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Produce an experimental write up of an acid to alkali titration using two solutions.

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Introduction

Introduction The aim of this assignment is to produce an experimental write up of an acid to alkali titration using two solutions. The solutions that are used are at a specific strength, this strength is 2mol. The solutions are hydrochloric acid (HCl) and sodium hydroxide (NaOH). The experimental aim was to find out what quantity of hydrochloric acid is needed to exactly neutralise the sodium hydroxide. Both the solutions are at different sides of the pH scale. The hydrochloric acid has a pH of 2, which would be classed as a very strong acid while the sodium hydroxide has a pH of 12, which is classed as very strong alkali. The pH scale is a rough scale to quickly show how acidic or alkali a solution can be. The 'p' means potenz (potential to be) while the 'H' means hydrogen. An element that contains lots of hydrogen atoms makes the solution acidic, but when there are hardly any or no hydrogen atoms in a solution this would make the solution an alkali. However when there is a right balance between the acids and the alkalis it results in the in between solution being neutral where the solution is not corrosive to any material or solution, a prime example of this is distilled water. ...read more.

Middle

Prepare the solution to be analysed by placing it in a clean Erlenmeyer flask or beaker Use the buret to deliver a stream of titrant to within a couple of cm3 of your expected endpoint. You will see the indicator change colour when the titrant hits the solution in the flask, but the colour change disappears upon stirring. Approach the endpoint more slowly and watch the colour of your flask carefully. Use a wash bottle to rinse the sides of the flask and the tip of the buret; to be sure all titrant is mixed in the flask. As you approach the endpoint, you may need to add a partial drop of titrant. You can do this with a rapid spin of a Teflon stopcock or by partially opening the stopcock and rinsing the partial drop into the flask with a wash bottle. Make sure you know what the endpoint should look like. If you think you might have reached the endpoint, you can record the volume reading and add another partial drop. Sometimes it is easier to tell when you have gone past the endpoint. When you have reached the endpoint, read the final volume in the buret and record it in your notebook. ...read more.

Conclusion

= 2 x 11.83 = 0.24mol 100 So working down to a simple expression it would turn out to be that 25cm3 of NaOH contains 0.05mol of NaOH. While 1 cm3 contains 0.05mol divided by 25 = 0.002mol of NaOH. Overall if I wanted to find out how much molar there is in 1 dm -3 (This basically means 1 litre) you would calculate 0.05 * 1000 = 2mol 25 Conclusion Overall the experiment could have been improved by including the use of a thermometer to measure the reaction between the two solutions. The experiment went well and the stronger calculations were easy to work out when the right results were shown. Overall the experimental hypothesis was proven to be successful without any problems to cause the results to be contaminated. The explanation of the titrant I believe has been fully explained to show people how to do this experiment without any scientific knowledge and allow them to get similar results to the ones found in the results page. However the experiment could of gone a lot better if the temperate of the room was a bit lower to rule out any tampering of the experiment. I am pleased with the overall results and I would like to do the experiment again with the additional equipment to help get use to the scientific area and atmosphere once again. ...read more.

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