2.5 g sample of `AgNO_(3)` is dissolved in 50 ml of water. It is titrated with 50 ml of KI solution. The Agi precipitate is filtered out. Excess KI in the filtrate is titrated with 50 ml M/10 `KIO_(3)` acidified with dilute `H_(2) SO_(4)`, 20 ml of the same stock solution of KI requires 30 ml of M/10 `KIO_(3)` under similar conditions. Calculate the percentage of `AgNO_(3)` in the sample.

Options

A. 85
B. 43
C. 90
D. 97

Answer

A

Explanation

`{:(I^(-)to(1)/(2)I_(2)+1e^(-)"]"xx5),(underset("....................................................")(IO_(3)^(-)+5e^(-)to(1)/(2)I_(2))),(5I^(-)+IO_(3)^(-)+6H^(+)to3H_(2)O+3I_(2)):}`
mmol of `KIO_(3)=30xx(1)/(10)=3` mmol =15 meqs
meqs of KI=5=mmol of KI `=20mlxxM_(KI)`
`M_(KI)=0.75M`
mmol of KI used `=50xx0.75=37.5` mmol
Excess `KI=50xx(1)/(10)xx5=25` meqs of `KIO_(3)=25` meqs of KI=25 mmol KI
mmol of KI reacted with `AgNO_(3)=37.5-25=12.5` mmol
mmol of `AgNO_(3)=12.5` mmol
Mass of `AgNO_(3)=12.5xx10-3xx170`
=2.125g`
Percent purity of
`AgNO_(3)=(2.125g)/(2.5)xx100=85%`