Chapter 24, Problem 1STP

Interpretation Introduction

Interpretation:

The final concentration of $\text{Potassium-40}$ is to be calculated.

Concept introduction:

To calculate the rate constant we use the formula which is given as

$\begin{array}{l}\text{K=}\frac{\text{0}\text{.693}}{{\text{t}}_{\text{1/2}}}\\ \text{where,}\text{K=rate constant}\\ {\text{t}}_{\text{1/2}}\text{=half life time}\end{array}$

And the final amount is to be calculated as

$\begin{array}{l}\text{K=}\frac{\text{0}\text{.2303}}{\text{t}}\text{log}\frac{\left[\text{A}{\text{°}}_{}\right]}{\left[\text{A}\right]}\\ \text{Where,}\left[\text{A}{\text{°}}_{}\right]\text{=Initial}\text{concentration}\\ \left[\text{A}\right]\text{=}\text{Final}\text{concentration}\\ \text{t =}\text{time}\text{period}\end{array}$

Answer to Problem 1STP

Correct answer:

Option B. The final concentration of $\text{Potassium-40}$ is $\text{2}{\text{.30×10}}^{\text{-7}}\text{g}$.

Explanation of Solution

Reason for correct option:

Given information:

$\begin{array}{l}\text{Half}\text{time}\text{period}\text{=}\text{1}{\text{.26×10}}^{\text{9}}\text{years}\\ \text{Time}\text{period}\text{=}\text{3}{\text{.15×10}}^{\text{8}}\text{years}\\ \text{Initial}\text{concentration}\text{=}\text{2}{\text{.73×10}}^{\text{-7}}\text{g}\end{array}$

To calculate the rate constant

$\begin{array}{l}\text{K=}\frac{\text{0}\text{.693}}{{\text{t}}_{\text{1/2}}}\\ \text{=}\frac{\text{0}\text{.693}}{\text{1}{\text{.26×10}}^{\text{9}}}\\ \text{=}\text{0}{\text{.55×10}}^{\text{-9}}\end{array}$

Now, we calculate the final concentration of $\text{Potassium-40}$

$\begin{array}{l}\text{K=}\frac{\text{2}\text{.303}}{\text{t}}\text{log}\frac{\left[{\text{A}}_{\text{°}}\right]}{\left[\text{A}\right]}\\ \text{Or}\\ \text{log}\frac{\left[{\text{A}}_{\text{°}}\right]}{\left[\text{A}\right]}\text{=}\frac{\text{2}\text{.303}}{\text{t}}\text{×K}\\ \end{array}$

By using the log property: $\begin{array}{l}\text{log}\frac{\left[{\text{A}}_{\text{°}}\right]}{\left[\text{A}\right]}\text{=log}\left[{\text{A}}_{\text{°}}\right]\text{-log}\left[\text{A}\right]\\ \end{array}$

The above equation becomes

$\begin{array}{l}\text{log}\left[{\text{A}}_{\text{°}}\right]\text{-log}\left[\text{A}\right]\text{=}\frac{\text{K×t}}{\text{2}\text{.303}}\\ \text{Or,}\text{log}\left[\text{A}\right]\text{=}\text{log}\left[{\text{A}}_{\text{°}}\right]\text{-}\frac{\text{K×t}}{\text{2}\text{.303}}\\ \text{=log}\left(\text{2}{\text{.703×10}}^{\text{-7}}\right)\text{-}\frac{\text{0}{\text{.55×10}}^{\text{-9}}\text{×3}{\text{.15×10}}^{\text{8}}}{\text{2}\text{.303}}\\ \text{=}\text{log2}\text{.703+log}\left({\text{10}}^{\text{-7}}\right)\text{-0}\text{.07523}\left(\text{by}\text{using}\text{log}\text{property}\right)\\ \text{=}\text{0}\text{.43185}\text{+}\left(\left(\text{-7}\right)\text{×1}\right)\text{-}\text{0}\text{.07523}\\ \text{log}\left[\text{A}\right]\text{=-6}\text{.6434}\\ \end{array}$

Now we take the antilog on right side to find out the final concentration

$\begin{array}{l}\text{log}\left[\text{A}\right]\text{= -6}\text{.6434}\\ \left[\text{A}\right]\text{=}\text{2}{\text{.30×10}}^{\text{-7}}\text{g}\end{array}$

The final concentration of $\text{Potassium-40}$ is $\text{2}{\text{.30×10}}^{\text{-7}}\text{g}$.