Titration experiment. Standardization of hydrochloric acid using sodium carbonate.
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Name : AMirul SYafiq Title : Standardization of hydrochloric acid using sodium carbonate. Variables : a) dependant : Volume of hydrochloric acid. b) independant : The colour of the solution. Results : 1) Preparation of the dilute HCl. a) Rough molarities:11M b) Volume in the measuring cylinder : 2.0 + 0.1cm³ c) Volume in the volumetric flask : 250 + 0.25cm³ 2) Preparation of the Na2CO3 solution. a) Mass of pure Na2CO3 : 1.329 + 0.0001g b) Volume in the volumetric flask : 250 + 0.25 cm³ c) ...read more.
3 Uncertainty : | 20.5-20.7 | + | 20.5-20.6 | + | 20.5-20.2 | x 100 = 2.9% 20.5 Gross volume of HCl used with percentage of uncertainty = 20.50 cm³ + 2.9% cm³. The molarities of HCl after dilution : M1V1 = M2V2 (2mL) (11M) = (X) (250mL) X = 0.088 M Uncertainty : V1 = 2.0 mL + 0.5 mL V2 = 250.0 mL + 0.5 mL = 0.5 x 100% = 0.5 x 100% 2.0 250.0 = 25 % = 0.2 % Then, add together to form general uncertainty = 25 % + 0.2 % = 25.2 %. ...read more.
of moles Volume 0.0125 = 0.05 M. 0.25 In this reaction, limiting reagent = Na2CO3 Theoretical moles of Na2CO3 solution : Moles = Volume x Molarities = (25cm³) (0.088M) 1000 = 0.0022 mole. Actual moles of Na2CO3 used when titration : Moles = 0.088 2 = 0.044 mole. So the theoretical volume of HCl needed to react with 0.025 mole of Na2CO3 : MV of Na2CO3 = (0.05) (25) = 1.25 (2M) = 2.5 . 1000 = 28.41 cm³. Percentage error : 28.41cm³-20.5cm³ x 100% = 27.8423% 28.41cm³ Conclusion : The actual volume of HCl needed to neutralize the Na2CO3 is 28.41cm³ for the titration. ...read more.
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