Map sapling learning In forming a chelate with a metal ion, a mixture of free ED

ID: 944333 • Letter: M

Question

Map sapling learning In forming a chelate with a metal ion, a mixture of free EDTA (abbreviated Y and metal chelate (abbreviated MYn 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 M n Y n -4 MY is governed by the equation M EDTA where Kr is the association constant of the metal and Y4 ay4 s the fraction of EDTA in the Y form, and [EDTA] is the total concentration of free (unbound) EDTA. K f is the "conditional formation constant." How many grams of Na2EDTA 2H20 (FM 372.23) should be added to 1.67 g of Ba(NO3)2 (FM 261.35) in a 500-mL volumetric flask to give a buffer with pBa 7.00 at pH 10.00? (log Kf for Ba-EDTA is 7.88 and ay at pH 10.00 is 0.30.) Number mass Na2EDTA. 2H20

Explanation / Answer

Moles of Ba(NO3)2=1.67g/261.35 g/mol=0.0064 moles

Concentration of Ba(NO3)2=0.0064 moles/0.5L=0.0128 mol/L or M

Initial conc of Ba2+=0.0128 M

Given pBa2+=7.00=-log [Ba2+]

[Ba2+]=10^-7 M (equilibrium concentration of Ba2+]

Kf=[MY^n-4]/{aY4- * [Mn+][EDTA]} [Mn+ here is Ba2+]

ICE table

[M^n+]

[EDTA]

[MY^n-4]

Initial

Change

-x

equilibrium

10^-7

0.0128 M-x=10^-7

Or ,x=0.0128-10^-7=0.0128 M

So, X-x=10^-7

X=10^-7+0.0128=0.0128 M (10^-7 <<< 0.0128 ignored]

[EDTA]= 0.0128 M

But , only a fraction of EDTA reacts

so , actual [EDTA] added * aY4-*=[ EDTA]reacted

actual [EDTA] added=[ EDTA]reacted/aY4-=0.0128/L/0.3= 0.0427mol/L

moles of EDTA added=0.0427 *0.5L=0.0213 mol

mass of EDTA added=0.0213 mol*232.64 g/mol=4.96 g (answer)

So, X=[EDTA]= 1.28*10^-5M=1.28*10^-5mol/L*0.5L =0.625*10^-5 mol=0.625*10^-5 mol*372.23g/mol=232.64 *10^-5 =0.00232 g(answer)