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9 Oct 2021
- During the titration of an HcIO4 solution with 0.1032 M NaOH, quantitative analysis student Alcohol became distracted and overshot the end point. A fellow student, I. M. Smart, suggested that he simply record the present volume of NaOH added and titrate the excess with a standard acid solution. If the original sample volume was 25.00 ml, the volume of NaOH added was 28.06 ml, and it took 3.47 ml of 0.1094 M HCl to back titrate the naOH, calculate the molar concentration of the original HclO4 solution.
- During the titration of an HcIO4 solution with 0.1032 M NaOH, quantitative analysis student Alcohol became distracted and overshot the end point. A fellow student, I. M. Smart, suggested that he simply record the present volume of NaOH added and titrate the excess with a standard acid solution. If the original sample volume was 25.00 ml, the volume of NaOH added was 28.06 ml, and it took 3.47 ml of 0.1094 M HCl to back titrate the naOH, calculate the molar concentration of the original HclO4 solution.
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