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# Chelate with a metal ion

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In forming a chelate with a metal ion, a mixture of free EDTA (abbreviated Y^4-) and metal chelate (abbreviated MY^n-4) can buffer the free metal ion concentration at values near the dissociation constant of the metal chelate, just as a weak acid and a salt can buffer the hydrogen ion concentration at values near the acid dissociation constant.

This equilibrium is governed by the equation

Where Kf is the association constant of the metal and Y^4-, αy^4- is the fraction of EDTA in the Y^4- form, and [EDTA] is the total concentration of free (unbound) EDTA. is the "conditional formation constant."

How many grams of Na_2EDTA·2H_2O (FM 372.23) should be added to 1.51 g of Mg(NO_3)_2·2H_2O (FM 184.35) in a 500-mL volumetric flask to give a buffer with pMg^2+ = 8.00 at pH 10.00?

(log K_f for Mg-EDTA is 8.79 and αy^4- at pH 10.00 is 0.30.)

Mass Na_2EDTA*H_2O = _________________g

Hint: after I missed the answer
[Mg^2+ ] = 10^-pMg2+ = 1.0 �- 10-8 M, so essentially all of the Mg^2+ in solution is in the form MgY^2-. The concentration of bound EDTA will therefore equal [MgY^2-]. Calculate [MgY^2-] from the mass of Mg(NO_3)_2·2H_2O and the volume of the volumetric flask. Then calculate the free [EDTA] from the equilibrium expression, and the quantity of EDTA needed from the sum of the bound and free EDTA.