Final answer:
The pH of the solution prepared by dissolving 1.26 g of pure nitric acid in water to give 0.475 L of solution is approximately 1.377. None of the choices are correct.
Step-by-step explanation:
To determine the pH of the solution, we need to calculate the concentration of hydronium ions (H3O+). Given that 1.26 g of pure nitric acid is dissolved in 0.475 L of solution, we can calculate the molarity of the solution:
Molarity (M) = moles of solute / volume of solution (L)
Moles of nitric acid = 1.26 g / molar mass of nitric acid
= 1.26 g / 63.01 g/mol
= 0.02 mol
Molarity = 0.02 mol / 0.475 L
= 0.0421 M
Since nitric acid is a strong acid, it completely ionizes in water:
HNO3(aq) → H+(aq) + NO3-(aq)
Therefore, the concentration of H3O+ ions is equal to the molarity of the solution.
pH = -log[H3O+]
pH = -log(0.0421)
pH = 1.377 (rounded to three decimal places)
Therefore the correct answer is none of the options.