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# Determination of the Amounts of Sodium Hydroxide Solution and Sodium Carbonate in a Mixed Solution

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Introduction

Experiment 2 Determination of the Amounts of Sodium Hydroxide Solution and Sodium Carbonate in a Mixed Solution Date of Experiment: 21/9/2005 Objective: To determine the percentage by mass of Sodium Hydroxide and Sodium Carbonate in a mixed solution. Introduction Sodium carbonate and sodium hydroxide are both alkaline in nature. When hydrochloric acid is added to the mixture, sodium hydroxide reacts like this: NaOH + HCl -----> NaCl + H2O ....................... (1) Sodium carbonate reacts with hydrochloric acid in a similar way but the reaction comes in 2 steps: Na2CO3 + HCl ----------->NaHCO3 + NaCl............. (2) NaHCO3 + HCl ----------->NaCl + H2O + CO2.......... (3) The pH conditions in which reactions (1) and (2) take place are approximately the same: they both occur at a relatively higher pH (at a pH value higher than 7). Reaction (3) is quite different from the other 2 reactions. It takes place at a lower pH (at a pH value lower than 7). So, using the 2 acid-base indicators, the end points for the reactions can be found. Phenolphthalein turns from pink to colorless as the pH value changes from 10 to 8. Methyl orange turns from yellow to orange when the pH value drops from about 4 to 3. ...read more.

Middle

So, there are also 5.033 X 10-4 moles of Na2CO3 in the experiment. Mass of Na2CO3 in the experiment: 5.033 X 10-4 X (23 X 2 + 12 + 16 X 3)= 0.5335g Mass of NaOH in the experiment: (1.3651 X10-3 - 5.033 X 10-4) X (23 + 17)= 0.03442g So, the percentages by mass of NaOH and Na2CO3 are 39.15% and 60.85% respectively. Method 2 A table showing the volume of HCl used for reacting both Na2CO3 and NaOH 1 2 3 Final reading 22.6 cm3 40.8 cm3 19.7 cm3 Initial reading 3.1 cm3 21.6 cm3 0.5 cm3 Volume of HCl used 19.5 cm3 19.2 cm3 19.2 cm3 (Due to time limitations, only 3 titrations could be done and the first titration is not rejected) Average volumes of HCl used = (19.5 + 19.2 X 2)/ 3 = 19.3 cm3 No of moles of HCl used = 19.3/1000 X 0.09868 = 1.905 X 10-3 moles A table showing the volume of HCl used for reacting NaOH. 1 2 3 Final reading 27.7 cm3 35.5 cm3 43.6 cm3 Initial reading 19.7 cm3 27.7 cm3 35.5 cm3 Volume of HCl used 8 cm3 7.8 cm3 8.1 cm3 (Due to time limitations, only 3 titrations could be done and the first titration is not rejected) ...read more.

Conclusion

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