A 11.2 g sample of an aqueous solution of nitric acid contains an unknown amount of the acid. If 23.2 mL of 0.873 M sodium hydroxide are required to neutralize the nitric acid, what is the percent by mass of nitric acid in the mixture?

1 Answer
Ernest Z.
May 2, 2018

The mass percent of nitric acid is 11.5 %.

Explanation:

The equation for the reaction is

#"HNO"_3 + "NaOH" → "NaNO"_3 + "H"_2"O"#

1. Calculate the moles of #"NaOH"#

#"Moles of NaOH" = 23.2 color(red)(cancel(color(black)("mL NaOH"))) × "0.873 mmol NaOH"/(1 color(red)(cancel(color(black)("mL NaOH")))) = "20.25 mmol NaOH"#

2. Calculate the moles of #"HNO"_3#.

#"Moles of HNO"_3 = 20.25 color(red)(cancel(color(black)("mmol NaOH"))) × "1 mmol HNO"_3/(1 color(red)(cancel(color(black)("mmol NaOH")))) = "20.25 mmol HNO"_3#

3. Calculate the mass of the #"HNO"_3#

#"Mass of HNO"_3 =20.25 color(red)(cancel(color(black)("mmol HNO"_3))) × ("63.01 mg HNO"_3)/(1 color(red)(cancel(color(black)("mmol HNO"_3)))) = "1286 mg HNO"_3 = "1.286 g HNO"_3"#

4. Calculate the mass percent of #"HNO"_3#

#"Mass %" = "Mass of sample"/"Total mass" × 100 % = (1.286 color(red)(cancel(color(black)("g"))))/(11.2 color(red)(cancel(color(black)("g")))) × 100 % = 11.5 %#

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