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# The Percentage Purity of an Unknown Acid.

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Introduction

Middle

in M) to calculate the acids concentration. * V1C1 = V2C2 * (25 ? 1000) ? (0.15) = (22.325 ? 1000) ? C2 * C2 = (25 ? 0.15) ? (22.325) * C2 = 0.168M The concentration of the acid is therefore 0.168M, which can then be used further to calculate the concentration of the acid in grams per dm3. * Mass = No. of Moles ? Relative Molecular Mass * Mass = (0.168) ? (163) ? 27.384 g/dm3 Though as he was only using 250cm3 then this value must be divided by four. * (27.384) ? (4) ? 6.846 g/dm3 From this I am then able to calculate the true percentage purity of the unknown acid. * (Calculated mass ? Original mass) ? 100 = ......? * ( 6.846 ? 7.1 ) ? 100 = 96.4? By this method the percentage purity was found to be 96.4? and not 99.1?. Apparatus error * Balance ? 0.1g ? (0.1 ? 7.1) ...read more.

Conclusion

The students titration technique was also at fault with him only filling the burette correctly twice out of five titrations. There is nothing mentioned either about him rinsing the burette or the pipette beforehand with the acid and sodium hydroxide respectively. This should always be done before attempting a titration to remove any impurities from the apparatus. After examining the students method I'm not that surprised my calculated percentage purity was a bit off as there were plenty of things that could have caused inaccuracy in the experiment. If the student was to attempt the whole thing again I would urge him to listen to the improvements listed below: * Use a balance that reads to two decimal places instead of one. * Weigh the bottle and acid together, then the bottle by itself and take the difference. * Make sure all washings are transferred to the volumetric flask. * When filling the burette make sure to fill it right to the top (0.00cm3) each time. ...read more.

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