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The rate of reaction between sodium thiosulfate and hydrochloric acid

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Introduction

DESIGN * Research question: Does the change in concentration of sodium thiosulfate and the fixed concentration of hydrochloric acid result a change in time taken for the yellow sulfur precipitate to form, thus lead to a change in time taken for the cross to disappear and the rate of reaction? * Variables: * Independent variable: The concentration of sodium thiosulfate / M. * Dependent variable: The time taken for the cross to disappear / second. * Controlled variables: * The concentration of hydrochloric acid / M. * The temperature in each conical (Erlenmeyer) flask prior to every reaction / oC. * The absence of unnecessary substances or ions. * The angle to view the cross. * Prediction: * For many reactions involving liquids or gases, increasing the concentration of the reactants increases the rate of reaction. In order for any reaction to happen, particles must first collide. This is true whether both particles are in solution, or whether one is in solution and the other a solid. If the concentration is higher, the chances of collision are greater [1]. * In the reaction between sodium thiosulfate solution and dilute hydrochloric acid, yellow sulfur (S(s)) ...read more.

Middle

each time to watch the cross disappearing. Making different angles could result uncertainties for when to stop the watch. DATA COLLECTION AND PROCESSING * Raw data table: Na2S2O3 (� 0.0500 ml) H2O (� 0.0500 ml) 1. Time taken (� 0.0100 s) 2. Time taken (� 0.0100 s) 3. Time taken (� 0.0100 s) 10.00 40.00 252.0 261.0 265.0 15.00 35.00 140.0 138.0 142.0 20.00 30.00 101.0 104.0 103.0 25.00 25.00 79.00 74.00 77.00 30.00 20.00 62.00 66.00 68.00 35.00 15.00 54.00 58.00 57.00 40.00 10.00 47.00 50.00 47.00 45.00 5.000 44.00 43.00 42.00 50.00 0.000 36.00 37.00 35.00 Table 2.1 shows the collected raw data table. * Processed data: * Calculating the concentration of sodium thiosulfate in each sample: * Formula: Concentration = 0.1 x (Volume of sodium thiosulfate 0.100 M divided by Volume of the solution). Multiplying 0.1 is because the given sodium thiosulfate is 0.1000 M. Volume of sodium thiosulfate 0.10 M (� 0.0500 ml) Volume of water (� 0.0500 ml) Volume of the solution (� 0.1000 ml ) Concentration / M 10.00 40.00 50.00 0.0200 15.00 35.00 50.00 0.0300 20.00 30.00 50.00 0.0400 25.00 25.00 50.00 0.0500 30.00 20.00 50.00 0.0600 35.00 15.00 50.00 0.0700 40.00 10.00 50.00 0.0800 45.00 5.000 50.00 0.0900 50.00 0.000 50.00 0.1000 Table 2.2 shows the processed different concentrations of sodium thiosulfate used. ...read more.

Conclusion

than other values. * There is irritating odour familiar as the smell of a just-struck match. Although it provides the qualitative observation that supports the hypothesis, the odour takes very long time to be completely removed by the air ventilator. * Improving the investigation: * The procedures can be partially replaced by computer data logging suggested by Laurence Rogers (1995) [6] to prevent uncertainties from human errors when stopping the watch. The experiment can be programmed to collect the data (Time taken for the cross to disappear) automatically. * The consequence of very low concentration of sodium thiosulfate (for instance, Na2S2O3 0.02 M and 0.03 M) is, these concentration value although still support the hypothesis (lower concentration of sodium thiosulfate -> more time taken for the cross to disappear), do not fit the expected very strong negative correlation between the concentration of sodium thiosulfate and the mean time taken for the cross to disappear. Attention therefore needs to be paid in the ranges from Na2S2O3 0.04 M to 0.1 M. The ranges are therefore can be altered to 7, from 0.04 M , 0.05 M , to 0.09 M, 0.1 M. * The enclosed bung can be used to cover the lip of the conical flask to prevent the release of sulfur dioxide gas. ...read more.

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