Question

In an isotope dilution experiment, chloride was determined by adding $5.0mg$of sodium chloride containing ${}^{38}\mathrm{Cl}\left(t_{1/2}=37.3\min \right)$to a sample. The specific activity of the added NaCl was localid="1646407073099" $3.6\times {10}^{4}cps/mg$. What was the total amount of chloride present in the original sample if localid="1646407080469" $400mg$of pure Agcl was isolated and if this material had a counting rate of localid="1646407089173" $38cps$above background localid="1646407097779" $150min$after the addition of the radiotracer?

Short answer

The chloride present in the original sample was $29g$.

Step-by-step solution

Given information

Mass of NaCl$=5.0mg$

Half life of ${}^{38}Cl=t_{1/2}=37.3min$

Specific activity of NaCl$=3.6\times {10}^{4}cps/mg$

Mass of AgCl$=400mg$

Counting rate of sample$=38cps$

Time$=150min$

Calculation

The tracer activity can be calculated as:

$$\begin{gathered} R_t=\frac{3.6\times {10}^4}{35.45/58.44} \\ {R}_t=5.9\times {10}^{4}cps/mg{\left(Cl\right)}^{-} \end{gathered}$$

Activity of isolated AgCl can be calculated as:

$$\begin{gathered} R_m=\frac{38}{400\times 35.45/143.32} \\ {R}_m=0.384cps/mg{\left(Cl\right)}^{-} \end{gathered}$$

The decay constant can be calculated by using the half time formula as:

$t_{1/2}=\frac{0.693}{\lambda }$

By rearranging the terms, we get,

$$\begin{gathered} \lambda =\frac{0.693}{t_{1/2}} \\ \lambda =\frac{0.693}{37.3} \\ \lambda =0.01858mi{n}^{-1} \end{gathered}$$

Now, the specific activity of the radioactive element can be calculated from the below equation as:

$A=A_0{e}^{-\lambda t}$

By rearranging the terms, the activity of isolated AgCl can be calculated as:

$$\begin{gathered} A_0=A{e}^{\lambda t} \\ {A}_0=0.384\times e^{0.01858\times 37.3} \\ {A}_0=6.233cps \end{gathered}$$

The amount of tracer in terms of chloride ion can be calculated as:

$$\begin{gathered} m_t=\frac{5.0\times 35.45}{58.44} \\ {m}_t=3.033mg{\left(Cl\right)}^{-} \end{gathered}$$

The total chloride will be precipitated as AgCl, hence,

$m_m=m_t=3.033mg{\left(Cl\right)}^{-}$

Now,

The mass of the chloride ion in the sample can be calculated from the below equation as:

$m_x=\frac{R_t}{R_m}\times m_m-m_t$

By substituting the calculated values in the above equation, we get,

$$\begin{gathered} m_x=\frac{5.9\times {10}^4}{6.233}\times 3.033-3.033 \\ {m}_x=28706.57mg \\ {m}_x=28.7g\approx 29g \end{gathered}$$

Final answer

The mass of the chloride ion is calculated as$29g$.