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A sample sodium carbonate (Na2CO3)was claimed trace amount of sodium hydrogen carbonate (NaHCO3). When 0.293g sample titrated with 0.113M hydrochloric acid (HCl).It was found that:(1) If phenolphthalein was used as indicator, 23.90 cm3 of the acid was required to reach the end-point.(2) If methyl orange... 顯示更多 A sample sodium carbonate (Na2CO3)was claimed trace amount of sodium hydrogen carbonate (NaHCO3). When 0.293g sample titrated with 0.113M hydrochloric acid (HCl). It was found that: (1) If phenolphthalein was used as indicator, 23.90 cm3 of the acid was required to reach the end-point. (2) If methyl orange was used as indicator, it required 48.40 cm3 of the acid to complete the reaction. Calculate the percentage of sodium carbonate and the sodium hydrogen carbonate in the sample Thank you!
最佳解答:
For detailed explanation of the titration, please refer to the website (6.2.7b): http://www.docbrown.info/page07/equilibria6a.htm The titration involves 2 stages, CO32-(aq) + H+(aq) ==> HCO3-(aq) ( Stage 1 ) HCO3-(aq) + H+(aq) ==> H2O(l) + CO2(g) ( Stage 2 ) Stage 1 has an end-point pH around 9 and Stage 2 has an end-point pH around 3. So (1) can detect sodium carbonate and (2) can detect both sodium carbonate and sodium hydrogencarbonate. Using (1), n(HCl) = 0.113 x 23.90/1000 = 0.00270mol = n(Na2CO3) Mass of Na2CO3 = 0.00270 x 106 = 0.286g Mass of NaHCO3= 0.293 - 0.286 = 0.00673g This gives 97.7% Na2CO3 and 2.3% NaHCO3 Using (2), n(HCl) = 0.113 x 48.40/1000 = 0.00547mol Let mass of Na2CO3= x g and mass of NaHCO3= (0.293- x) g n(Na2CO3) + n(NaHCO3) = n(HCl) x/106 + [ x/106 + (0.293- x)/84 ] = 0.00547 On solving, x = 0.284 Mass of Na2CO3= 0.284g Mass of NaHCO3= 0.00913g This gives 96.9% Na2CO3 and 3.1% NaHCO3 Average the results, we have 97.3% Na2CO3 and 2.7% NaHCO3. 2009-12-21 18:39:26 補充： It is so tedious to deal with so many decimal places and thus I might hv rounded off the ans all the way down. So the resultant ans might be slightly deviated. But anyway, the sample has roughly 97% Na2CO3 and 3% NaHCO3. After all, the concept involved here is more important than the exact values.
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Titration calculation (chem)發問:
A sample sodium carbonate (Na2CO3)was claimed trace amount of sodium hydrogen carbonate (NaHCO3). When 0.293g sample titrated with 0.113M hydrochloric acid (HCl).It was found that:(1) If phenolphthalein was used as indicator, 23.90 cm3 of the acid was required to reach the end-point.(2) If methyl orange... 顯示更多 A sample sodium carbonate (Na2CO3)was claimed trace amount of sodium hydrogen carbonate (NaHCO3). When 0.293g sample titrated with 0.113M hydrochloric acid (HCl). It was found that: (1) If phenolphthalein was used as indicator, 23.90 cm3 of the acid was required to reach the end-point. (2) If methyl orange was used as indicator, it required 48.40 cm3 of the acid to complete the reaction. Calculate the percentage of sodium carbonate and the sodium hydrogen carbonate in the sample Thank you!
最佳解答:
For detailed explanation of the titration, please refer to the website (6.2.7b): http://www.docbrown.info/page07/equilibria6a.htm The titration involves 2 stages, CO32-(aq) + H+(aq) ==> HCO3-(aq) ( Stage 1 ) HCO3-(aq) + H+(aq) ==> H2O(l) + CO2(g) ( Stage 2 ) Stage 1 has an end-point pH around 9 and Stage 2 has an end-point pH around 3. So (1) can detect sodium carbonate and (2) can detect both sodium carbonate and sodium hydrogencarbonate. Using (1), n(HCl) = 0.113 x 23.90/1000 = 0.00270mol = n(Na2CO3) Mass of Na2CO3 = 0.00270 x 106 = 0.286g Mass of NaHCO3= 0.293 - 0.286 = 0.00673g This gives 97.7% Na2CO3 and 2.3% NaHCO3 Using (2), n(HCl) = 0.113 x 48.40/1000 = 0.00547mol Let mass of Na2CO3= x g and mass of NaHCO3= (0.293- x) g n(Na2CO3) + n(NaHCO3) = n(HCl) x/106 + [ x/106 + (0.293- x)/84 ] = 0.00547 On solving, x = 0.284 Mass of Na2CO3= 0.284g Mass of NaHCO3= 0.00913g This gives 96.9% Na2CO3 and 3.1% NaHCO3 Average the results, we have 97.3% Na2CO3 and 2.7% NaHCO3. 2009-12-21 18:39:26 補充： It is so tedious to deal with so many decimal places and thus I might hv rounded off the ans all the way down. So the resultant ans might be slightly deviated. But anyway, the sample has roughly 97% Na2CO3 and 3% NaHCO3. After all, the concept involved here is more important than the exact values.
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