Final answer:
To calculate the pH of a 1.36 g HNO₃ solution in 500 ml, we first determine its molarity, which is 0.0432 M. Since HNO₃ is a strong acid, the concentration of H⁺ ions equals the molarity. The pH is then calculated as -log(0.0432), resulting in a pH of approximately 1.36.
Step-by-step explanation:
To calculate the pH of a 1.36 g HNO₃ solution in 500 ml of water, first, we need to determine the molar concentration (Molarity) of HNO₃. Nitric acid (HNO₃) is a strong acid, which means it completely dissociates into H⁺ and NO₃⁻ ions in water. The molar mass of HNO₃ is approximately 63.01 g/mol. So, to find the number of moles of HNO₃ we have:
Number of moles = mass (g) / molar mass (g/mol) = 1.36 g / 63.01 g/mol ~ 0.0216 moles
Next, to find the molarity (M) of the solution, we divide the number of moles by the volume of the solution in liters:
Molarity (M) = Number of moles / Volume (L) = 0.0216 moles / 0.5 L = 0.0432 M
Since HNO₃ is a strong acid and dissociates completely, we can assume that the concentration of H⁺ ions is equal to the molarity of the acid. Therefore, the concentration of H⁺ ions ([H⁺]) is 0.0432 M. The pH is the negative base-10 logarithm of the H⁺ ion concentration:
pH = -log([H⁺]) = -log(0.0432) ~ 1.36
Thus, the pH of the solution is approximately 1.36. This is a highly acidic solution, as expected from a strong acid such as nitric acid.