It is still possible to develop analytical                                                        methods for species that...
KA1 / KA2   = 7.5x10-3 / 6.0x10-8                  = 1.25 x 105      KA2 / KA3   = 6.0x10-8 / 4.8x10-13                  =...
[OH- ]2              0.1000 - [OH-]                                              mlHCl    % titration1st eq.pt.    pH     ...
12                                             10                                             8                           ...
12                                                        10                                                        8     ...
Neither the phenolphthalein or methyl orange endpoint detection is stellar.An alternate approach is to use methyl red  Thi...
Let’s take a look at the following two acids:                               Sulfurous               Tartaric              ...
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09 complex acid-base_titrations

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09 complex acid-base_titrations

  1. 1. It is still possible to develop analytical methods for species that contain multiple acid or base groups. This unit will review the types of calculations and titration curves involved. Some specific examples will also be included. KA1 KA2 [H3O+] [HA-] [H2A] [H3O+] [A2-] [HA-] KIt can be shown that if KA1 > 103, then KA1 KA2 KA3 A2 H3PO4 H2PO4- HPO42- PO43- the effects of the second dissociation is negligible during the titration of H2A. [H3O+][H2PO4-] The same test can be used for other weak [H3PO4] acids with more than two titratible acid groups [H3O+][HPO42-] and bases. [H2PO4-] You must test K values in sequence. To [H3O+][PO43-] compare KA1 to KA3 has no meaning. [HPO42-]
  2. 2. KA1 / KA2 = 7.5x10-3 / 6.0x10-8 = 1.25 x 105 KA2 / KA3 = 6.0x10-8 / 4.8x10-13 = 1.25 x 105 This indicates that it is possible to treat each step separately during a titration.Just because you have a material where KA1/KA2 We’ll now review the calculations involved for is not greater than 1000 does not mean it the titration of a ‘well behaved’ diprotic acid. can’t be titrated. It is still possible to determine equilibrium This really is nothing more that doing two concentrations at any point during a titration. separate sets of calculations. However, when KA1/KA2 < 1000, you don’t get a sharp ‘first’ endpoint. You can Triprotic acids (or higher) simply amount to repeating the process as many times as Titrate to the ‘final endpoint’ necessary. Titrate to a form where Kx/KY > 1000.Construct a curve for the titration of 25.00 ml of 0.1000 M sodium carbonate (Na2CO3) with 0.1000 M HCl. [ OH- ] [ CO32- ] [ OH- ] [ CO32- ] KB1 = 2.00 x 10-4 = [HCO3- ] KB1 = 2.00 x 10-4 = [HCO3- ] - - KB2 = 2.51 x 10-8 = [ OH ] [HCO3 ] [ H2CO3 ] [ OH- ] = [ CO32- ] KB1 / KB2 = 8.00 x 103 so our base is ‘well behaved’ and we can treat [ HCO3- ] = 0.1000M - [ CO32- ] each equilibrium separately.
  3. 3. [OH- ]2 0.1000 - [OH-] mlHCl % titration1st eq.pt. pH 0.0 0 11.65 2.5 10 11.25 5.0 20 10.90 7.5 30 10.67 10.0 40 10.48 12.5 50 10.30 [CO32] 15.0 60 10.12 [HCO3-] 17.5 70 9.93 20.0 80 9.70 22.5 90 9.35 12 10 8pH 6 4 2 0 0 5 10 15 20 25 pKA1 + pKA2 pH = ml HCl 2
  4. 4. 12 10 8 pH 6 4 2 0 0 5 10 15 20 25 ml HCl mlHCl % titration1st eq.pt. pH 25.0 0 8.35 27.5 10 7.35 30.0 20 7.00 32.5 30 6.77 35.0 40 6.58 37.5 50 6.40 40.0 60 6.22 42.5 70 6.03 45.0 80 5.80 47.5 90 5.45 12 10 8 [H3O+] [HCO3-]pH 6 [H2CO3] 4 2 0 0 10 20 30 40 50 ml HCl
  5. 5. 12 10 8 pH 6 4 [H3O+ ]2 2 KA1 = 0.03333 0 0 10 20 30 40 50 pH = 3.94 ml HCl 12 10 8 pH 6 4 2 0 0 20 40 60 80 ml HClHaving used carbonate as an example, we should point out some of the specifics of bicarbonate titrations. Carbonates and bicarbonate/carbonate mixtures are commonly measured. They are relatively common in nature. The recommended titrant is HCl. Indicator used is based on the type of sample.
  6. 6. Neither the phenolphthalein or methyl orange endpoint detection is stellar.An alternate approach is to use methyl red This indicator can only be used to detect the second endpoint.The transition is from yellow to red and is still gradual unless some additional steps are taken. carbonate only mixed system bicarbonate onlyWhat happens when the K values do not differ A few points are relatively easy to estimate. by 1000 or more? At 50% conversion of any specific form, pH will be relatively close to the pKA value. Can you still titrate them? The equivalence point pH of intermediate What would the curves look like? forms can be estimated based on the average of the pKA values.First, lets outline how to predict what a titration You should also be able to calculate the pH curve would look. when over titrating the sample.
  7. 7. Let’s take a look at the following two acids: Sulfurous Tartaric 1st buffer region 1.9 3.0 1st eq. point 4.6 3.7 pKA1 pKA2 2nd buffer region 7.2 4.4 Sulfurous 1.9 7.2 Tartaric 3.0 4.4 The basic shape of a titration curve should still be more or less the same. We’ll assume that we have 100 ml of a 0.10 M for each and we are using 0.10 M NaOH as our A change of about 2 pH units from 10 to 90% titrant. titration - assuming no overlap of the curves. 13 11 You can see that 9 sulfurous we still have a usable 2nd equivalence point 7 in both cases. 5 For tartaric acid, it 3 would be difficult to tartaric detect the 1st eq. point. 1 0 100 200 300So when working with diprotic acids (or bases) where K2 is < 1000 K1, you can only work with the final equivalence point. Usable results can still be obtained but you can’t work with mixed systems. The same is true if you have more than 2 titratible groups. However, you can obtain a good endpoint when any adjacent K values differ by 1000.

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