ( ).88 (.98 ) M.88 ph.1 Second Buffer Region : addition of 5.50 ml of NaH [M CNa M ( ) 0.00/ /50.50 M M C NaHM [5.00( ) 0.00/ /50.50 M (.5/50.50) /(0.050/50.50).89 5 M << CNa M and C NaHM is valid and ph log Just Prior to Second Equivalence Point at adding 9.90 and 9.99 ml of titrant, M >> HM at adding 9.90 ml C HM 1.5 and C M 0. M H HM H [H M [H ( b1 7 a [M (0.0 [H ) [H ) 1.00 [H (1.5 b1 ) [H 0.0 b1 0 [H. 6 M ph 5.9 ph at adding 9.90 ml: [H M ph 9.6 Second Equivalence Point: addition of ml of NaH [Na M 0.0 M M H H HM a [H M 1 b 1.69 [M [H M, [M 0.0 [H 0.0 [H / , [H.8 5 ph 1.00 ( log.8 5 ) 9.8 8

9 ph Just Beyond the Second Equivalence Point: add ml of NaH c M / M excess [H / M [M c M M 0.0 M, [H 1. 5 M b1 5 [H M M (1. M ) [M 0.0M M (1. 5 b1 ) M 0.0 b1 0 M M [H M ph.50 and ph 1 ph 9.50 add 50. ml of NaH ph.1 ph Beyond the Second Equivalence Point: add ml of NaH [H / ph 1.00 ( log 1. ) 11.1 Volume of 0.00 M NaH, ml Fig. 15 Titration curve for 5.00 ml of 0.00 M maleic acid, H M, ith 0.00 M NaH. Volume of 0.00 M NaH, ml Fig. 15 Curve for the titration of 0.00 M H P (A), 0.00M oxalic acid (B), and 0.00 M H S (C). 15F Titration Curves for Polyfunctional Bases Volume of 0.00 M HCl, ml Fig. 155 Curve for the titration of 5.00 ml of 0.00M Na C ith 0.00 M HCl. C H HC H b a.69 HC H H C H b

10 15G Titration Curves for Amphiprotic Species In NaH P solution: can be titrate ith a standard base solution H P H HP H 8 6. H P H H H P a b 1.1 In Na HP solution: can be titrate ith a standard acid solution HP H P H 1.5 a 1 HP H H H P b a 6. 15H The Composition of Solutions of a Polyprotic Acid as a Function of ph α 0 M C T α 1 T M C α [M C CT M M [M α0 α1 α 1 T α 0 α a α1 a 1 a a 1 a a Fig 156 Composition of HM solution as a function of ph. Fig. 157 Fig. 157 Titration of 5.00 ml of 0.00 M maleic acid ith 0.00 M NaH. The solid curves are plots of alpha values as a function of volume. The broken curve is a plot of ph as a function of volume. 1

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