Question

A 13.5 g sample of an aqueous solution of nitric acid contains an unknown amount of the acid. If 23.5 mL of 0.434 M barium hydroxide are required to neutralize the nitric acid, what is the percent by mass of nitric acid in the mixture

Answer 1

The percent by mass of nitric acid in the mixture is 3.36 %

2 HNO3 + Ba(OH)2 --->Ba(NO3)2 + 2H2O

so 1 mole Ba(OH)2 neutralizes 2 mol HNO3

moles of Ba(OH)2 present = molarity of base* volume of base

= 0.229 M * 15.6 ml

= 3.5724 milli mols

mols of HNO3 neutralized = 2* mols of Ca(OH)2 used

= 2* 3.5724 milli mols = 7.1448 x10^-3 mols

molar mass of HNO3 = 63.01 g/mol

mass of HNO3 present = molar mass * mols of HNO3

= 63.01 g/mol * 7.1448 x10^-3 mols

=0.4502 g

mass of sample = 13.4 g

mass % = mass of HNO3/mass of sample * 100

= 0.4502/13.4 *100

= 3.36 %

Nitric acid is a colorless, fuming, and distinctly corrosive liquid that could be a not unusual laboratory reagent and an important commercial chemical for the manufacture of fertilizers and explosives.

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