Question

A solution of natural water contains 1.00 x 10-3 M Mg2+. Calculate the value of pMg2+ after the following additions of 0.00200 M EDTA to an initial volume of 10.00 mL of the Mg2+ solution. A borax buffer is used to maintain the pH at 10.70 for all the titration.

a) 0.00 mL b) 2.00 mL c) 3.50 mL d) 5.00 mL e) 7.50 mL f) 10.00 mL

Answer 1

The magnesium-EDTA titration is a complexometric titration where Mg²⁺ forms a complex with EDTA. EDTA is a multidentate (hexadentate ligand). Therefore the complex formation takes place in steps. Therefore we need to consider the conditional formation constant for the complex formation.

If K_f is the formation constant for the metal EDTA complex given by

MgY-

But we need to consider the conditional formation constants $\alpha$_Y^4- and $\alpha$_Mg²⁺

$\alpha$_Y^4- = $\frac{[Y^{4-}]}{C_{EDTA}}$

$\alpha$_Mg²⁺ = $\frac{[Mg^{2+}]}{C_{Mg^{2+}}}$

Therefore, we get K_f" = K_f$\ast$$\alpha$_Y^4-$\ast$$\alpha$_Mg^2+.

For pH 10 $\alpha$_Y^4- = 3.8*10^(-8) and $\alpha$_Mg²⁺ = 0.984

Equivalence volume is given by V=( =5mL

001*1 0.002

i) For 0 mL of EDTA added,

[Mg²⁺] is 0.001.

$\Rightarrow$pMg = -log([Mg²⁺ ])

[Mg²⁺ ] = $\alpha$_Mg²⁺ * C_Mg²⁺

            = 0.984 * 0.001= 0.0984

pMg= 3.007

ii)For 2mL

Amount of Mg²⁺ consumed = amount of EDTA added

Amount of Mg²⁺ left = =0.0005 M

(0.001*10-0.002-2)

Using $\alpha$_Mg²⁺ , [Mg²⁺] = $\alpha$_Mg²⁺ * C_Mg²⁺ = 0.0005*0.984= 0.000492

pMg = -log([Mg²⁺ ]) = 3.30

iii)For 3.5mL

Amount of Mg²⁺ left = = 0.00022

(0.001-10-0.002+3.5)

[Mg²⁺ ]= $\alpha$_Mg²⁺ * C_Mg²⁺ = 0.984 * 0.00022 = 0.0002186

pMg = 3.66

iv)For 5 mL

Here 5 mL is the equivalence volume.

Therefore,

amount of Mg²⁺ left =( Initial moles of Mg²⁺) ÷(Total Volume)

                            = (0.001*10)/(15) =0.00066

[Mg²⁺ ]= 0.00066*0.984 = 0.000656

pMg = 3.18