Question

A $2.5-mL$aliquot of a solution that contains $6.4$ppm iron(III) is treated with an appropriate excess of $KSCN$to form the $Fe{\left(SCN\right)}^{2+}$complex and diluted to $50.0mL$. What is the absorbance of the resulting solution at $580$nm in a $2.50-mL$cell? See Problem $13.8$for absorptivity data.

Short answer

The absorbance is$0.100$

Step-by-step solution

Step 1. Given information

Concentration is$6.4ppm$at$580nm$

Step 2. Formula used

he Beer's law, also known as the Lambert-Beers law, asserts that the absorbance (A)of a solution is proportional to the thickness or path length (b)of the solution and the concentration (c) of the absorbing species for monochromatic light. This law is mathematically expressed as follows when the concentration is in moles per litre (M) and the route length is in centimetres:

$A=\epsilon bc.....\left(1\right)$

$\epsilon$is a proportionality constant called molar absorptivity.

$A=-{\log }_{10}\left(T\right)....\left(2\right)$

Step 3. Converting Units

Change ppm to mg/L as,

$$\begin{gathered} 6.4\mathrm{ppm}=6.4\mathrm{ppm}\times 1\frac{\mathrm{mg}/\mathrm{L}}{\mathrm{nnm}} \\ =6.4mg/L \end{gathered}$$

Now convert the weight of iron into concentration as,

$$\begin{gathered} 1\mathrm{mgFe}={10}^{-3}\mathrm{gFe} \\ 6.4\mathrm{mg}/\mathrm{L}=6.4\mathrm{mg}/\mathrm{L}\times {10}^{-3}\frac{\mathrm{g}}{\mathrm{mg}}\times \frac{1\mathrm{mol}}{55.85\mathrm{g}}\times \frac{2.50\mathrm{ml}}{50.0\mathrm{ml}} \\ =5.730\times {10}^{-6}\mathrm{mol}{\mathrm{L}}^{-1} \end{gathered}$$

Step 4. Finding absorbance

Now substitute all values in the equation $\left(1\right)$,

$$\begin{gathered} A=\left(7.00\times {10}^3\mathrm{L}{\mathrm{mol}}^{-1}{\mathrm{cm}}^{-1}\right)(2.50\mathrm{cm})\left(5.730\times {10}^{-6}\mathrm{M}\right) \\ =0.100 \end{gathered}$$