20 mL of 0.1 M H_(3)BO_(3) solution on complete neutralisation requires x mL of 0.05 M NaOH solution. The value of x will be :

20 mL of 0.1 M H_(3)BO_(3) solution on complete neutralisation require

1.	20 mL of 0.1 M H_(3)BO_(3) solution on complete neutralisation requires x mL of 0.05 M NaOH solution. The value of x will be :
Answer» 20 ML 40 mL 120 mL 80 mL Solution :Boric acid is monobasic acid. `H_(3)BO_(3)+NAOH to Na[B(OH)_(4)]` `(M_(1)V_(1))/(n_(1))=(M_(2)V_(2))/(n_(2))` `(0.1xx20)/(1)=(0.05xx x)/(1)` x=40 mL