Monday, January 6, 2014

Acid Base Titration

Chemis probe lab report 28th of Sep 2009/ Monday Aya Naji Shnawa try 1.3 An acid base titration *data collection and processing * determination and evaluation DATA COLLECTION AND PROCESSING: Readings initiative spark invoke 2nd trace3rd trailquaternary trailAverage Initial ±0.05cm 30.00.00.00.00.0 Final ±0.05cm 320.521.521.021.121.0 Table 1.1: a first appearance of processed data in all the trails. ________________________________________ The investigatesThe remnant 1st to 2nd 1.0 2nd to 3rd 0.5 3rd to 4th 0.1 Table1.2: the oddment of opinion of the concluding readings between the experiments ________________________________________ The average Standard excursus 21.0 0.4 Table1.3: the residuum between average and standard deviation. ________________________________________ The sum of the difference The average 1.60.5 Table 1.4 the average of the difference. PRESENTATIONN OF DATA: Chart 1.1: the last(a) reading s of each of the four trails. ________________________________________ Chart 1.2: the difference between the readings in each of the trails. Chart 1.3: the difference between the sum and the average.
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mop up picture AND EVALUATION CONCLUSION: Equivalence point is the point at which the moles of H+ is equal to the moles of OH+,an indicator is used to guide the comparison point during a titration. in a titration the regularity is around totaling one reactant from the burette (regularly the acid),to a known hatful of the some other reactant in a conical flask(regularly the base) . In order to hap the concentration of NaOH we need to tag on the following stairs: a-note down the balanced chemical equating ! for the reply C8H5O5K+NaOH?C8H4O4KNa+H2O b-pull out applicable information from the experiment: C8H5O5K V=0.025 dm3 C=0.2M NaOH V=0.021 dm3 C= ?? c-calculate number of moles n(C8H5O5K)=VxC = 0.2x0.025 = 0.005 moles . d- bet moles ratio from the equation : 1:1 e-find moles...If you demand to set forth a full essay, order it on our website: OrderEssay.net

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