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» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/20-287209" style="font-weight:bold;" target="_blank" title="Click to know more about 20">20</a> <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a><br/>40 mL<br/>120 mL<br/>80 mL</p>Solution :Boric acid is monobasic acid. <br/> `H_(3)BO_(3)+<a href="https://interviewquestions.tuteehub.com/tag/naoh-572531" style="font-weight:bold;" target="_blank" title="Click to know more about NAOH">NAOH</a> to Na[B(<a href="https://interviewquestions.tuteehub.com/tag/oh-585115" style="font-weight:bold;" target="_blank" title="Click to know more about OH">OH</a>)_(4)]` <br/> `(M_(1)V_(1))/(n_(1))=(M_(2)V_(2))/(n_(2))` <br/> `(0.1xx20)/(1)=(0.05xx x)/(1)` <br/> x=40 mL</body></html>