`V_(1)` mL of xN NaOH and `V_(2)` mL of yN Ba`(OH)_(2)` were together sufficient to neutralise 100mL of 0.1N HCl. The ratio of `V_(1)` : `V_(2)` is 1:4 and x:y is 4:1 . What is the fraction of acid neutralised by barium hydroxide solution?

Explanation

Number of m. eq of HCl =M.eq of NaOH +m.eq. of `Ba(OH)_(2)` `=100xx0.1=10=V_(1)x+V_(2)Y`
By dividing with `V_(2)Y`. We get
`(10)/(V_(2)y)=(V_(1)x)/(V_(2)y)+1=(1)/(4)xx(4)/(1)+1=2`
`V_(2)y=5`
number of m.eq.of `Ba(OH)_(2))=5`
Fraction of acid neutralised by `Ba(OH)_(2)=(5)/(10)=(1)/(2)=0.5`