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Updated: June 18, 2025
When the content of manganese is large the solution may be divided into two equal portions, one of which is first to be roughly titrated to ascertain its content approximately, after which the whole is to be mixed together and the titration completed, which can thus be performed with greater speed and certainty.
Some care is necessary to keep the volume of the solutions to be titrated approximately uniform in standardization and in analysis, and this volume should not in general exceed 125-150 cc. for the best results, since the compounds formed by the indicator undergo changes in very dilute solution which lessen its sensitiveness.
The iodine liberated by the reaction 2FeCl + 2HI = 2HCl + 2FeCl + I is reduced by the addition of 50 cc. of sodium thiosulphate solution and the excess thiosulphate is titrated with standard iodine and requires 7.85 cc. 45 cc. I solution = 45.95 cc. Na S O solution; 45 cc. As O solution = 45.27 cc. I solution. 1 cc. arsenite solution = 0.005160 gram As O . !Answer!: 23.77%.
!Answer!: 74.18 cc. One gram of crude ammonium salt is treated with strong potassium hydroxide solution. The ammonia liberated is distilled and collected in 50 cc. of 0.5 N acid and the excess titrated with 1.55 cc. of 0.5 N sodium hydroxide. Calculate the percentage of NH in the sample. !Answer!: 41.17%.
A mixture of pure lithium chloride and barium bromide weighing 0.6 gram is treated with 45.15 cubic centimeters of 0.2017 N silver nitrate, and the excess titrated with 25 cc. of 0.1 N KSCN solution, using ferric alum as an indicator. Calculate the percentage of bromine in the sample. !Answer!: 40.11%.
The precipitate formed is transferred to the filter and well washed with water containing NH Cl and NH O, then dissolved in hydrochloric acid and reprecipitated with ammonia, filtering and washing as before. It is again dissolved in HCl and titrated with uranium solution, or decomposed by tin, as noted below, and the manganese precipitated as binoxide with chlorine, and determined.
!Answers!: 6.67% NaOH; 84.28% Na CO . A sample of sodium carbonate containing sodium hydroxide weighs 1.179 grams. It is titrated with 0.30 N hydrochloric acid, using phenolphthalein in cold solution as an indicator and becomes colorless after the addition of 48.16 cc. Methyl orange is added and 24.08 cc. are needed for complete neutralization. What is the percentage of NaOH and Na CO ?
It is the indicator most generally employed for the titration of organic acids. In general, it may be stated that when a strong acid, such as hydrochloric, sulphuric or nitric acid, is titrated against a strong base, such as sodium hydroxide, potassium hydroxide, or barium hydroxide, any of these indicators may be used, since very little hydrolysis ensues.
In the analysis of a sample of steel weighing 1.881 grams the phosphorus was precipitated with ammonium molybdate and the yellow precipitate was dissolved, reduced and titrated with KMnO . If the sample contained 0.025 per cent P and 6.01 cc. of KMnO were used, to what oxide was the molybdenum reduced? 1 cc. KMnO = 0.007188 gram Na C O . !Answer!: Mo O . !Answer!: 0.009385 gram.
Sample II. On titrating with standard acid, it required 15.26 cc. for a change in color, using phenolphthalein, and 17.90 cc. additional, using methyl orange as an indicator. Sample III. The sample was titrated with hydrochloric acid until the pink of phenolphthalein disappeared, and on the addition of methyl orange the solution was colored pink.
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