Question

Cu Determination (5 Points: 5 Points titration, 10 points error/statistics) Copper ions in solution were directly determined with an EDTA titration at a pH of 5.50 using glycinecresol red indicator. Assume a 25.00 mL aliquot of tap water in this titration. Cu Titration Trial VEDTA (mL) 16.98 16.48 16.92 16.29 18.12 16.41 17.12 Pb2 Determination (15 Points: 5 Points Tiration, 10 Points Erroristatistics) Lead was determined through back-titration using Eriochrome Black T indicator at a pH of 10.00. A 50.00 mL aliquot sample of tap water was reacted with 25.00 mL of your EDTA solution in excess. The resultant solution was titrated with the standard Mg2+ solution to the endpoint. Pb Titration V Mg (mL) Trial 19.71 19.2 19.15 19.11 19.23 19.17
This is for an analytical chemistry. Please show all work and excel sheets.

Answer 1

EDTA is a chelating agent and chelates metal ions, i.e. attaches metal ions to itself.

For the Cu titration,

taking the average value of the 7 trials for the volume of EDTA consumed, we get:

V EDTA (avg) = 16.90 mL

Assume that the tap water has a Cu²⁺ conc. of x moles/L

Volume of aliquot = 25 mL = 0.025 L

Moles of Cu²⁺ in this aliquot = 0.025*x moles

These many moles of EDTA will also be required for titration.

Lets say the standard EDTA solution used in this experiment has EDTA conc. of 'y' moles/L

Volume of EDTA consumed = 16.90 mL = 0.0169 L

Thus, moles of EDTA used = 0.0169*y moles

Equating we get:
0.0169*y = 0.025*x

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For the Pb²⁺ titration:
Lets assume that tap water has Pb²⁺ conc of 'x' moles/L, standard EDTA solution has conc 'y' moles/L and standard Mg²⁺ solution has conc. 'z' moles/L.

Moles of EDTA supplied = Volume*conc = 0.025*y

Moles of Pb²⁺ present = 0.050*x

Moles of EDTA left after the reaction of EDTA with Pb²⁺ = 0.025*y - 0.050*x

These many moles of Mg²⁺ are supplied from outside to react with EDTA.

Calculating the average volume of Mg solution used in titration from the 6 trials, we get;

V Mg (average) = 19.26 mL

Thus, moles of Mg²⁺ = 0.01926*z

Thus, equating we get:

0.01926 = 0.025*y - 0.050*x

Putting 'y' and 'z', we can calculate 'x'