A 0.110-L sample of an unknown $HNO_3$ solution required 52.7 mL of 0.100 M $Ba(OH)_2$ for complete neutralization. What was the concentration of the $HNO_3$ solution?

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Answer 1 · 2016-11-15T08:09:08

Approx. $0.1*mol*L^-1$

We need a stoichiometrically balanced equation:

$Ba(OH)_2(aq) + 2HNO_3(aq) rarr Ba(NO_3)_2(aq) + 2H_2O(l)$

Thus 2 equiv of nitric acid were necessary for each equiv barium hydroxide.

$"Concentration of nitric acid"$

$=$ $(52.7xx10^-3Lxx0.100*mol*L^-1xx2)/(0.110*L)=??*mol*L^-1$

A 0.110-L sample of an unknown HNO_3HNO_3 solution required 52.7 mL of 0.100 M Ba(OH)_2Ba(OH)_2 for complete neutralization. What was the concentration of the HNO_3HNO_3 solution?