Just curious:
The melting point of pure benzoic acid and pure naphthol are 122.5 *C and 123 *C respectively. You are given a pure sample that is known to be either pure benzoic acid or 2- naphthol. Describe a procedure you will use to determine the identity of the sample.
To determine the identity of the sample (whether it is pure benzoic acid or 2-naphthol), you can utilize a technique called melting point determination. Here's a procedure you can follow:
1. Prepare two separate capillary tubes: One for benzoic acid and another for 2-naphthol. Make sure both capillary tubes are clean and dry.
2. Take a small amount of the unknown sample and carefully fill both capillary tubes to the same level. Ensure that you have a sufficient amount of the sample in each capillary tube.
3. Set up a melting point apparatus or a melting point apparatus adapter on a hot plate or heating device according to its instructions. Make sure to calibrate the apparatus.
4. Insert the capillary tube containing the unknown sample into the melting point apparatus, ensuring it is securely held in place. The capillary tube should be positioned so that the end with the sample is closest to the heat source.
5. Gradually heat the apparatus while observing the sample through the magnifying lens or the apparatus's thermometer (if it has one). Note the temperature range at which the sample melts or undergoes a phase change.
6. Repeat steps 4 and 5 for each of the reference capillary tubes containing known samples of benzoic acid and 2-naphthol. Note the melting temperature range for both of these known substances.
7. Compare the melting point range of the unknown sample with the melting points of benzoic acid and 2-naphthol. If the unknown sample melts within the range of 122.5-123°C, it is likely to be 2-naphthol. If it melts at 123°C or above, it is likely to be benzoic acid.
Remember that this technique only provides an indication of the possible identity of the substance. To conclusively confirm the identity, additional analytical techniques, such as spectroscopy, may be necessary.