Titration of EDTA with metal ion. Use Equation 12-12 to reproduce the results of Exercise 12-C

Amount of EDTA required for titration = 50 mL of 0.04 M EDTA

Concentration of Cu(NO3)2required for titration =0.08 M

pH maintained =5

EDTA is one type of chelating agent also known as Ethylenediaminetetraacetic acid. It is used to treat heavy metal toxicity, neurotoxicity, lead poisoning, coronary artery disease etc. EDTA has applications in different industries like textile, pulp and paper etc. It is used in biochemistry and molecular biology lab.

Equation 12-12 need to be used for the fraction of titration of metal ion with EDTA is as follows

$\varphi =\frac{{\mathrm{C}}_{\mathrm{M}}{\mathrm{V}}_{\mathrm{M}}}{{\mathrm{C}}_{\mathrm{L}}{\mathrm{V}}_{\mathrm{L}}}=\frac{{\mathrm{K}}_{\mathrm{f}}\left[\mathrm{M}\right]+{\displaystyle \frac{\left[\mathrm{M}\right]+{\mathrm{K}}_{\mathrm{f}}{\left[\mathrm{M}\right]}^2}{{\mathrm{C}}_{\mathrm{L}}}}}{1+{\mathrm{K}}_{\mathrm{f}}\left[\mathrm{M}\right]-{\displaystyle \frac{\left[\mathrm{M}\right]+{\mathrm{K}}_{\mathrm{f}}\left[\mathrm{M}\right]}{{\mathrm{C}}_{\mathrm{M}}}}}$

Other formulas need to be used

$\left[\mathrm{M}\right]={10}^{-\mathrm{pM}}$

Here instead of formation constant (Kf), conditional formation constant$\left({\mathrm{K}}_{\mathrm{f}}\right)$should be taken in the above formula stated.

${\mathrm{K}}_{\mathrm{f}}={\mathrm{a}}_{{\mathrm{Y}}^4}{\mathrm{K}}_{\mathrm{f}}$

From table 12-1 the value ofrole="math" localid="1663425796200" ${\mathrm{a}}_{{\mathrm{Y}}^2}$for EDTA at 250C and pH 5 is 2.9×10-7

From table 12-2 we get log Kf for Cu2+= 18.78

Spreadsheet developed is shown below

The curves for both the metals Cd and Cu are shown below