Question

Solution A has a pH of 12.32. Solution B has [H \(^{+}\) ] three times as large as that of solution A. Solution C has a pH half that of solution \(\mathrm{A}\). (a) What is \(\left[\mathrm{H}^{+}\right]\) for all three solutions? (b) What is the pH of solutions \(\mathrm{B}\) and \(\mathrm{C}\) ? (c) Classify each solution as acidic, basic, or neutral.

Short answer

Question: Calculate the pH values of solutions B and C, and classify each solution as acidic, basic, or neutral. Answer: For solution B: $$\mathrm{pH}_B = -\log_{10}(3\times10^{-12.32}) \approx 11.48$$ For solution C: $$\mathrm{pH}_C = -\log_{10}(10^{\frac{-1}{2} \times 12.32}) \approx 6.16$$ Solution A is basic with a pH of 12.32. Solution B is basic with a pH of 11.48. Solution C is acidic with a pH of 6.16.

Step-by-step solution

Calculate \(\left[\mathrm{H}^{+}\right]\) for Solution A

We are given that the pH of solution A is 12.32. We can use the formula \(\mathrm{pH} = -\log_{10} \left[\mathrm{H}^{+}\right]\) to calculate the concentration of \(\mathrm{H}^{+}\) ions in solution A: $$\left[\mathrm{H}^{+}\right]_A = 10^{-\mathrm{pH}_A} = 10^{-12.32}$$

Calculate \(\left[\mathrm{H}^{+}\right]\) for Solution B

We are given that the concentration of \(\mathrm{H}^{+}\) ions in solution B is three times as large as that of solution A. Using the result from Step 1, we can calculate the concentration of \(\mathrm{H}^{+}\) ions in solution B: $$\left[\mathrm{H}^{+}\right]_B = 3\left[\mathrm{H}^{+}\right]_A = 3 \times 10^{-12.32}$$

Calculate \(\left[\mathrm{H}^{+}\right]\) for Solution C

We are given that the pH of solution C is half that of solution A. First, we can calculate the pH of solution C: $$\mathrm{pH}_C = \frac{1}{2} \mathrm{pH}_A = \frac{1}{2} \times 12.32$$ Then, we can use the formula \(\mathrm{pH} = -\log_{10} \left[\mathrm{H}^{+}\right]\) to calculate the concentration of \(\mathrm{H}^{+}\) ions in solution C: $$\left[\mathrm{H}^{+}\right]_C = 10^{-\mathrm{pH}_C} = 10^{\frac{-1}{2} \times 12.32}$$

Calculate the pH of Solutions B and C

Now that we have the concentrations of \(\mathrm{H}^{+}\) ions for solutions B and C, we can use the formula \(\mathrm{pH} = -\log_{10} \left[\mathrm{H}^{+}\right]\) to calculate their pH values: $$\mathrm{pH}_B = -\log_{10}\left[\mathrm{H}^{+}\right]_B = -\log_{10}(3\times10^{-12.32})$$ $$\mathrm{pH}_C = -\log_{10}\left[\mathrm{H}^{+}\right]_C = -\log_{10}(10^{\frac{-1}{2} \times 12.32})$$

Classify Each Solution

We can now classify each solution as acidic, basic, or neutral using their pH values: - Solution A: With a pH of 12.32, it is basic (pH > 7). - Solution B: We need to evaluate \(\mathrm{pH}_B\). If \(\mathrm{pH}_B > 7\), it is basic; if \(\mathrm{pH}_B < 7\), it is acidic; if \(\mathrm{pH}_B = 7\), it is neutral. - Solution C: We need to evaluate \(\mathrm{pH}_C\). If \(\mathrm{pH}_C > 7\), it is basic; if \(\mathrm{pH}_C < 7\), it is acidic; if \(\mathrm{pH}_C = 7\), it is neutral. Now that we have calculated the values for the concentrations of \(\mathrm{H}^{+}\) ions and the pH of solutions B and C, and classified each solution as acidic, basic, or neutral, we have answered all the questions in the exercise.