Chapter 15: Q. 15.11 (page 386)

Iron(II) ions catalyze the oxidation of luminol by H₂O₂. The intensity of the resulting chemiluminescence has been shown to increase linearly with iron(II) concentration from 10^-10 to 10^-8 M.
Exactly 1.00 mL of water was added to a 2.00-mL aliquot of an unknown Fe(II) solution, followed by 2.00 mL of a dilute H₂O₂ solution and 1.00 mL of an alkaline solution of luminol. The chemiluminescence from the mixture was integrated over a 10.0-s period and found to be 12.7. To a second 2.00-mL aliquot of the sample was added 1.00 mL of a 3.27 x 10^-5 M Fe(II) solution followed by the same volume of H₂O₂ and luminol. The integrated intensity was 27.9. Find the concentration of Fe(II)
in the sample.

Short Answer

Expert verified

The concentration of Fe(II) in the sample is 2.73 x 10^-5M.

Step by step solution

Given Information

The concentration of Fe(II) in the sample should be determined.

Explanation

The expression of luminescent intensity of first sample is:

$L_{1} = \frac{k c_{x} V_{x}}{V_{1}}$ .......... (I)

Here, the proportionality constant is k , the volume of the sample is $V_{x}$ , the concentration of original sample is $c_{x}$ and the final volume of solution is $V_{1}$ .

The expression of luminescent intensity of second sample is:

$L_{2} = \frac{k c_{x} V_{x}}{V_{1}} + \frac{k c_{s} V_{s}}{V_{1}}$ .......... (II)

Here, the concentration of standard solution is $c_{s}$ and the volume of the standard solution is $V_{s}$ .

Divide Equation (II) by Equation (I).

localid="1646727380308" $\frac{L_{2}}{L_{1}} = \frac{k c_{x} V_{x} + k c_{s} V_{s}}{k c_{x} V_{x}} \frac{L_{2}}{L_{1}} = \frac{c_{x} V_{x} + c_{s} V_{s}}{c_{x} V_{x}} L_{1} c_{s} V_{s} = L_{2} c_{x} V_{x} - L_{1} c_{x} V_{x} L_{1} c_{s} V_{s} = c_{x} (L_{2} - L_{1}) V_{x} c_{x} = \frac{L_{1} c_{s} V_{s}}{(L_{2} - L_{1}) V_{x}}$

.....................(III)

Substitute 12.7 s for L₁, 3.27 x 10^-5 M for $c_{s}$ , 1 mL for $V_{s}$ , 27.9 s for L₂ and 1 mL for $V_{x}$ in Equation (III).

$c_{x} = \frac{(12.7 s) (3.27 \times 10^{- 5} M) (1 mL)}{(27.9 s - 12.7 s) (1 mL)} = \frac{(12.7 s) (3.27 \times 10^{- 5} M)}{(15.2 s)} = 2.73 \times 10^{- 5} M$