`25 mL` of a solution of barium hydroxide on titration with `0.1 "molar"` solution of hydrochloric acid give a titre value of `35 mL`. The molarity of barium hydroxide is:

Options

A. `0.07`
B. `0.14`
C. `0.28`
D. `0.35`

Answer

1

Explanation

The reaction involved is
`Ba(OH)_(2)+2HCl rarr BaCl_(2)+2H_(2)O`
According to law of equivalence
Eqs of `Ba(OH)_(2)= Eqs of HCl`
`N_(Ba(OH)_(2)) V_(Ba(OH)_(2)) =N_(HCl) V_(HCl)`
Since
Normality `=n_("factor")M`
and `n_("factor")` for `BA(OH)_(2)=2`
`n_("factor")` for HCl=1
we can write
`2M_(Ba(OH)_(2)) V_(Ba(OH)_(2))=M_(HCl) V_(HCl)`
thus
`M_(Ba(OH)_(2))=(M_(HCl) V_(HCl))/(2 V_(Ba(OH)_(2)))`
`=((0.1 M)(35 mL))/(2(25 mL))`
`=0.07 mol L^(-1)`