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Investigation into the volume of acid needed to neutralise an alkali.

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Introduction

Investigation into the volume of acid needed to neutralise an alkali In this investigation I am going to investigate the volume of different concentrations of Sulphuric acid (H2SO4) needed to neutralise 25cm� of 0.1M Sodium Hydroxide (NaOH). I will do this by using titration. Background Knowledge When a substance/solution is alkaline it contains hydroxide ions, OH . When a substance/solution is acidic it contains hydrogen ions, H . Neutralization is when the OH ions neutralize the acid by removing the H ions and turning them into water. The equation for this is: A salt is always the end product of neutralization. The general equations for this are: Acid+ metal oxide-->metal salt+ water Acid+ metal hydroxide--> metal salt + water Acid+ metal carbonate--> metal salt + water+ carbon dioxide Acid+ metal hydrogen carbonate--> metal salt+ water + carbon dioxide The general equation for the reaction of my experiment is: Acid+ metal hydroxide--> metal salt+ water And the specific equation is: 2NaOH + H2SO4 --> Na2SO4 + 2H2O In order to be able to tell the exact point at which the solution becomes neutral, a colour change is needed and so an indicator is used. An indicator is a substance, which has a different colour in acid and alkali. For this experiment, I will be using Methyl Red. Methyl Red is yellow in alkali and bright pink in acid. I'm using this because it gives a good colour change. To neutralize an acid and alkali you can use a process called titration. Titration enables you to find out the exact volume of acid required to neutralize a certain volume of alkali. Using your results from titration, you can then use the equation to make a prediction. no. of moles = concentration (mol/dm�) x volume(v) (cm�) 1000 V to convert cm� into dm�. 1000 The ionic equation of the reaction gives a mole ratio of acid (H ): alkali ( OH ) as 1:1 so at the exact point of neutralization no. moles of acid= no. ...read more.

Middle

1000 because: cm� x mol so mol x 1000 dm� Another prediction I could make would be to predict the volume of acid needed to neutralize the solution and check the accuracy of my experiment. I can do this because I know the volume and concentration of alkali (which equals the number of moles) and so I could work out the number of moles for the acid. That would be 1/2 the number of moles of alkali because 2NaOH: H2SO4 2:1 The volume of the alkali is 25cm� and the concentration is 0.1M and so I worked out the volume of acid needed to neutralize the solution: No. moles of alkali No. moles of acid Concentration of acid mol/dm� Volume of acid needed/ cm� 25x0.1 = 2.5x 10�� 2.5x 10�� = 1.25 x10�� 0.1 1.25x 10�� =v x 0.1 12.50 25x0.1 = 2.5x 10�� 2.5x 10�� = 1.25 x10�� 0.08 1.25x 10�� =v x 0.08 15.63 25x0.1 = 2.5x 10�� 2.5x 10�� = 1.25 x10�� 0.06 1.25x 10�� = v x 0.06 20.83 25x0.1 = 2.5x 10�� 2.5x 10�� = 1.25 x10�� 0.04 1.25x 10 �� = v x 0.04 31.25 25x0.1 = 2.5x 10�� 2.5x 10�� = 1.25 x10�� 0.02 1.25x10 �� = v x 0.02 62.50 To help with my method I carried out some preliminary experiments. I titrated the highest and lowest concentrations to get an idea of the range of concentrations I wanted to use. I did this by: Method 1. Rinse the burette with a small amount of acid, then close the tap and fill above the 0.00mark. Run out the solution until there are no air bubbles and close the tap. Put the burette in the clamp stand and read the volume to 2decimal places. Record this reading as '1st burette reading' in the results table. 2. Rinse the pipette with a small amount of alkali. Then, depending on the pipette filler, press and hold 1 or A and press the bulb to expel the air, then let go. ...read more.

Conclusion

For the larger volume the percentage of error would be for example: 0.2 x 100 = 0.31% 63.50 Therefore I am more likely to get anomalous results for the smaller volumes of acid added because there is a larger percentage of error. My results were mostly accurate due to using the same solutions, which meant that the concentrations were consistent and using the same apparatus, which meant my experiment, was a fair test. My results on my 1st graph show the accuracy of the experiment because they are very close to the line of best fit. I thought my procedure was suitable. If I did the experiment again I would do repeat readings for the anomalous results I got. I would use a pipette or burette instead of a measuring cylinder to measure volumes when making up concentrations of acid because they would be more accurate and so the results would be more accurate. I would also give myself more time to do the experiment so I wouldn't feel like I was rushing and therefore make silly mistakes, which could contribute to anomalous results. I might use a different indicator to see if this made a difference to my results. Therefore I would change the equipment, not the method. These changes would make my results more reliable and accurate and therefore make my experiment better. To get additional evidence for my conclusion I could use smaller acid concentration intervals e.g. 0.1M, 0.95M, 0.9M. This would mean I had more points to plot on the graph and so the curve would be more definite and therefore emphasise my conclusion that as the acid concentration increases, the volume needed to neutralize the solution decreases. I could also do more concentrations (0.07M, 0.05M, 0.03M) to make the curve more definite or even higher concentrations to see if the curve leveled off on the graph e.g. 2M, 3M, 4M. I could use another acid, for example nitric acid, to see if my ionic equation for neutralization is correct. Rhiannon Knowles U5A Chemistry Coursework 2 October '03 ...read more.

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