How do I solve this AP Chemistry problem on HNO3?

Question

The percentage by weight of nitric acid HNO3 in a sample of concentrated nitric acid is to be determined.

 

A. Initially a NaOH solution was standardized by titration with a sample of potassium hydrogen phthalate, KHC8H4O4, a monoprotic acid often used as a primary standard. A sample of pure KHC8H4O4 weighing 1.518 grams was dissolved in water and tritrated with the NaOH solution. To reach the equivalence point, 26.90mL of base was required. Calculate the molarity of the NaOH

 

B. A 10.00mL sample of the concentrated nitric acid was diluted with water to a total volume of 500.00mL. Then 25.00mL of the diluted acid was tritrated with the standardized NaOH solution prepared in part A. The equivalence point was reached after 28.35mL of the base had been added. Calculate the molarity of the concentrated nitric acid.

 

C. The density of the concentrated nitric acid used in this experiment was determined to be 1.42 grams per milliliter. Determine the percentage by weight of HNO3 in the original sample of concentrated nitric acid.

Answer 1

A.  molar mass KHC8H4O4 = 204.22 g/mole

moles KHC8H4O4 = 1.518 g x 1 mole/204.22 = 0.007433 moles

[NaOH] = 0.007433 moles/0.02690 L = 0.2763 M

 

B.  (0.02835 L)(0.2763 mol/L) = 0.007834 moles NaOH = 0.007834 moles HNO3 in 25 ml

[HNO3] = 0.007834 moles/0.025 L = 0.3134 M

[Concentrated HNO3] = 0.3134 M x 500/10 = 15.67 M

 

C.  molar mass HNO3 = 63.01 g/mole

mass of 1 liter HNO3 = 1 ml x 1.42 g/ml x 1000 ml = 1420 g

mass of HNO3 in the liter = 15.67 mol/L x 63.01 g/mol = 987.2 g

mass % = 987.2 g/1420 g (x100%) = 69.52%