Question

Find \(\left[\mathrm{H}^{+}\right]\) and the \(\mathrm{pH}\) of the following solutions. (a) A 456-mL sample of a \(12.0 \%\) (by mass) solution of \(\mathrm{HNO}_{3}(d=1.00 \mathrm{~g} / \mathrm{mL})\). What is the \(\mathrm{pH}\) of \(10.0 \mathrm{~mL}\) of the same sample? (b) A solution made up of \(1.0 \mathrm{~g}\) of \(\mathrm{HCl}\) dissolved in enough water to make \(1.28 \mathrm{~L}\) of solution. What is the \(\mathrm{pH}\) of the solution? If the same mass of \(\mathrm{HCl}\) is dissolved in enough water to make \(128 \mathrm{~mL}\) of solution, what would the \(\mathrm{pH}\) be?

Short answer

Answer: The pH values of the solutions are approximately: (a) 0.22, (b) 0.22, (c) 1.67, and (d) 0.67.

Step-by-step solution

Calculate the moles of HNO3

Firstly, we need to calculate the moles of HNO3 in the solution by using the given mass percent and density. First, we determine the mass of the HNO3 in the 456-mL sample. Mass of HNO3 = (12.0%)(456 mL)(1.00 g/mL) = 54.72 g Now, calculate the moles of HNO3: Moles of HNO3 = \(\frac{54.72\,\text{g}}{63.01\,\text{g/mol}}\) = 0.8687 mol

Calculate the concentration of HNO3

Now that we have the moles of HNO3, we can calculate the concentration of HNO3 in the solution. Concentration of HNO3 = \(\frac{0.8687\,\text{mol}}{0.456\,\text{L}}\) = 1.904 M Since HNO3 is a strong acid which dissociates completely in the water, the concentration of H+ ions is equal to the concentration of the acid. \(\left[\mathrm{H}^{+}\right]\) = 1.904 M

Calculate the pH of the solution

Now that we have found the concentration of H+ ions in the solution, we can calculate the pH. pH = \(-\log{\left[\mathrm{H}^{+}\right]}\) = \(-\log(1.904)\) ≈ 0.22 The pH for the 456-mL sample of HNO3 is approximately 0.22. For the 10.0 mL sample, we don't need to recalculate the concentration because it remains the same as the 456 mL sample. The pH for the 10.0 mL sample is also approximately 0.22.

Calculate the moles of HCl

For part (b), first, find the moles of HCl. Moles of HCl = \(\frac{1.0\,\text{g}}{36.46\,\text{g/mol}}\) ≈ 0.0274 mol

Calculate the concentration of HCl in 1.28 L solution

Use the moles of HCl to find the concentration in the 1.28 L solution. Concentration of HCl = \(\frac{0.0274\,\text{mol}}{1.28\,\text{L}}\) ≈ 0.0214 M Since HCl is a strong acid which dissociates completely in the water, the concentration of H+ ions is equal to the concentration of the acid. \(\left[\mathrm{H}^{+}\right]\) = 0.0214 M

Calculate the pH of the 1.28 L solution

Now that we have found the concentration of H+ ions in the 1.28 L solution, we can calculate the pH. pH = \(-\log{\left[\mathrm{H}^{+}\right]}\) = \(-\log(0.0214)\) ≈ 1.67 The pH of the 1.28 L solution is approximately 1.67.

Calculate the concentration of HCl in 128 mL solution

Next, use the moles of HCl to find the concentration in the 128 mL solution. Concentration of HCl = \(\frac{0.0274\,\text{mol}}{0.128\,\text{L}}\) ≈ 0.214 M Since HCl is a strong acid which dissociates completely in the water, the concentration of H+ ions is equal to the concentration of the acid. \(\left[\mathrm{H}^{+}\right]\) = 0.214 M

Calculate the pH of the 128 mL solution

Finally, calculate the pH of the 128 mL HCl solution. pH = \(-\log{\left[\mathrm{H}^{+}\right]}\) = \(-\log(0.214)\) ≈ 0.67 The pH of the 128 mL HCl solution is approximately 0.67.