Introduction: The aim of the experiment is to find the exact concentration of the hydrochloric acid (HCl) solution by reacting (titrating) it with the standard base sodium hydroxide (NaOH) solution. Titration is the name of the analytical process that is used to determine the concentration of an unknown solution by reacting it with a standard solution (known concentration). The known concentration solution is called the titrant, and the unknown concentration solution is called the analyte. The purpose of using HCl and NaOH is because they are strong acid and base, which means they will fully dissociate in water (will separate to ions). This is important because when the reaction reaches the equivalence point H and OH ions will combine together
Figure 2 (a), (b) and (c) Show the average particle sizes of HAp synthesized via co-precipitation, sol-gel and solid state reaction method respectively. The particle size found for solid state reaction method is 94.1 nm which very low comparing to the other two chemical methods. This could be the reaction at high temperature were leads to agglom-erations for the chemical methods, because of the chemical precursors reduced HAp powder tends to agglomeration. But comparing to these the solid state derived HAp is far better that has not got ag-glomerated. 3.2 UV ANALYSIS The spectra of ultraviolet light for the three synthe-sized of HAp solutions are measured by UV spec-trometer and the results are shown in Fig.2.
(2) Titration of Acetic Acid with Sodium Hydroxide 10ml of distilled water was added into a 250ml Erlenmeyer flask. 20ml of diluted acetic acid was then added into it, followed by setting up a titration system with 0.100M of NaOH in buret. A pH meter was used to monitor the pH of the solution as the base (NaOH) was added. Sodium hydroxide was added in an increment of 1ml until the solution pH reached 4.8. After the solution pH reached, the base was added in an increment of 0.2ml until the equivalence point was passed.
Degradation study of Product 01 using Aqueous 1N NaOH solution .The mechanism is operated by hydrolysis. The hydroxyl group (-OH) of NaOH attacks an electrophilic carbon of >C=O group which an removal of tertiary Nitrogen gives 4-MBA and PD as by products. Degradation study of Product 02 using Aqueous 1N NaOH solution . The mechanism is operated by hydrolysis. The hydroxyl group (-OH) of NaOH attacks an electrophilic carbon of >N-C=O which as rearrangement gives carbonial .
If the pH is less than 6.0, it must be boiled for 15 minutes and cooled at room temperature. (c) Sodium hydroxide solution (0.02 N): 1N sodium hydroxide solution is prepared by dissolving 40 g of L.R grade NaOH in distilled water and diluting it to 1000 ml. For the preparation of 0.02 N NaOH solution, take 20 ml of IN NaOH and dilute it to 1000 ml. 0.02 N NaOH can be standardize against standard potassium acid phthalate. 1.0 ml 0.02 N NaOH = 1.0 mg CaCO3 (d) Phenolphthalein indicator: Dissolve 0.5 g of phenolphthalein in 100 ml alcohol and water.
Oxidation Number One of the less known examples of electron transfer is when hydrogen and oxygen combine to form water. All of us know the reaction Even though it is not a very common reaction but we can notice that the H atom in the neutral state (zero) in H2 converts to positive state after the formation of H2O. Similarly, the oxygen atom in O2 is present in its zero
This ensure one to have a healthier lifestyle. This experiment determines the amount of sodium hydroxide, NaOH that needed to neutralize a certain measurement of acidity in fatty acid of a two gram of refined oil sample. NaOH played the role as a catalyst in this experiment. A catalyst is used to reduce the amount of the remains and help produce a larger yield of product. Furthermore, isopropanol that was added into the flask containing the refined oil sample is used to dissolve the oil.
10ml of the sample was added to the 0.05 ml of 0.5% phenolphthalein indicator. It was mixed and allowed to stand for few minutes and was neutralized with 0.1M NaOH to the standard pink colour. 2 ml of formalin was added, mixed and allowed to stand for few minutes. The new acidity produced was titrated against 0.1M NaOH for the same pink colour. Then 2ml of the formalin and 10ml of H2O were titrated separately against 0.17M NaOH as blank.