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Solubility of Ca(OH)2

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IB Chemistry IA: Solubility of Ca(OH)2 Data Collection and Processing Table of results to show the amount of HCl(0.053mol/dm3) needed to neutralize 25cm3 of Ca(OH)x Rough Titration 1 Titration 2 Titration 3 Initial Volume of HCl +0.05(cm3)(2dp) 0.00 20.10 0.00 23.25 Final Volume of HCl +0.05 (cm3)(2dp) 20.10 42.50 23.25 45.75 Difference +1.00 (cm3)(2dp) 20.10 22.40 23.25 22.50 Average value of HCl needed to neutralize 25cm3 of Ca(OH)x: (22.40 + 23.25 + 22.50)/3 = 22.72 cm3 To obtain the amount of HCl required to neutralize the limewater, a qualitative observation was observed. The solution of the Ca(OH)2 with the pink phenolphthalein indicator turned colorless when a certain amount of HCl (0.053mol/dm3) ...read more.


Mass = moles x molar mass Mass of Ca(OH)2 = 0.0006 x [40.1+2(16+1)] = 0.0006 x 74.1 = 0.0445g Since this is the value of the mass of Ca(OH) 2 in 25cm3, and we want to compare our value with a literature value that is in g/100cm3 t we multiply this value by 4. Concentration of Ca(OH)2: 0.0445 x 4 = 0.1778g/100cm3 = 0.18g/100cm3 (2d.p) Percentage Errors Concentration of Ca(OH)2: [(Estimated value - literature value) /literature value] x 100 = (0.18 - 0.17)/0.17 x 100 = 5.9% (only 1 d.p. since % error is >2%) ...read more.


= 5.9 + 0.60 + 0.66 = + 7.16% Range of uncertainty on the Ca(OH)2:7.16% of 0.18 = 0.012 The concentration of Ca(OH)2 should be 0.18 + 0.012 = 0.192g/100cm3 or 0.18 - 0.012 = 0.168g/100cm3 The concentration can therefore be expressed as 0.18 + 0.012 g/100cm3 Conclusion The aim of this experiment was to find the solubility of Ca(OH)2 from a titration with HCl(0.053mol/dm3.) The value of Ca(OH) 2 obtained in the experiment was 0.18 + 0.012 g/100cm3, which is approximately 0.1g off from the literature value of Ca(OH)2 that is 0.17g/100cm3. The total percentage error was relatively significant with a value of +7.16%. Inability to attain the exact literature value could have been due to several chemical, human, and equipment errors. ...read more.

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