A difinite amount of `BaCl_(2)` was dissolved in HCl solution of unknown normality. 20 " mL of " this solution was treated with 21.4 " mL of " 01 N NaOH, for complete neutralisation. Further 20 " mL of " the solution was added to 50 " mL of " 0.1 N `Na_(2)CO_(3)` and the precipitate was filtered off. The filtrate reacted with 10.5 " mL of " 0.08 N `H_(2)SO_(4)` using phenolphthalein as indicator. Calculate the strength of `BaCl_(2)` in mixture.

Explanation

`m" Eq of "NaOH=m" Eq of "HCl`
`21.4xx0.1=20xxN`
`impliesN_(HCl)=(21.4)/(200)`
`m" Eq of "H_(2)SO_(4)=10.5xx0.08`
`=(1)/(2)m" Eq of "Na_(2)CO_(3)`
(Phenolphthalein indicator)
`thereforem" Eq of "unused Na_(2)CO_(3)=2xx10.5xx0.08`
`m" Eq of "used Na_(2)CO_(3)=m" Eq of "BaCl_(2)+m" Eq of "HCl`
`(50xx0.1)-(2xx10.5xx0.08)=m" Eq of "BaCl_(2)+N_(HCl)xxV_(HCl)`
`3.32=m" Eq of "BaCl_(2)+[(21.4)/(200)xx20]`
`m" Eq of "BaCl_(2)=3.32-2.14`
`=(1.18)/(20mL)`
`=59(mEq)/(1000mL)`
Weight of `BaCl_(2)=59xx10^(-3)xx(208)/(2)=6.136gL^(-1)`