1. Compleximetric Problems
Tutorial 8
Compleximetric Problems

2. Objectives: Quick overview on type of EDTA titrations.
Solving EDTA problems.

3. Types of EDTA titrations A. Direct titration:
Many metal ions can be titrated using EDTA taking into account choosing a suitable indicator and pH medium.
This type has a very important application in the determination of: hardness of water (hard water can’t form foam with soap)

4. B. Back titration:
1.Back titrations are useful for the determination of cations that form stable EDTA complexes and for which no suitable indicator is available as in case of determining Thallium. 2. The method is also useful for cations as Cr(III) and Co(III) that react only slowly with EDTA. A measured excess is added and after the reaction , the excess EDTA is titrated against standard zinc or magnesium. 3. It can also be used when the analyte forms precipitates at the required pH of its titration.

5. C. Displacement titration:
In displacement titrations ,an unmeasured excess of a solution containing the magnesium or zinc complex of EDTA is introduced into the analyte solution.
The analyte which forms more stable complex than that of magnesium or zinc will undergo the following displacement reaction:
MgY2- + M2+ --- MY2- + Mg2+
Where M2+ can be Hg(II), Pd(II), Ti(II), Mn(II) and V(II).
The liberated Mg2+ or Zn 2+ can be then titrated with standard EDTA solution.
Displacement titrations are used on lacking sharp end point or when no suitable indicator is available.

6. Remember that Number of moles = MV = weight molec.weight
(MV)edta =(MV)metal ion
Wt (g)= M x V(ml)x Mwt = No. of moles x Mwt
1000
Strength (g/L)= M x Mwt
Water Hardness (ppm) = Strength x 1000

7. Questions
1- A 0.5 g sample containing Ca and Mg carbonates was dissolved in dil HCl and completed with distilled water to 250 ml. 10 ml of the resulting solution were titrated with 0.01M EDTA solution. Using EBT indicator, 19 ml of EDTA were consumed, while on using murexide indicator, 8 ml of EDTA were consumed. Calculate the percentage of both Ca and Mg carbonates in the sample.
Using EBT, Volume of EDTA is equivalent to Ca2+ + Mg2+ = 19 ml
Using Murexide, V EDTA is equivalent to Ca2+ only = 8ml
Volume of EDTA equivalent to Mg2+ = 19-8 = 11 ml
(MV)EDTA = (MV)Ca2+ , 0.01x8 = Mx10, MCa2+ = 0.008mol/l
(MV)EDTA = (MV)Mg2+, 0.01x11=Mx10, MMg2+ = mol/l
wCaCO3 = 0.008x100x250/1000= 0.2 g
% CaCO3 =(0.2/0.5)x100 = 40%
wMgCO3 = 0.011x84x250/1000= 0.231g,
% MgCO3=(0.231/0.5)x100=46.2%

8. (MV)EDTA = (MV)Mg2+ = (MV)Mn2+
2- To 10 ml of Mn(II) solution, 25 ml of 0.1 M Mg-EDTA is added. The liberated Mg2+ ions were titrated with 14.4 ml of M EDTA solution. Calculate the molarity of the Mn(II) solution.
(MV)EDTA = (MV)Mg2+ = (MV)Mn2+
(14.4 x 0.027) = M x 10
MMn2+ = mol /l

9. 3- An EDTA solution was prepared by dissolving approximately 4g of disodium salt in approximately 1 L of water. An average of ml of this solution was required to titrate 50 ml aliquots of a standard that contained g of MgCO3 per liter. Titration of 25 ml sample of mineral water at pH required 18.8 ml of the EDTA solution. A 50 ml aliquot of the mineral water was rendered strongly alkaline to precipitate the magnesium as Mg(OH) Titration with a calcium specific indicator required ml of the EDTA solution. Calculate: a- The Molarity of the EDTA solution. b- The ppm of CaCO3 in the mineral water. c- The ppm of MgCO3 in the mineral water.

10. We can only add or subtract number of moles.
Answer
Molarity of MgCO3=Strength/molecular mass
=0.7682/84
=9.14×10-3 M
(MV)EDTA= (MV) Mg
M×42.35=9.14×10-3×50
MEDTA=0.01 M
At pH 10:
(MV)Ca+ (MV) Mg= (MV) EDTA
MCa×25+MMg×25=0.01× (1)
At strongly alkaline medium:
(MV)Ca= (MV) EDTA
MCa×50=0.01×31.54
MCa=6.81×10-3 M (2)
Substitute from (2) in (1)
MMg=1.31×10-3 M
We can never add or subtract concentrations, and we can never calculate total concentration of a mixture.
We can only add or subtract number of moles.

11. For CaCO3:
Strength×1000=Molarity × molecular weight×1000
=6.81×10-3×100×1000
=681 ppm
For MgCO3:
Strength×1000=Molarity×molecular weight×1000
=1.31×10-3×84×1000
= ppm

12. 4- The Tl in a 9.76 g sample of rodenticide was oxidized to the trivalent state and treated with an unmeasured excess of Mg/EDTA solution. The reaction is Tl3+ + MgY TlY- + Mg2+ Titration of the liberated required ml of M EDTA. Calculate the % of Tl2SO4 (504.8 g/mol) in the sample.

13. Answer (MV)Mg= (MV) EDTA= (MV) Tl=No. of Tl moles
No. of Tl mmoles=0.0356×13.34= mmoles
No. of Tl2SO4 moles=No. of Tl moles/2= /2 X 1000
= x 10-3 moles
Mass of Tl2SO4=No. of moles × mass of one mole
= ×10-3×504.8
=0.119 grams
Percentage of Tl2SO4= (0.119/9.76)×100=1.228%
No of Moles equal ½ that of Tl ions as each mole of molecule has 2 atoms of Tl

14. 5- A g sample of brass (containing lead, zinc, copper and tin) was dissolved in nitric acid. The sparingly soluble SnO2.4H2O was removed by filtration and the rest of the solution was then diluted to ml. A 10 ml aliquot was suitably buffered, titration of the lead , zinc and copper in this aliquot required ml of M EDTA. The copper in a 25 ml aliquot was masked with thiosulphate; the lead and zinc were then titrated with ml of The EDTA solution. Cyanide ion was used to mask Copper and Zinc in a 100 ml Aliquot; ml of EDTA solution was needed to titrate Lead ion. Determine the composition of the brass sample and evaluate percentage of tin by difference.

15. Strategy of answer:
The main idea of the problem is that, we have aliquots of samples of different volumes, 10, 25, 100 ml, and we have to put this into consideration, we can’t subtract volumes used by EDTA, unless the samples we are working on have the same volumes.
Also, The percentage of the last element will be obtained by subtraction and this is not done in all cases EXCEPT it is CLEARLY REQUESTED or the sample is indicated TO BE PURE.

16. In the first titration: (1)
(MV)Pb+ (MV) Zn+ (MV) Cu= (MV) EDTA
MPb×10+MZn×10+MCu×10=0.0025×37.56
In the second titration: (2)
(MV)Pb+ (MV) Zn= (MV) EDTA
MPb×25+MZn×25=0.0025×27.67
In the third titration:
(MV)Pb= (MV) EDTA
MPb×100=0.0025×10.8
MPb=2.7×10-4 M (3)
Substitute from (3) in (2)
MZn=2.497×10-3 M
Substitute from (2) in (1)
MCu=6.623×10-3 M

17. Weight% of Lead=
[(2.7×10-4×0.5×207.2)/0.3284]×100=8.52%
Weight% of Zinc=
[(2.497×10-3×0.5×65.39)/0.3284]×100=24.86%
Weight% of copper=
[(6.623×10-3×0.5×63.546)/0.3284]×100=64.08%
Weight% of tin=100-( ) =2.54%
( WE CANT DO THIS ALL THE TIME UNLESS IT IS STATED THAT THE SAMPLE IS 100% PURE)

18. Extra problems

19. 1-Calamine, used for the relief of skin irritation, is a mixture of zinc and iron oxides. A g sample of dried calamine was dissolved in acid and diluted to ml. Potassium fluoride was added to a 10.0 ml aliquot of the diluted solution to mask the iron; after suitable adjustment of the pH, Zn2+ consumed ml of M EDTA. A second 50.0 ml aliquot was suitably buffered and titrated with 2.40 ml of M ZnY2- solution : Fe3+ + ZnY FeY- + Zn2+ Calculate the % of ZnO and Fe2O3 in the sample.

20. Answer (MV)Zn= (MV) EDTA (MV)Fe= (MV) ZnY M×10=0.01294×38.71
MZn=0.05 M
Weight of ZnO=
0.05×250x81.39 / = g
% of ZnO= ( /1.022)×100=99.73%
(MV)Fe= (MV) ZnY
M×50= ×2.4
MFe=1.3×10-4 M
No. of Fe2O3 moles= No. of Fe moles/2
= (1.3×10-4×250)/ (2)
= mmoles
Weight of Fe2O3= x /1000
=2.61×10-3 g
% of Fe2O3=
[(2.61×10-3)/1.022]×100= 0.256%
No of Moles equal ½ that of Fe ions as each mole of molecule has 2 atoms of Fe

21. 2-A 50 ml sample containing 0. 45g of MgSO4 (Mwt 120. 37 g/mole) in 0
2-A 50 ml sample containing 0.45g of MgSO4 (Mwt g/mole) in 0.5 L required 37.6 ml of EDTA for titration. How many milligrams of CaCO3 ( g/mole) will react with 1.00 ml of this EDTA solution?
First calculate the concentration of MgSO4 as molar concentration
Strength=M x molec.wt.
0.9 g/l=Mx120.37
M= M
Calculate the Molarity of EDTA
(MV)EDTA=(MV)Mg++
M x 37.6 = x 50
M= M
Find out the weight Of CaCO3 reacting with 1 ml of this EDTA
(MV)EDTA= wt
mol.wt
x 1 = wt
100.09
Wt of CaCO3 = 0.98 mg