Question

An aqueous solution at 25 ∘C is 11.3 % HNO3 by mass and has a density of 1.041 g/mL. What is the pH of the solution? pH=

Answer 1

To calculate the molarity of the HNO3 solution, divide the moles of HNO3 by the volume of the solution in liters. The pH of the solution can be calculated using the formula pH = -log[H3O+]. The molarity of the HNO3 solution is 16.49 M and the pH of the solution is -1.22.

Step-by-step explanation:

To calculate the molarity of the HNO3 solution, we need the amount of HNO3 in moles and the volume of the solution in liters. First, we convert the mass percent of HNO3 to grams by assuming a 100g solution. Then, we calculate the amount of HNO3 in moles using its molar mass. Next, we divide the moles of HNO3 by the volume of the solution in liters to obtain the molarity of the solution.

The pH of the solution can be calculated using the formula pH = -log[H3O+]. Since HNO3 is a strong acid, it completely ionizes in water, so the concentration of H3O+ ions is equal to the concentration of HNO3. Therefore, we can use the molarity of the HNO3 solution to calculate the pH using the same formula.

Let's calculate the molarity of the HNO3 solution first:

1. Calculate the mass of HNO3 in grams: 11.3% of 100g = 11.3g
2. Calculate the moles of HNO3: moles = mass / molar mass = 11.3g / 63.01 g/mol = 0.179 mol
3. Calculate the volume of the solution in liters: density = mass / volume, volume = mass / density = 11.3g / 1.041 g/mL = 10.85 mL = 0.01085 L
4. Calculate the molarity of the solution: Molarity = moles / volume = 0.179 mol / 0.01085 L = 16.49 M

The molarity of the HNO3 solution is 16.49 M. Now let's calculate the pH:

1. Calculate the concentration of H3O+ ions in mol/L: 16.49 M
2. Calculate the pH using the formula pH = -log[H3O+]: pH = -log(16.49) = -1.22

The pH of the solution is -1.22.

Answer 2

The pH of the given aqueous solution at 25 °C is approximately 0.27.

Firstly, determine the molarity of HNO3. The mass percent is 11.3%, implying that in 100 g of solution, 11.3 g is HNO3. Convert this mass to moles using the molar mass of HNO3 (63.02 g/mol).

Moles of HNO3 = Mass of HNO3 ÷ Molar Mass of HNO3

Moles of HNO3 = 11.3 g/63.02 g/mol

Moles of HNO3 ≈ 0.179mol

As the density is 1.041 g/mL, this solution has a volume of 100 g / 1.041 g/mL ≈ 96.01 mL. Convert this volume to liters.

Volume (L) = Volume (mL)/1000

Volume (L) = 96.01 mL/1000

Volume (L) ≈ 0.09601L

Molarity (M) = Moles of solute ÷ Volume of solution (L)

Molarity (M) = 0.179 mol/0.09601 L

Molarity (M) ≈ 1.86M

Since HNO3 is a strong acid, it completely dissociates, and [H+] is equal to the molarity of the solution.

[H+] =1.86 M

Now, apply the formula pH = -log₁₀([H+]) to find the pH.

pH=−log₁₀(1.86)

pH ≈ 0.27

In conclusion, the detailed calculation shows that the pH of the given aqueous solution is approximately 0.27. This result indicates a highly acidic nature due to the presence of nitric acid.