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The purpose of this experiment is to find the composition of a sample of sodium carbonate mixture by titration.

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Name : CHAN CHI HEI Class: 6LS Class no :1 Title: Double titration Date of experiment : Aim The purpose of this experiment is to find the composition of a sample of sodium carbonate mixture by titration. Theory Double indicator titration is used in checking the composition of a mixture of sodium carbonate and sodium hydrogen carbonate. First, no of mol of Na2CO3 in mixture can be directly calculated in the data of first titration. For the calculation of no of mol of NaHCO3 need to involve the data from both 1st and 2nd data , because the first titration will give out NaHCO3 also and it will contaminated with the NaHCO3 in the mixture. Requirements watch glass weighing bottle spatula sodium carbonate Na2CO3 250 cm3 beaker, wash bottle of distilled water glass rod 250 cm3 volumetric flask filter funnel, dropper Procedure 1. 10.00 g of the sodium carbonate is transferred onto the watch glass and weighted in the nearest 0.01g 2. A standard solution is prepared. 3. Burette is rinsed with HCL and then filled with the acid. ...read more.


used(I)/cm3 7.28 Methyl Orange Trial 1st 2nd 3rd Final Burette Reading(II)/cm3 49.4 34.0 34.2 34.0 Initial Burette Reading(II)/cm3 30.1 12.5 12.5 12.05 Volume used(II)/cm3 19.3 21.5 21.7 21.95 Mean Volume of HCl(aq) used(II)/cm3 21.7 First stage of titration Na2CO3(aq) + HCl(aq) >> NaHCO3(aq) + NaCl(aq) Since HCL : Na2CO3 = 1:1 No of mol of Na2CO3 = Mean volume 1 of HCL used x Concentration of HCL = 7.28 /1000 x 0.5 = 3.64 x 10-3 mol Second stage of titration NaHCO3(aq) + HCl(aq) >> NaCl(aq) + H2O(l) + CO2(g) No of mole of NaHCO3 = Mean volume 2 of HCL used x Concentration of HCL = 21.7/1000 x 0.5 =0.01085 mol Calculation From above , No of mole of Na2CO3 in 250 cm3 solution = 3.64 x 10-3 mol No of mole of NaHCO3 in the 250cm3 solution = (No of mole of NaHCO3 in second stage) - (No of mole of NaHCO3 in first stage) = 0.01085 - 3.64 x 10-3 = 7.21 x 10-3 mol Mass of Na2CO3 = 0.0364 x 106 = 3.8584 g Which is around 3.9 g Mass of NaHCO3 = 0.0721 x 84 = 6.0564 g ...read more.


After that , methyl orange is used because NaHCO3 is a weak base . It changes from red below pH 3.1 to yellow above pH 4.4. Therefore , in the second part of titration, when the indicator turn from yellow to orange, the solution is in acidic but not exactly neutral . Therefore , the result obtained from above is not accurate. If we want to get a more accurate result , ph meter should be used but not indicator. The calculation above has a little bit complicated. Since Na2CO3 only involve in the first titration , no of mol in Na2CO3 will exactly equal to the no of mol of HCL used. After that , we need to calculate No of mole of NaHCO3 in the mixture by using the (No of mole of NaHCO3 in second stage) - (No of mole of NaHCO3 in first stage) because the No of mol of NaHCO3 in second stage contain both NaHCO3 from the mixture and that from the first titration. Thus , we need to minus the no of mole of NaHCO3 in first stage to obtain the no of mol of NaHCO3 in the mixture. ...read more.

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