Question

You wish to measure the iron content of the well water on the new property you are about to buy. You prepare a reference standard Fe3 solution with a concentration of 5.17 Ă— 10-4 M. You treat 13.0 mL of this reference with HNO3 and excess KSCN to form a red complex, and dilute the reference to 45.0 mL. The diluted reference is placed in a cell with a 1.00-cm light path. You then take 30.0 mL of the well water, treat with HNO3 and excess KSCN, and dilute to 100.0 mL. This diluted sample is placed in a variable pathlength cell. The absorbance of the reference and the sample solutions match when the pathlength is 4.78 cm. What is the concentration of iron in the well water? For each solution, the zero is set with a blank.

Answer 1

1.04 ⨯ 10$^{-4}$ M

Explanation

A = ε $\cdot l \cdot c$ by the Beer-Lambert law, where

• A the absorbance,
• $l$ the path length,
• ε the molar absorptivity of the solute, and
• $c$ concentration of the solution.

A and ε are the same for both solutions. Therefore, $l \cdot c$ is constant; $l$ is inversely proportional to $c$. The 100 mL sample would have a concentration 1/4.78 times that of the 45.0 mL reference.

The 13.0 mL standard solution has a concentration of 5.17 ⨯ $10^{-4}$ M. Diluting it to 45.0 mL results in a concentration of $5.17 \times 10^{-4} \times \frac{13.0}{45.0} =$ 1.494 M.

$c$ is inversely related to $l$ for the two solutions. As a result, c₂ = $c_1 \cdot \frac{l_1}{l_2} = 1.494 \times 10^{-4} \times \frac{1}{4.78} =$ 3.126 M.

The 30.0 mL sample has to be diluted by 30.0 / 100.0 times to produce the 100.0 mL solution being tested. The 100.0 mL solution has a concentration of 3.126 M. Therefore, the 30.0 mL solution has a concentration of $3.126 \times \frac{100.0}{30.0} =$ 1.04 ⨯ $10^{-4}$ M.