Question

When I was a boy, Uncle Wilbur let me watch as he analyzed the iron content of runoff from his banana ranch. A 25.0 - mLsample was acidified with nitric acid and treated with excess KSCN to form a red complex. (KSCN itself iscolorless.) The solution was diluted to 100.0mLand put in a variable-pathlength cell. For comparison, a 10.0 - mLreference sample of$\mathbf{6}\mathbf{.}\mathbf{80}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{M}\mathbf{}{\mathbf{Fe}}^{\mathbf{3}\mathbf{+}}$was treated with ${\mathbf{HNO}}_{\mathbf{3}}$andand diluted to. The reference was placed in a cell with a 1.00 - mL light path. The runoff sample exhibited the same absorbance as the reference when the path length of the runoff cell was 2.48cm. What was the concentration of iron in Uncle Wilbur’s runoff?

Short answer

The concentration${\mathrm{C}}_{\mathrm{s}}$ of iron in Uncle Wilbur’s runoff is role="math" localid="1663651601222" $2.19\cdot {10}^{-4}\mathrm{M}$.

Step-by-step solution

The Concentration of a Solution.

The concentration of a solution is the amount of solute (small proportion) dissolved in the solvent (larger proportion) or solution.

A concentrated solution has a relatively large amount of dissolved solute.

A dilute solution is one that has a relatively small amount of dissolved solute.

The concentration of iron in the reference cell.

$$\begin{gathered} \left[\mathrm{Fe}\right]=\frac{10}{50}\cdot 6.8\cdot {10}^{-4}\mathrm{M} \\ \left[\mathrm{Fe}\right]=1.36\cdot {10}^{-4}\mathrm{M} \end{gathered}$$

The concentration of iron in Uncle Wilbur’s runoff.

As,

$$\begin{gathered} {\epsilon }_s-c_s·b_s={\epsilon }_r·c_r·b_r \\ {c}_5·b_s=c_r·b_r \end{gathered}$$

Hence,

$$\begin{gathered} {\mathrm{c}}_{\mathrm{s}}=\frac{{\mathrm{c}}_{\mathrm{r}}\times {\mathrm{b}}_{\mathrm{r}}}{{\mathrm{b}}_{\mathrm{s}}} \\ {\mathrm{c}}_{\mathrm{s}}=\frac{1.36\cdot {10}^{-4}\mathrm{M}\times 1\mathrm{cm}}{2.48\mathrm{cm}} \\ {\mathrm{c}}_{\mathrm{s}}=5.48\cdot {10}^{-5}\mathrm{M} \end{gathered}$$

The concentration of iron in Uncle Wilbur’s runoff after the dilution.

If the dilution is done $\frac{1}{4}{c}_s$,

$$\begin{gathered} {\mathrm{c}}_{\mathrm{s}}=5.48\cdot {10}^{-5}\mathrm{M}\times 4 \\ {\mathrm{c}}_{\mathrm{s}}=2.19\cdot {10}^{-4}\mathrm{M} \end{gathered}$$