Question

Your book describes a treatment for -Thalassemia, a disease where the body cannot make the form of hemoglobin. Transfusions are necessary to provide the body with external sources ofhemoglobin but that procedure causes a potentially fatal buildup of Fe in the bofy. To eliminate Fe2+, desferrioxamine B (DB) is given to patients. DB is a chelating agent for Fe2+ that has three hydroxamate groups which complex with the Fe or H. The hydroxamate group in its uncomplexed basic form is

   O- O

    |   ||

R-N-C-R’

In this formula, R and R’ stand for other carbon-containing parts. When dB is administered to patients, it chelates with the Fe2+ in three spots forming a Fe2+-DB complex with a formation constant of 5 x 1030. The Fe2+-DB complex can then be excreted from the body.

a) Assume the body has a blood volume of 4.8 liters, the concentration of Fe2+ in the blood is 3.6 x 10-5 M, and the blood pH is 7.1. Assume that 200 mgm of DB is administered to a patient (DB has a molecular weight of 555) and the DB goes completely into the patient’s blood system. Determine the concentration of uncomplexed Fe that remains after the administration. The acid dissociation constants for DB are

K1 = 5 x 10-5

K2 = 6 x 10-6

K3 = 8 x 10-8

b) How many grams of Fe2+ will be removed from the patient’s body if all the Fe2+ - D is excreted? (atomic weight of Fe = 55)

c) If the number of grams determined in part b) are removed each day, how many days will it take to remove the 4 grams of iron that typically accumulate in a person’s body over 1 year?

Answer 1

In blood, uncomplexed basic Db must be undergoing two competing reactions: reaction with Fe²⁺ to form the Fe²⁺-DB complex and protonation by water (blood is aqueous in nature). The three K_a values are given, we can calculate the Pk_a values by noting that pK_a = -log₁₀K_a.

Ka	pKa
K1 = 5*10-5	pK1 = 4.30
K2 = 6*10-6	pK2 = 5.22
K3 = 8*10-8	pK3 = 7.10

The pH of blood is 7.1 and hence, we must work with pK₃. Since, uncomplexed DB is a base, the three step-wise de-protonation reactions are

H₃DB³⁺ <=====> H⁺ + H₂DB^{+ 2} ……K₁

H₂DB⁺² <======> H⁺ + HDB⁺ ……K₂

HDB⁺ <=====> H⁺ + DB …..K₃

a) The problem says that we administered 200 mg DB (mol. wt. 555 g/mol) to a patient. Moles of DB administered = (200 mg)*(1 g/1000 mg)*(1 mole/555 g) = 3.603*10^-4 mole.

Volume of blood = 4.8 L; therefore, initial molar concentration of DB given = (3.603*10^-4 mole)/(4.8 L) = 7.506*10^-5 M.

The equilibrium constant for the third reaction above is

K₃ = [H⁺][DB]/[HDB⁺] = (x).(x)/(7.506*10^-5 – x) where x is the equilibrium concentration of DB in blood.

K₃ is low; hence, x is lower than the initial concentration of HDB⁺ and hence,

K₃ = 8*10^-8 = x²/7.506*10^-5

====> x² = 8*10^-8*7.506*10^-5 = 6.0048*10^-12

====> x = 2.450*10^-6

The equilibrium concentration of DB is 2.45*10^-6 M.

The reaction of DB with Fe²⁺ is

Fe²⁺ + DB ------> Fe²⁺-DB

We must assume a 1:1 molar ration of reaction; hence the concentration of Fe²⁺-DB must be 2.45*10^-6 M.

The concentration of uncomplexed Fe²⁺ in blood is given as

[Fe²⁺]_rem = [Fe²⁺]_initial – [Fe²⁺-DB]_eq = (3.6*10^-5) M – (2.45*10^-6) M = 3.355*10^-5 M (ans).

b) Fe²⁺-DB <=====> Fe²⁺ + DB

There is a 1:1 molar ratio, hence 1 mole of Fe²⁺-DB gives 1 mole of Fe²⁺. Moles of Fe²⁺-DB in blood = (2.45*10^-6 mole/L)*(4.8 L) = 1.176*10^-5 mole.

Therefore, moles of Fe²⁺ produced = (1.176*10^-5 mole)*(55 g/1 mole) = 6.468*10^-4 gm.

c) As per the problem, 6.468*10^-4 g Fe²⁺ is removed per day. Therefore, days required to remove 4.0 g Fe²⁺ = (4.0 g/6.468*10^-4 g/day) = 6184.29 days ≈ 6184 days (since we cannot have fractional days) = (6184 days)*(1 year/365 days) = 16.94 years = 16.9 years (ans).