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Photograph illustrating a lab setup with various glassware containing colored solutions, each labeled with different concentrations (1 ppm, 2 ppm, 5 ppm, 10 ppm, 20 ppm, 30 ppm) derived from a 250.0 ppm stock solution for a spectrophotometry experiment.

Standard Solution Preparation for Spectrophotometric Analysis

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Question: Prepare the following 100 ml standard solutions from 250.0ppm stock solution: 1,2,5,10,20 and 30 ppm. You will be issued with a solution of unknown Fe2+ and/or Cu2+ concentrations. A demonstrator will measure the absorbance of each solution for you. Measurements will be done at 324.7 Cu and 248.3 nm for Fe. WRITE IN PAST TENSE

Asked by mahlatsi

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To prepare 100 ml standard solutions of 1, 2, 5, 10, 20, and 30 ppm from a 250.0 ppm stock solution, I followed these steps: 1. **Preparation of 1 ppm Solution:** - I calculated the required volume of the 250.0 ppm stock solution using the dilution formula: \( C_1V_1 = C_2V_2 \). - For 1 ppm: \( V_1 = \frac{(1 \text{ ppm})(100 \text{ ml})}{250 \text{ ppm}} = 0.4 \text{ ml} \). - I pipetted 0.4 ml of the 250.0 ppm stock solution into a 100 ml volumetric flask. - I filled the flask up to the mark with distilled water and mixed thoroughly. 2. **Preparation of 2 ppm Solution:** - For 2 ppm: \( V_1 = \frac{(2 \text{ ppm})(100 \text{ ml})}{250 \text{ ppm}} = 0.8 \text{ ml} \). - I pipetted 0.8 ml of the 250.0 ppm stock solution into a 100 ml volumetric flask. - I filled the flask up to the mark with distilled water and mixed thoroughly. 3. **Preparation of 5 ppm Solution:** - For 5 ppm: \( V_1 = \frac{(5 \text{ ppm})(100 \text{ ml})}{250 \text{ ppm}} = 2 \text{ ml} \). - I pipetted 2 ml of the 250.0 ppm stock solution into a 100 ml volumetric flask. - I filled the flask up to the mark with distilled water and mixed thoroughly. 4. **Preparation of 10 ppm Solution:** - For 10 ppm: \( V_1 = \frac{(10 \text{ ppm})(100 \text{ ml})}{250 \text{ ppm}} = 4 \text{ ml} \). - I pipetted 4 ml of the 250.0 ppm stock solution into a 100 ml volumetric flask. - I filled the flask up to the mark with distilled water and mixed thoroughly. 5. **Preparation of 20 ppm Solution:** - For 20 ppm: \( V_1 = \frac{(20 \text{ ppm})(100 \text{ ml})}{250 \text{ ppm}} = 8 \text{ ml} \). - I pipetted 8 ml of the 250.0 ppm stock solution into a 100 ml volumetric flask. - I filled the flask up to the mark with distilled water and mixed thoroughly. 6. **Preparation of 30 ppm Solution:** - For 30 ppm: \( V_1 = \frac{(30 \text{ ppm})(100 \text{ ml})}{250 \text{ ppm}} = 12 \text{ ml} \). - I pipetted 12 ml of the 250.0 ppm stock solution into a 100 ml volumetric flask. - I filled the flask up to the mark with distilled water and mixed thoroughly. Once the standard solutions were prepared, the demonstrator measured the absorbance of each solution at 324.7 nm for Cu and 248.3 nm for Fe. These measurements were used to determine the concentrations of Fe2+ and Cu2+ in the unknown solution by comparing the absorbance values with those of the standard solutions.

Answered by gpt-4o

Aug. 7, 2024, 12:59 p.m.

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This article was generated in part by one of OpenAI’s large-scale language-generation model, answering user mahlatsi's question. Images were generated with OpenAI's AI model, DALL·E 3. The Internet takes ultimate responsibility for the content of this publication.
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Published: Wednesday, August 7, 2024

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