Question

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

Answer 1

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

Now, for phenolphthalein end point

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

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

$\frac{1×20}{50×1000}=\frac{1}{2500}M equivalent$

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

But,

For Methyl Orange end point

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

or 1M equivalent of monoacidic base formed

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

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

$\frac{2×1000}{2500×100}=\frac{2}{250}=\frac{1}{125}N or 0.008N$