Help me please In forming a chelate with a metal ion, a mixture of free EDTA (abbreviated Y*-) and metal chelate (abbreviated MY") canbuffer the free metal ion concentration at values near the dissociation constant of the metal chelate, just as a weak acid and a saltcan buffer the hydrogen ion concentration at values near the acid dissociation constant.The equilibriumM+ + Y*- MY"4is governed by the equationMY )[M**] [EDTA]K = aya · Kr =where K, is the association constant of the metal and Y, ays is the fraction of EDTA in the form Y, and [EDTA] is the totalconcentration of free (unbound) EDTA. K' is the conditional formation constant.How many grams of Na, EDTA 2 H,0 (FM 372.23 g/mol) should be added to 1.64 g of Ba(NO,), (FM 261.35 g/mol) in a500. mL volumetric flask to give a buffer with pBa+= 7.00 at pH 10.00? log K, for Ba-EDTA is 7.88 andat pH 10.00is 0.30.

The moles of Ba2+ is calculated as follows: Moles of Ba2+ = (Mass of Ba(NO3)₂ / Molar mass of…

In forming a chelate with a metal ion, a mixture of free EDTA (abbreviated Y) and metal chelate (abbreviated MY") 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. The equilibrium Mt + Y = MY" is governed by the equation MY K = ays - Kr = [M*] [EDTA] where K, is the association constant of the metal and Y, aya is the fraction of EDTA in the form Y, and [EDTA] is the total concentration of free (unbound) EDTA. K' is the conditional formation constant. How many grams of Na, EDTA - 2H,0 (FM 372.23 g/mol) should be added to 1.64 g of Ba(NO,), (FM 261.35 g/mol) in a 500. ml. volumetric flask to give a buffer with pBa+ = 7.00 at pH 10.00? log K, for Ba-EDTA is 7.88 and aya at pH 10.00 is 0.30.