Ethylenediaminetetraacetic acid (H 4 EDTA) is a tetraprotic acid. Its salts are

Question

Ethylenediaminetetraacetic acid (H4EDTA) is a tetraprotic acid. Its salts are used to treat toxic metal poisoning through the formation of soluble complex ions, which are then excreted. Because EDTA4- also binds essential calcium ions, it is often administered as the calcium disodium salt. For example, when Na2Ca(EDTA) is given to a patient, the [Ca(EDTA)]2- ion reacts with circulating Pb2+ ions and exchanges metal ions:

[Ca(EDTA)]2-(aq) + Pb2+(aq) = [Pb(EDTA)]2-(aq) + Ca2+(aq) K = 2.5 x 10^7

A child is found to have a dangerously high level of 120g/100mL of lead ions. If the child is administered 10.0 mL of 2.0 M Na2Ca(EDTA), what will be the final concentration of Pb2+ (in mol/L) in her blood? Assume a total blood volume of 2.5 liters.

Explanation / Answer

total Pb2+in the child's blood=2.5*1000ml*120ug/100ml=3000ug

total moles of Pb2+=3000ug/207.2 g/mol=3000*10^-6 g/207.2g/mol=1.4*10^-5 moles

moles of Na2Ca(EDTA) added=0.010L*2.0mol/L=0.02 moles

moles of Pb2+ reacted=1.4*10^-5 moles

As keq is very large the rxn will go to completion.

     ICE table             [Ca(EDTA)]2-                        [Pb2+] [Pb(EDTA)]2    [Ca2+]               

initial concentration   0.02 moles                   1.4*10^-5 moles                           0                  0

completion                 0.02-1.4*10^-5              1.4*10^-5 -1.4*10^-5 =0                1.4*10^-5        1.4*10^-5

change           -x                                         -x                                    +x                 +x

final concentration     0.02-1.4*10^-5 -x                        x                               1.4*10^-5 + x        1.4*10^-5 +x

the rxn reverses at equilibrium

keq= 1/kf= [Ca(EDTA)]2- * [Pb2+]/ = [Pb(EDTA)]2 [Ca2+]=(0.02-1.4*10^-5 -x)*x/(1.4*10^-5 + x )^2

or,1/2.5*10^-7=keq=(0.02 -x)*x/(1.4*10^-5+ x )^2

or,0.4*10^-7 =(0.02)*x/(1.4*10^-5 )^2         [x<<<as very less dissociation occurs]

solving for x,

x=39.2*10^-17 moles

so, pb2+ at eq=x=39.2*10^-17 moles

[Pb2+]=39.2*10^-17 moles/2.5L=15.7*10^-17 M=(15.7*10^-17 moles*207.2g/mol)/L=3248.9 *10^-17 g/1000ml=3248.9 *10^-17*10^6 ug/1000ml=3248.9 *10^-12 ug/100ml

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