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20 mL of 0.1 M H(3)BO(3) solution on com...

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 :

A

20 mL

B

40 mL

C

120 mL

D

80 mL

Text Solution

Verified by Experts

The correct Answer is:
B
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
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