Question

An 8.89 g sample of an aqueous solution of nitric acid contains an unknown amount of the acid. If 27.1 mL of 0.581 M potassium hydroxide is 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 11.1 %

Step-by-step explanation:

Step 1: Data given

Mass of HNO3 = 8.89 grams

Volume of KOH = 27.1 mL = 0. 0271 L

Molarity of KOH = 0.581 M

Step 2: The balanced equation

HNO 3 + KOH → KNO 3 + H 2 O

Step 3: Calculate the moles of KOH

Moles of KOH = molarity KOH * volume

Moles KOH = 0.581 M * 0.0271 L

Moles KOH = 0.0157 moles

Step 4: Calculate moles of HNO3

For 1 mol of KOH we need 1 mol of HNO3

For 0.0157 moles of KOH we need 0.0157 moles of HNO3

Step 5: Calculate mass of HNO3

Mass KOH = moles KOH * molar mass KOH

Mass KOH = 0.0157 moles * 63.01 g/mol

Mass KOH = 0.989 grams

Step 6: Calculate mass % HNO3 in sample

mass % = (0.989 grams / 8.89 grams)*100%

mass % = 11.1 %

The percent by mass of nitric acid in the mixture is 11.1 %