Determine the following concentrations when 7.48 g of nitric acid (HNO3) is dissolved in 125 mL of aqueous solution.

Question

Determine the following concentrations when 7.48 g of nitric acid (HNO₃) is dissolved in 125 mL of aqueous solution.

Moles of HNO₃ dissolved per liter = 

[H⁺]_{in soln.}= 

[NO₃^–]_{in soln.}= 

Answer 1

Molar mass HNO₃ = 63.0 g / mol

moles HNO₃ present = 7.48 g x 1 mol / 63.0 g = 0.119 mols

moles HNO₃ per liter = 0.119 mols / 125 mls x 1000 mls / L = 0.952 mols/liter

HNO₃ is a strong acid and will ionize completely to produce H⁺ and NO₃^-

[H⁺] = 0.952 mols/liter

[NO₃^-] = 0.952 mols/liter