Question

Q1. A water sample is not alkaline to phenolphthalein. However, 100 ml of the sample on titration with N/50 HI required 17.5 ml to obtain end point using methyl orange as indicator. What is the type and amount of alkalinity present in the sample?

Answer 1

Answer:

Here diacidic base is present which give two end point one with phenalphthein and second with methy orange

For phenolphthalein end point-

\frac{1}{2}\ Meq\ of\ Diacidic\ base = Meq.\ of\ of \ 8ml \ \frac{N}{50}\ HCl

2

1

Meq of Diacidic base=Meq. of of 8ml

50

N

HCl

=\frac{1\times 8}{50\times1000}=\frac{1}{6225}\ Meq\\

50×1000

1×8

=

6225

1

Meq

Meq.\ of\ Diacidic\ base=2\times Meq.\ of\ HCl=\frac{2}{6225}\ Meq.Meq. of Diacidic base=2×Meq. of HCl=

6225

2

Meq.

For methyl orange end point-

Meq.\ of\ monoacidic\ base\ formed=Meq.\ of\ 1ml of\ extra\ HCl\ usedMeq. of monoacidic base formed=Meq. of 1mlof extra HCl used

=\frac{1\times1}{50\times 1000}=\frac{1}{50000\ Meq.}=

50×1000

1×1

=

50000 Meq.

1

Normality of diacidic alkalinity present in 100 ml water = \frac{2\times 1000}{6225\times 100}=\frac{20}{6225}=\frac{4}{1245}\ N

6225×100

2×1000

=

6225

20

=

1245

4

N