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double indicator titration calculation
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25 cm3 of a solution containing sodium hydroxide and sodium carbonate was titrated with 0.05 M hydrochloric acid and phenolphthalein was used as the indicator. After 18.50 cm3 of tha acid was added, phenolphthalein turned colourless. Methyl orange was then added. As the titration continued, a ferther 10.00 cm3 of... 顯示更多 25 cm3 of a solution containing sodium hydroxide and sodium carbonate was titrated with 0.05 M hydrochloric acid and phenolphthalein was used as the indicator. After 18.50 cm3 of tha acid was added, phenolphthalein turned colourless. Methyl orange was then added. As the titration continued, a ferther 10.00 cm3 of acid was needed to turn methyl orange to red colour. Calculate the concentration of sodium hydroxide and sodium carbonate in the original solution.
Assume that there are x mol of NaOH and y mol of Na2CO3 in 25 cm3 of the solution. ========== At the end point when phenolphthalein indicator is used, NaOH is completely neutralized, and Na2CO3 is changed to NaHCO3. NaOH + HCl → NaCl + H2O ...... (*) Na2CO3 + HCl → NaHCO3 + NaCl ...... (**) In (*), mole ratio NaOH : HCl = 1 : 1 No. of moles of NaOH used = x mol Hence, no moles of HCl used = x mol In (**), mole ratio Na2CO3 : HCl : NaHCO3 = 1 : 1 : 1 No. of moles of Na2CO3 used = y mol Hence, no. of moles of HCl used = y mol and, no of moles of NaHCO3 formed = y mol Total number of moles of HCl used : x + y = 0.05 x (18.5/1000) x + y = 9.25 x 10-4 ...... (1) ========== At the end point of further titration using methyl orange as indicator, all NaHCO3 left in the solution is changed to CO2. NaHCO3 + HCl → NaCl + H2O + CO2 Mole ratio NaHCO3 : HCl = 1 : 1 From above, no. of moles of NaHCO3 reacted = y mol Hence, no. of moles of HCl used = y mol Since 10 cm3 of 0.05 M HCl is used, y = 0.05 x (10/1000) y = 5 x 10-4 (mol) Substitute y into (1) x = 4.25 x 10-4 (mol) ========== Concentration of NaOH = mol / vol = (4.25 x 10-4) / (25/1000) = 0.017 M Concentration of Na2CO3 = mol / vol = (5 x 10-4) / (25/1000) = 0.02 M
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double indicator titration calculation
發問:
25 cm3 of a solution containing sodium hydroxide and sodium carbonate was titrated with 0.05 M hydrochloric acid and phenolphthalein was used as the indicator. After 18.50 cm3 of tha acid was added, phenolphthalein turned colourless. Methyl orange was then added. As the titration continued, a ferther 10.00 cm3 of... 顯示更多 25 cm3 of a solution containing sodium hydroxide and sodium carbonate was titrated with 0.05 M hydrochloric acid and phenolphthalein was used as the indicator. After 18.50 cm3 of tha acid was added, phenolphthalein turned colourless. Methyl orange was then added. As the titration continued, a ferther 10.00 cm3 of acid was needed to turn methyl orange to red colour. Calculate the concentration of sodium hydroxide and sodium carbonate in the original solution.
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最佳解答:Assume that there are x mol of NaOH and y mol of Na2CO3 in 25 cm3 of the solution. ========== At the end point when phenolphthalein indicator is used, NaOH is completely neutralized, and Na2CO3 is changed to NaHCO3. NaOH + HCl → NaCl + H2O ...... (*) Na2CO3 + HCl → NaHCO3 + NaCl ...... (**) In (*), mole ratio NaOH : HCl = 1 : 1 No. of moles of NaOH used = x mol Hence, no moles of HCl used = x mol In (**), mole ratio Na2CO3 : HCl : NaHCO3 = 1 : 1 : 1 No. of moles of Na2CO3 used = y mol Hence, no. of moles of HCl used = y mol and, no of moles of NaHCO3 formed = y mol Total number of moles of HCl used : x + y = 0.05 x (18.5/1000) x + y = 9.25 x 10-4 ...... (1) ========== At the end point of further titration using methyl orange as indicator, all NaHCO3 left in the solution is changed to CO2. NaHCO3 + HCl → NaCl + H2O + CO2 Mole ratio NaHCO3 : HCl = 1 : 1 From above, no. of moles of NaHCO3 reacted = y mol Hence, no. of moles of HCl used = y mol Since 10 cm3 of 0.05 M HCl is used, y = 0.05 x (10/1000) y = 5 x 10-4 (mol) Substitute y into (1) x = 4.25 x 10-4 (mol) ========== Concentration of NaOH = mol / vol = (4.25 x 10-4) / (25/1000) = 0.017 M Concentration of Na2CO3 = mol / vol = (5 x 10-4) / (25/1000) = 0.02 M
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