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?

V_(1) mL of xN NaOH and V_(2) mL of yN Ba(OH)_(2) were together suffic

1.	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?
Answer» Solution :Number of m. eq of HCl =M.eq ofNaOH +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`