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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 mLB. 40 mLC. 120 mLD. 80 mL |
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Answer» Correct Answer - 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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