Answer to Question #201694 in Physical Chemistry for Gaurav Dhiran

Question #201694

A sample of water was alkaline both to phenolphthalein and methyl orange. 100 mL of this water sample required 20 mL of N/50 HCl for phenolphthalein end point and another 15 mL for complete neutralization. Elucidate the type and strength of alkalinity present in the water sample


1
Expert's answer
2021-06-02T06:04:23-0400

Here,diacidic base is present which gives two end points one with phenolphthalein and another with methyl orange.

Now, for phenolphthalein end point

12M\frac{1}{2}M equivalent to diacidic base=1M=1M equivalent of 15mLN50HCl\frac{N}{50}HCl

or 12M\frac{1}{2}M equivalent of diacidic base=


1×2050×1000=12500Mequivalent\frac{1×20}{50×1000}=\frac{1}{2500}M equivalent


or, 1M equivalent of diacidic base=2×12500MequivalentofHCl2×\frac{1}{2500}M equivalent of HCl


But,

For Methyl Orange end point

1M equivalent of monoacidic base formed=1M equivalent of 15mLN50HCl\frac{N}{50}HCl

or 1M equivalent of monoacidic base formed

=1×1550×1000=310,000MofHCl=\frac{1×15}{50×1000}=\frac{3}{10,000}M of HCl


Thus ,normality of diacidic alkalinity present in 100mL sample of water is given by :-

2×10002500×100=2250=1125Nor0.008N\frac{2×1000}{2500×100}=\frac{2}{250}=\frac{1}{125}N or 0.008N


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Comments

Gaurav
04.06.21, 05:58

Thanks ☺️

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