Question

Titration of metal ion with EDTA. Use Equation 12-11 to compute curves (pM versus mL of EDTA added) for the titration of 10.00 mL of 1.00 mM M2+ (= Cd2+ or Cu2+) with 10.0 mM EDTA at pH 5.00. Plot both curves on one graph.

Short answer

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

Step-by-step solution

Information given

Titration of 10.0 mL of 1.0mM M2+ with 10 mM EDTA. The titration should be carried out at pH=5.00.

Equation need to be used to develop spreadsheet

Using Equation 12-11 the equation for the fraction of titration of metal ion with EDTA is as follows

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

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.

From table 12-1 the value of ${\mathrm{a}}_{{\mathrm{Y}}^4}$for EDTA at 250C and pH 5 is 2.9×10-7

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

From table 12-2 we get log Kf for Cd2+= 16.5

Spreadsheet

Spreadsheets developed are shown below

Graph Plot

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