Question

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

Answer 1

Solution.

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.