1. 1 Indicators The indicator is usually a weaker chelate forming ligand. The indicator has a color when free in solution and has a clearly different color in the chelate. The following equilibrium describes the function of an indicator (H 3 In) in a Mg 2+ reaction with EDTA: MgIn - (Color 1) + Y 4-  MgY 2- + In 3- (Color 2)

2. 2 Problems Associated with Complexometric Indicators There could be some complications which may render some complexometric titrations useless or have great uncertainties. Some of these problems are discussed below: 1. Slow reaction rates In some EDTA titrations, the reaction is not fast enough to allow acceptable and successful determination of a metal ion. An example is the titration of Cr 3+ where direct titration is not possible. The best way to overcome this problem is to perform a back titration. However, we are faced with the problem of finding a suitable indicator that is weaker than the chelate but is not extremely weak to be displaced at the first drop of the titrant.

3. 3 2. Lack of a suitable indicator This is the most sever problem in EDTA titrations and one should be critical about this issue and pay attention to the best method which may be used to overcome this problem. First let us take a note of the fact that Mg 2+ -EDTA titration has excellent indicators that show very good change in color at the end point. Look at the following situations:

4. 4 a. A little of a known standard Mg 2+ is added to the metal ion of interest. Now the indicator will form a clear cut color with magnesium ions. Titration of the metal ion follows and after it is over, added EDTA will react with Mg-In chelate to release the free indicator, thus changing color. This procedure requires performing the same titration on a blank containing the same amount of Mg 2+.

5. 5 b. A blank experiment will not be necessary if we add a little of Mg-EDTA complex to the metal ion of interest. The metal ion will replace the Mg 2+ in the Mg-EDTA complex thus releasing Mg 2+ which immediately forms a good color with the indicator in solution. No need to do any corrections since the amount of EDTA in the added complex is exactly equal to the Mg 2+ in the complex.

6. 6 c. If it is not easy to get a Mg-EDTA complex, just add a little Mg 2+ to the EDTA titrant. Standardize the EDTA and start titration. At the very first point of EDTA added, some Mg 2+ is released forming a chelate with the indicator and thus giving a clear color. It is wise to consult the literature for suitable indicators of a specific titration. There are a lot of data and information on titrations of all metals you may think of. Therefore, use this wealth of information to conduct successful EDTA titrations.

7. 7 Example Find the concentrations of all species in solution at equilibrium resulting from mixing 50 mL of 0.200 M Ca 2+ with 50 mL of 0.100 M EDTA adjusted to pH 10.  4 at pH 10 is 0.35. k f = 5.0x10 10 Solution Ca 2+ + Y 4-  CaY 2- mmol Ca 2+ added = 0.200 x 50 = 10.0 mmol EDTA added = 0.100 x 50 = 5.00 mmol Ca 2+ excess = 10.0 – 5.00 = 5.00 [Ca 2+ ] excess = 5.00/100 = 0.050 M mmol CaY 2- formed = 5.00 [CaY 2- ] = 5.00/100 = 0.050 CaY 2-  Ca 2+ + Y 4-

8. 8 C T = [H 4 Y] + [H 3 Y - ] + [H 2 Y 2- ] + [HY 3- ] + [Y 4- ] K f = [CaY 2- ]/[Ca 2+ ]  4 C T [Ca 2+ ] = C T Using the same type of calculation we are used to perform, one can write the following:

9. 9 K f = [CaY 2- ]/[Ca 2+ ][Y 4- ] 5.0x10 10 = (0.05 – x)/((0.050 + x)*  4 x) assume that 0.05>>x x = 5.6x10 -11 Relative error will be very small value The assumption is valid [Ca 2+ ] = 0.050 + x = 0.050 M [CaY 2- ] = 0.050 – x = 0.050 M [Y 4- ] = 0.35 * 5.6x10 -11 = 1.9x10 -11 M

10. 10 Example Calculate the pCa of a solution at pH 10 after addition of 100 mL of 0.10 M Ca 2+ to 100 mL of 0.10 M EDTA.  4 at pH 10 is 0.35. k f = 5.0x10 10 Solution Ca 2+ + Y 4- = CaY 2- mmol Ca 2+ = 0.10 x 100 = 10 mmol EDTA = 0.10 x 100 = 10 mmol CaY 2- = 10 [CaY 2- ] = 10/200 = 0.05 M Therefore, Ca 2+ will be produced from partial dissociation of the complex

11. 11 CaY 2-  Ca 2+ + Y 4- C T = [H 4 Y] + [H 3 Y - ] + [H 2 Y 2- ] + [HY 3- ] + [Y 4- ] K f = [CaY 2- ]/[Ca 2+ ]  4 C T [Ca 2+ ] = C T 5.0x10 10 = 0.05/([Ca 2+ ] 2 x 0.35) [Ca 2+ ] = 1.7x10 -6 M pCa = 5.77 Using the same type of calculation we are used to perform, one can write the following:

12. 12 K f = [CaY 2- ]/[Ca 2+ ][Y 4- ] 5.0x10 10 = (0.05 – x)/(x*  4 x) assume that 0.05>>x x = 1.7x10 -6 Relative error = (1.7x10 -6 /0.05) x 100 = 3.4x10 -3 % [Ca 2+ ] = 1.7x10 -6 M pCa = 5.77

13. 13 Example Calculate the titer of a 0.100 M EDTA solution in terms of mg CaCO 3 (FW = 100.0) per mL EDTA Solution The EDTA concentration is 0.100 mmol/mL, therefore, the point here is to calculate the mg CaCO 3 reacting with 0.100 mmol EDTA. We know that EDTA reacts with metal ions in a 1:1 ratio. Therefore 0.100 mmol EDTA will react with 0.100 mmol CaCO 3. Mg CaCO 3 = 0.100 mmol x 100.0 mg/mmol = 10.0 Therefore, the titer og EDTA in terms of CaCO 3 is 10.0 mg CaCO 3 /mL EDTA

14. 14 Example An EDTA solution is standardized against high purity CaCO 3 by dissolving 0.3982 g of CaCO 3 in HCL and adjusting the pH to 10. The solution is then titrated with EDTA requiring 38.26 mL. Find the molarity of EDTA. Solution EDTA reacts with metal ions in a 1:1 ratio. Therefore, mmol CaCO 3 = mmol EDTA mg/FW = Molarity x V mL 398.2/100.0 = M x 38.26 M EDTA = 0.1041

15. 15 Example Find the concentration of Ca 2+ in a 20 mL of 0.20 M solution at pH 10 after addition of 100 mL of 0.10 M EDTA.  4 at pH 10 is 0.35. k f = 5x10 10 Solution Initial mmol Ca 2+ = 0.20 x 20 = 4.0 mmol EDTA added = 0.10 x 100 = 10 mmol EDTA excess = 10 – 4.0 = 6.0 C T = 6.0/120 = 0.050 M mmol CaY 2- = 4.0 [CaY 2- ] = 4.0/120 = 0.033 M Ca 2+ + Y 4-  CaY 2- k f = 5.0x10 10

16. 16 K f = [CaY 2- ]/[Ca 2+ ][Y 4- ] 5x10 10 = (0.033 – x)/(x*  4 (0.050 + x) ) assume that 0.033>>x x = 3.9x10 -11 The assumption is valid by inspection of the values and no need to calculate the relative error. [Ca 2+ ] = 3.9x10 -11 M pCa = 10.41

17. 17 Example Find pCa in a 100 mL solution of 0.10 M Ca 2+ at pH 10 after addition of 0, 25, 50, 100, 150, and 200 mL of 0.10 M EDTA.  4 at pH 10 is 0.35. k f = 5x10 10 Solution Again, we should remember that EDTA reactions with metal ions are 1:1 reactions. Therefore, we have: Ca 2+ + Y 4-  CaY 2- k f = 5.0x10 10 1. After addition of 0 mL EDTA [Ca 2+ ] = 0.10 pCa = 1.00

18. 18 2. After addition of 25 mL EDTA Initial mmol Ca 2+ = 0.10 x 100 = 10 mmol EDTA added = 0.10 x 25 = 2.5 mmol Ca 2+ left = 10 – 2.5 = 7.5 [Ca 2+ ] left = 7.5/125 = 0.06 M In fact, this calcium concentration is the major source of calcium in solution since the amount of calcium coming from dissociation of the chelate is very small. However, let us calculate the amount of calcium released from the chelate: mmol CaY 2- formed = 2.5 [CaY 2- ] = 2.5/125 = 0.02 M

19. 19 K f = [CaY 2- ]/[Ca 2+ ][Y 4- ] 5x10 10 = (0.02 – x)/((0.06 + x)*  4 x) assume that 0.02>>x x = 1.9x10 -11 The assumption is valid even without verification. [Ca 2+ ] = 0.06 + 1.9x10 -11 = 0.06 M pCa = 1.22

20. 20 3. After addition of 50 mL EDTA mmol EDTA added = 0.10 x 50 = 5.0 mmol Ca 2+ left = 10 – 5.0 = 5.0 [Ca 2+ ] left = 5.0/150 = 0.033 M We will see by similar calculation as in step above that the amount of Ca 2+ coming from dissociation of the chelate is exceedingly small as compared to amount left. However, for the sake of practice let us perform the calculation: Mmol CaY 2- formed = 5.0 [CaY 2- ] = 5.0/150 = 0.033 M

21. 21 K f = [CaY 2- ]/[Ca 2+ ][Y 4- ] 5x10 10 = (0.033 – x)/((0.033 + x)*  4 x) assume that 0.033>>x x = 5.7x10 -11 The assumption is valid even without verification. [Ca 2+ ] = 0.033+ 5.7x10 -11 = 0.033 M pCa = 1.48

22. 22 4. After addition of 100 mL EDTA mmol EDTA added = 0.10 x 100 = 10 mmol Ca 2+ left = 10 – 10 = ?? This is the equivalence point. The only source for Ca 2+ is the dissociation of the Chelate mmol CaY 2- formed = 10 [CaY 2- ] = 10/200 = 0.05 M Ca 2+ + Y 4-  CaY 2- k f = 5.0x10 10

23. 23 K f = [CaY 2- ]/[Ca 2+ ][Y 4- ] 5x10 5 = (0.05 – x)/(x*  4 x) assume that 0.05>>x, x = 1.7x10 -6 Relative error = (1.7x10 -6 /0.05) x 100 = 3.4x10 -3 % [Ca 2+ ] = 1.7x10 -6 M, pCa = 5.77 5. After addition of 150 mL EDTA mmol EDTA added = 0.10 x 150 = 15 mmol EDTA excess = 15 – 10 = 5.0 C T = 5.0/250 = 0.02 M Mmol CaY 2- = 10 [CaY 2- ] = 10/250 = 0.04 M Ca 2+ + Y 4-  CaY 2- k f = 5.0x10 10

24. 24 K f = [CaY 2- ]/[Ca 2+ ][Y 4- ] 5x10 10 = (0.04 – x)/(x*  4 (0.02 + x) ) assume that 0.02>>x x = 1.1x10 -10 The assumption is valid [Ca 2+ ] = 1.1x10 -10 M pCa = 9.95

25. 25 6. After addition of 200 mL EDTA mmol EDTA added = 0.10 x 200 = 20 mmol EDTA excess = 20 – 10 = 10 C T = 10/300 = 0.033 M mmol CaY 2- = 10 [CaY 2- ] = 10/300 = 0.033 M Ca 2+ + Y 4-  CaY 2- k f = 5.0x10 10

26. 26 K f = [CaY 2- ]/[Ca 2+ ][Y 4- ] 5x10 10 = (0.033 – x)/(x*  4 (0.033 + x) ) assume that 0.033>>x x = 5.7x10 -11 The assumption is undoubtedly valid [Ca 2+ ] = 5.7 x10 -11 M pCa = 10.24