Chapter 25, Problem 25.AE

Interpretation Introduction

Interpretation:

The concentration of A in the unknown solution has to be calculated.

Answer to Problem 25.AE

The concentration of A is $\text{0}\text{.418}\text{mg/mL}$.

Explanation of Solution

Peak area of A is $5.97$.

Peak area of B is $6.38$.

$\begin{array}{l}\frac{{\text{Area}}_{\text{A}}}{\text{[A]}}\text{=}\text{F}\left(\frac{{\text{Area}}_{\text{B}}}{\text{[B]}}\right)\\ \frac{10.86}{\left[1.03\right]}=\text{F}\left(\frac{4.37}{\left[1.16\right]}\right)\\ F=\frac{10.86}{\left[1.03\right]}\times \frac{\left[1.16\right]}{4.37}\\ =\frac{12.5976}{4.5011}\\ =2.789\\ =2.79\end{array}$

Total volume of the unknown solution is given as $\text{25}\text{.00}\text{mL}$.

Weight of B added is $\text{12}\text{.49}\text{g}$.

Total volume of A added is $\text{10}\text{mL}$.

Concentration of B in the unknown solution can be found as shown below,

$\text{12}\text{.49}\text{g/25}\text{.00}\text{mL}\text{=}\text{0}\text{.4996}\text{g/mL}$

The concentration of A can be found as shown below,

$\begin{array}{l}\frac{{\text{Area}}_{\text{A}}}{\text{[A]}}\text{=}\text{F}\left(\frac{{\text{Area}}_{\text{B}}}{\text{[B]}}\right)\\ \frac{\text{5}\text{.97}}{\text{[A]}}\text{=}\text{2}\text{.79}\left(\frac{\text{6}\text{.38}}{\text{[0}\text{.4996]}}\right)\\ \left[\text{A}\right]\text{=}\frac{\text{5}\text{.97}}{\text{2}\text{.79}}\text{×}\frac{\left[\text{0}\text{.4996}\right]}{\text{6}\text{.38}}\\ \text{=}\frac{\text{2}\text{.982612}}{\text{17}\text{.8002}}\\ \text{=}\text{0}\text{.1675}\text{g/mL}\end{array}$

The concentration of A in original unknown solution is found as,

$\begin{array}{l}\left[\text{A}\right]\text{in original unknown}\text{=}\frac{\text{25}\text{.00}}{\text{10}\text{.00}}\left(\text{0}\text{.167}\text{mg/mL}\right)\\ \text{=}\text{0}\text{.418}\text{mg/mL}\end{array}$

Therefore, the concentration of A in the unknown is found as $\text{0}\text{.418}\text{mg/mL}$.

Conclusion

The Concentration of A in the unknown was calculated as $\text{0}\text{.418}\text{mg/mL}$.