Ethylenediaminetetraacetic acid (H4EDTA) is a tetraprotic acid. It's salts are used to treat toxic metal poisoning through the formation of soluble complex ion that 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 [CaEDTA]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 107 A child is found to have a dangerously high lead level of (1.64x10^2) µg/ 100 mL. If the child is administered (1.000x10^2) mL of (1.1x10^-1) M Na2Ca(EDTA), what will be the final concentration of Pb2+ in the blood (in mol/L)? Assume the child has a blood volume of 1.5 L.
Please clearly show steps and explain
The equilibrium is
The lead level is (1.64x10^2) µg/ 100 mL, or 164 µg/ 100 mL, or 1640 µg/L, or
The molar mass of lead is 207.2 g/moll
The molar concentration of lead is
The child has administered (1.000x10^2) mL of (1.1x10^-1) M Na2Ca(EDTA), or 100 mL of 0.11 M solution. This solution was mixed with child's blood, making the concentration
Let's assume that the equilibrium concentration of [Pb(EDTA)]2- complex is x; then the concentration of Ca2+ ions is x as well, the concentration of Pb2+ is (7.915*10-6-x), the concentration of [Ca(EDTA)]2- is (6.875*10-3-x). Plug in the values into the equilibrium expression:
The solution of this equation is it means that the final concentration of Pb2+ is about zero. But to be precise, we will find it as
This is the final concentration of lead ions in blood.
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