Theoretical+Background+for+Isotopic+Dilution

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Isotopic dilution analysis is a powerful method of determining small amounts of material in a form where 100% separation and recovery of the material is difficult. In principle a quantity of radioactive material of the same chemical form and of known specific activity is added to the unknown amount of inactive material. Determination of the specific activity of the mixture then allows calculation of the quantity of inactive material present. Suppose we have x grams of a radioactive material of activity A 1 dpm and a molecular weight M 1. The **specific activity (S)** of this material is:

math S_1=\frac{A_1\cdot M_1}{x}\, dpm\, mol^{-1} math

If y grams of chemically identical, but inactive, material is added the final specific activity becomes: math S_2=\frac{A_1\cdot M_1}{x+y}\,dpm\,mol^{-1} math If a chemical reaction is now performed, resulting in a change in the chemical form of the material (e.g. S*O 4 2- → S* 2-, where the * indicates the radiolabelled element), 9 then the specific activity of the new chemical form will still be S 2 , even though some of the material may have been lost in performing the reaction. Thus, if the weight of material recovered is z grams, then:

math s_2=\frac{A_2\cdot M_2}{z}\, dpm \, mol^{-1} math Where A 2 is the activity of the recovered material and M 2 is the molecular weight of the new chemical form. If y had been the unknown quantity, its value may now be determined, since: math S_2=\frac{A_1\cdot M_1}{x+y}=\frac{A_2\cdot M_2}{z} math Therfore: math \frac{x}{A_1\cdot M_1}+\frac{y}{A_1 \cdot M_1}=\frac{1}{S_2} math Therfore: math \frac{y}{A_1 \cdot M_1}=\frac{1}{S_2}-\frac{1}{S_1} math And Since:

math A_1 M_1=S_1\cdot X math

math y=S_1\cdot X \left( \frac{1}{S_2}-\frac{1}{S_1}\right) math therfore:

math y=X\cdot\left( \frac{S_1}{S_2}-1\right) \, grams math In this experiment, a variation on the above is required since the weight of the radioactive material is unknown. You are provided with a solution containing y g dm -3 of phosphate mixed with an unspecified amount of chloride impurity. A measured volume of the solution is mixed with radiolabelled 32 P-phosphate and a volume of the mixture, V 1 is evaporated and counted. If the activity of the material is A 1 dpm, then the specific activity of the phosphate (molecular weight M 1 ) is:

math S_1=\frac{A_1\cdot M_1}{y \cdot V_1}\, dpm \, mol^{-1} math

The mixed solution is then reacted with ammonium molybdate so that a precipitate of ammonium (12) molybdosphosphate ((NH 4 ) 3 PO 4 .12MoO 3 ) is formed. A known mass of this precipitate (W 2 grams) is counted, and the resulting activity, A 2, used to determine the specific activity of the molybdosphosphate (molecular weight M 2 ).

math S_2=\frac{A_2\cdot M_2}{W_2}\, dpm\, mol^{-1} math

As before the specific activities are the same, so that: math \frac{A_1\cdot M_1}{y\cdot V_1}= \frac{A_2\cdot M_2}{W_2} math and math y=\frac{A_1 \cdot M_1 \cdot W_2}{A_2 \cdot M_2 \cdot V_1}\, g \, dm^{-3} math

Thus, we have determined the phosphate concentration of the solution provided for assay. Note that the mass of phosphate in the original carrier free 32 P-phosphate solution has been neglected in this calculation. In fact, the mass per cm 3 of 32 P-phosphate in this solution would be in the 10 -9 – 10 -12 g range, and can justifiably be neglected in comparison with y. For the calculation you will need to calculate the molecular weights of the phosphate (as Na 2 HPO 4 ) and the molybdophosphate (as (NH 4 ) 3 PO 4 .12MoO 3 ).