`V_(1)` ml of NaOH of molarity X and `V_(2)` ml of `Ba(OH)_(2)` of molarity `(y)/(2)` are mixed together. Mixture is completely neutralized by 100 ml `(0.1)/(2)"M H"_(2)SO_(4)`/. If `(V_(1))/(V_(2))=(1)/(4)` and `(x)/(y)=4`, what fraction of acid is neutralized by `Ba(OH)_(2)`?

Options

A. `0.5`
B. `0.25`
C. `0.33`
D. `0.67`

Answer

D

Explanation

`NaOH = XV_(1)`
total no. of milli moles of acid used `=100 xx 0.1 = 10`
Let K milli moles of acid used for neutralisation `Ba(OH)_(2) :. 2YV_(2) = K`
milli moles of acid used for neutralising
`NaOH :. x V_(1) = (10-K)`
Fraction of acid used for neutralising `Ba(OH)_(2)`
`= (K)/(K+(10-K)) = =(2YV_(2))/(XV_(1)+2YV_(2))`
given that `(X)/(y) = 4, (V_(1))/(V_(2)) = (1)/(4)`
`rArr X = 4Y, V_(1) = (V_(2))/(4)`
Fraction of acid used for neutralising `Ba(OH)_(2)`
`=(2YV_(2))/(4Yxx (V_(2))/(4)+2YV_(2)) = (2)/(3) = 0.67`