Problem 48 - Chemistry A Molecular Approach

Question

Calculate \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) in each aqueous solution at \(25^{\circ} \mathrm{C},\) and classify each solution as acidic or basic. a. \(\left[\mathrm{OH}^{-}\right]=1.1 \times 10^{-9} \mathrm{M}\) b. \(\left[\mathrm{OH}^{-}\right]=2.9 \times 10^{-2} \mathrm{M}\) c. \(\left[\mathrm{OH}^{-}\right]=6.9 \times 10^{-12} \mathrm{M}\)

Step-by-Step Solution

Answer

\(a) [\mathrm{H}_3\mathrm{O}^{+}] = 9.09 \times 10^{-6} \,\mathrm{M}\), basic; \(b) [\mathrm{H}_3\mathrm{O}^{+}] = 3.45 \times 10^{-13} \,\mathrm{M}\), basic; \(c) [\mathrm{H}_3\mathrm{O}^{+}] = 1.45 \times 10^{-3} \,\mathrm{M}\), acidic.

Step 1: Understanding the Relation between \(\mathrm{H}_3\mathrm{O}^{+}\) and \(\mathrm{OH}^{-}\) concentrations

Use the ion product of water at \(25^\circ\mathrm{C}\), which is \(K_w = 1.0 \times 10^{-14}\). Since \(K_w = [\mathrm{H}_3\mathrm{O}^{+}][\mathrm{OH}^{-}]\), you can calculate \(\mathrm{H}_3\mathrm{O}^{+}\) by rearranging this equation to \(\mathrm{H}_3\mathrm{O}^{+} = \frac{K_w}{[\mathrm{OH}^{-}]}\).

Step 2: Calculate \(\mathrm{H}_3\mathrm{O}^{+}\) concentration for (a)

Plug in the given \(\mathrm{OH}^{-}\) concentration into the equation from Step 1: \[\left[\mathrm{H}_3\mathrm{O}^{+}\right] = \frac{1.0 \times 10^{-14}}{1.1 \times 10^{-9}} = 9.09 \times 10^{-6} \,\mathrm{M}\].

Step 3: Classify the solution for (a)

Since \(\left[\mathrm{H}_3\mathrm{O}^{+}\right] < \left[\mathrm{OH}^{-}\right]\), the solution is basic.

Step 4: Calculate \(\mathrm{H}_3\mathrm{O}^{+}\) concentration for (b)

Again use the equation from Step 1 with the given \(\mathrm{OH}^{-}\) value: \[\left[\mathrm{H}_3\mathrm{O}^{+}\right] = \frac{1.0 \times 10^{-14}}{2.9 \times 10^{-2}} = 3.45 \times 10^{-13} \,\mathrm{M}\].

Step 5: Classify the solution for (b)

Here \(\left[\mathrm{H}_3\mathrm{O}^{+}\right] < \left[\mathrm{OH}^{-}\right]\) again, indicating a basic solution.

Step 6: Calculate \(\mathrm{H}_3\mathrm{O}^{+}\) concentration for (c)

Using the provided \(\mathrm{OH}^{-}\) concentration: \[\left[\mathrm{H}_3\mathrm{O}^{+}\right] = \frac{1.0 \times 10^{-14}}{6.9 \times 10^{-12}} = 1.45 \times 10^{-3} \,\mathrm{M}\].

Step 7: Classify the solution for (c)

In this case, \(\left[\mathrm{H}_3\mathrm{O}^{+}\right] > \left[\mathrm{OH}^{-}\right]\), so the solution is acidic.

Key Concepts

Aqueous Solution Chemistry

Aqueous solution chemistry revolves around the behaviors and properties of substances when dissolved in water, the most common solvent. Water's polarity allows it to dissolve a wide range of substances, resulting in solutions with varying degrees of acidity or basicity. In these solutions, water can react with solutes to produce hydronium ions \(\mathrm{H}_3\mathrm{O}^{+}\) and hydroxide ions \(\mathrm{OH}^{-}\).
An essential part of understanding aqueous solutions is the ability to calculate the concentrations of these ions, which determine the solution's acid-base character. The exercise provided deals with calculating hydronium ion concentration based on the concentration of hydroxide ions, a key skill in aqueous solution chemistry.

pH and pOH

pH and pOH are logarithmic measures of the concentrations of hydronium ions \(\mathrm{H}_3\mathrm{O}^{+}\) and hydroxide ions \(\mathrm{OH}^{-}\), respectively, in a solution. The pH scale typically ranges from 0 to 14, with pH values less than 7 indicating acidity (higher \(\mathrm{H}_3\mathrm{O}^{+}\) concentration), pH values greater than 7 indicating basicity (higher \(\mathrm{OH}^{-}\) concentration), and a pH of 7 indicating a neutral solution.
The calculations in the exercise involve understanding that the product of \(\mathrm{H}_3\mathrm{O}^{+}\) and \(\mathrm{OH}^{-}\) concentrations at a given temperature, which for pure water is equal to \(1.0 \times 10^{-14}\) at \(25^\circ\mathrm{C}\). By knowing the concentration of one ion, you can calculate the other and then determine the pH or pOH of the solution. A crucial point for students is that pH and pOH always add up to 14, which helps classify the acidity or basicity of solutions.

Ion Product of Water

The ion product of water \(K_w\) is a constant value that represents the equilibrium concentration of hydronium and hydroxide ions in water. At \(25^\circ\mathrm{C}\), \(K_w = 1.0 \times 10^{-14}\). It's derived from the self-ionization of water, a process where water molecules react with each other to form hydronium and hydroxide ions.
Let's apply this concept to the exercise: the provided \(\mathrm{OH}^{-}\) concentration lets us use the ion product of water to determine \(\mathrm{H}_3\mathrm{O}^{+}\), following the relationship \(K_w = [\mathrm{H}_3\mathrm{O}^{+}][\mathrm{OH}^{-}]\). Calculating the hydronium ion concentration from hydroxide ion concentration is a fundamental concept in acid-base chemistry, allowing one to deduce the solution's nature—acidic or basic. This understanding is key in many chemistry applications, from industrial processes to biological systems.