Question

Phosphorus -32 is a commonly used radioactive nuclide in biochemical research, particularly in studies of nucleic acids. The half-life of phosphorus-32 is 14.3 days. What mass of phosphorus-32 is left from an original sample of 175 mg of Na332PO4 after 35.0 days

Answer 1

Answer: The mass of P-32 left from the original sample is 32.07 mg

Step-by-step explanation:

All radioactive decay processes follow first-order reactions.

Calculating rate constant for first order reaction using half life:

`t_(1/2)=(0.693)/(k)` .....(1)

`t_(1/2)` = half life period = 14.3 days

k = rate constant = ?

Putting values in equation 1:

`k=(0.693)/(14.3days)\\\\k=0.0485days^(-1)`

The integrated rate law equation for first-order kinetics:

`k=(2.303)/(t)\log (a)/(a-x)` ......(2)

Given values:

a = initial concentration of reactant = 175 mg

a - x = concentration of reactant left after time 't' = ? mg

t = time period = 35.0 days

Putting values in equation 2:

`0.0485days^(-1)=(2.303)/(35.0 days)\log ((175)/(a-x))\\\\\log ((175)/(a-x))=(0.0485* 35.0)/(2.303)\\\\\log ((175)/(a-x))=0.737\\\\(175)/(a-x)=10^(0.737)\\\\a-x=(175)/(5.457)=32.07mg`

Hence, the mass of P-32 left from the original sample is 32.07 mg