Chapter 18: Q14 P (page 456)

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 $6.80 \times 10^{- 4} M {Fe}^{3 +}$ was treated with ${HNO}_{3}$ 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?

Expert verified

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

Step by step 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.

$[Fe] = \frac{10}{50} \cdot 6.8 \cdot 10^{- 4} M [Fe] = 1.36 \cdot 10^{- 4} M$

As,

$ε_{s} - c_{s} \cdot b_{s} = ε_{r} \cdot c_{r} \cdot b_{r} c_{5} \cdot b_{s} = c_{r} \cdot b_{r}$

Hence,

$c_{s} = \frac{c_{r} \times b_{r}}{b_{s}} c_{s} = \frac{1.36 \cdot 10^{- 4} M \times 1 cm}{2.48 cm} c_{s} = 5.48 \cdot 10^{- 5} M$

If the dilution is done $\frac{1}{4} c_{s}$ ,

$c_{s} = 5.48 \cdot 10^{- 5} M \times 4 c_{s} = 2.19 \cdot 10^{- 4} M$

Unlock Step-by-Step Solutions & Ace Your Exams!

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.