Chapter 14: Q. 14.1 (page 353)

A $25.0 - m L$ aliquot of an aqueous quinine solution was diluted to $50.0 m L$ and found to have an absorbance of $0.636$ at $348$ nm when measured in a $2.50 - m L$ cell. A second $25.0 - m L$ aliquot was mixed with $10.00$ mL of a solution containing $23.1 - m L$ ppm of quinine; after dilution to $50.0$ mL, this solution had an absorbance of $0.903 (2.50 - c m$ cell). Calculate the concentration of quinine in parts per million in the sample.

Short Answer

Expert verified

The percentage of copper is $0.0795 %$

Step by step solution

step 1. Theory

In analytical chemistry, the standard addition method is most commonly employed, in which the standard is introduced directly to sample aliquots of equal volume. It's used to figure out how much of an unknown solution there is. Before being measured, each solution is diluted to a certain volume. It's very useful for examining complicated samples. Complex samples are exemplified by the matrix effect. The following is a representation of the standard addition method:

$c_{x} = \frac{A_{x} c_{s} V_{s}}{(A_{x + s} - A_{x}) V_{x}}$

Here, $A_{x}$ represents the absorbance of an unknown concentration $(c_{x}), c_{s}$ represents the concentration of a known solution, $V_{x}$ represents the volume of an unknown solution with a concentration of $(c_{x}), V_{s}$ represents the variable volume, and $A_{x + s}$ represents the absorbance of a known concentration $(c_{s})$

Step 2. Explanation

The absorbance of unknown solution is,

$A_{x} = \frac{ε b c_{x} V_{x}}{V_{1}} . . . . . (1)$

Here, $A_{x}$ represents the absorbance of an unknown concentration $(c_{x}), V_{x}$ represents the volume of an unknown solution with a concentration of $(c_{x}), V_{1}$ represents the entire volume of the solution, and $c_{x}$ represents the concentration of an unknown solution.

The following is the absorbance of a given solution at a known concentration:

$A_{x + s} = \frac{ε b (c_{x} V_{x} + c_{x} V_{s})}{V_{1}} . . . . . . (2)$

Here, $A_{x + s}$ is equal to the absorbance of known concentration $(c_{x}), V_{x}$ is the volume of unknown solution with unknown concentration $(c_{x}), V_{1}$ is the total volume of the solution, $c_{x}$ is concentration of unknown solution, $c_{s}$ is the concentration of known solution, $V_{s}$ is the variable volume.

Divide equation $(1)$ by equation $(2)$ ,

$\frac{A_{x}}{A_{x + s}} = \frac{c_{x} V_{x}}{c_{x} V_{x} + c_{s} V_{s}} . . . . . . (3)$

Rearrange the above equation,

$c_{x} = \frac{A_{x} c_{s} V_{s}}{(A_{x + x} - A_{x}) V_{x}} . . . . . . (4)$

Step 3. Solution

$c_{{Cu}^{2 *}} = \frac{(0.671) (2.50 ppm) (1.00 mL)}{(0.849 - 0.671) (5.00 mL)} = 1.8848 p p m$ Substitute the values in equation $(4)$ ,

Now convert it into percent as,

$% Z = (x mg / L) (10^{- 3} \frac{g}{mg}) (10^{- 3} \frac{L}{mL}) (\frac{100 %}{m}) % Cu = (200.0 ml) (1.8848 mg / L) (10^{- 3} \frac{g}{mg}) (10^{- 3} \frac{L}{ml}) (\frac{100 %}{0.474 g}) = 0.0795 %$