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Dilute hydrochloric acid was titrated with sodium carbonate solution. • 10.0cm3 of 0.100mol/dm3 hydrochloric acid were placed in a conical flask. • A few drops of methyl orange indicator were added to the dilute hydrochloric acid. • The mixture was titrated with sodium carbonate solution. • 16.2cm3 of sodium carbonate solution were required to react completely with the acid. Calculate how many moles of hydrochloric acid were used |
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Answer» Given: 10 ml of 0.1 molar Dilute HCL is titrated with 16.2 ml of SODIUM CARBONATE solution . To FIND : How many moles of HCL will be used. Solution: NA2CO3 (aq) + 2 HCl (aq) → 2 NaCl (aq) + CO2 (g) + H2O (l) We Know that , Molarity = [n/v]*100 For HCL solution 0.1 = [n/10]*100 n = 0.01 moles For 1 mole of sodium carbonate 2 moles of HCL is required => For 1 mole of HCL , 1/2 mole of sodium carbonate is required => For 0.01 mole of HCL , 0.005 mole of sodium carbonate is required Hence, 0.01 mole of HCL & 0.005 mole of sodium carbonate are used |
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