InterviewSolution
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InterviewSolution
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A solution containing `Na_(2)CO_(3)` and NaOH requires 300 mL of 0.1 N HCl using phenolphthalein as an indicator. Methyl orange is then added to above titrated solution when a further 25 mL of 0.2 M HCl is required. The amount of NaOH present in the original solution isA. 0.5 gB. 1gC. 2gD. 4g |
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Answer» Correct Answer - B 300 ml of 0.1 N HCl will neutralize all the NaOH and half of `Na_(2)CO_(3)` (with phenolphthalein as indicator). The remaining half will be neutralized by 25 ml of 0.2 N HCl (with methyl orange as indicator). Thus, 0.2 N HCl required for half neutralization of `Na_(2)CO_(3) = 25 ml` or 0.1 N HCl required will be 50 ml. Hence, 0.1 N HCl used up for NaOH = 300 - 50 = 250 ml . Gram equivalents present in 250 ml of 0.1 N HCl `=(0.1)/(1000)xx250=0.025`. It will neutralize 0.025 g eq. of NaOH, i.e., `0.025xx40g` = 1 g. |
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