Problem 45 - Chemistry An Atoms First Approach

Question

Calculate the \(\left[\mathrm{OH}^{-}\right]\) of each of the following solutions at \(25^{\circ} \mathrm{C} .\) Identify each solution as neutral, acidic, or basic. a. \(\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-7} \mathrm{M}\) c. \(\left[\mathrm{H}^{+}\right]=12 \mathrm{M}\) b. \(\left[\mathrm{H}^{+}\right]=8.3 \times 10^{-16} \mathrm{M}\) d. \(\left[\mathrm{H}^{+}\right]=5.4 \times 10^{-5} \mathrm{M}\)

Step-by-Step Solution

Answer

a. \(\left[\mathrm{OH}^{-}\right] = 1.0 \times 10^{-7} \mathrm{M}\) - Neutral b. \(\left[\mathrm{OH}^{-}\right] = 1.20 \times 10^{-1} \mathrm{M}\) - Basic c. \(\left[\mathrm{OH}^{-}\right] = 8.33 \times 10^{-16} \mathrm{M}\) - Acidic d. \(\left[\mathrm{OH}^{-}\right] = 1.85 \times 10^{-10} \mathrm{M}\) - Acidic

Step 1: Calculate the concentration of hydroxide ions

To calculate the concentration of hydroxide ions, use the ion product of water equation: \(K_w = \left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]\) Solve for \(\left[\mathrm{OH}^{-}\right]\): \(\left[\mathrm{OH}^{-}\right] = \frac{K_w}{\left[\mathrm{H}^{+}\right]} = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-7}} = 1.0 \times 10^{-7} \mathrm{M}\)

Step 2: Identify the solution

The solution is neutral because \(\left[\mathrm{H}^{+}\right] = \left[\mathrm{OH}^{-}\right]\). b. \(\left[\mathrm{H}^{+}\right]=8.3 \times 10^{-16} \mathrm{M}\)

Step 3: Calculate the concentration of hydroxide ions

To calculate the concentration of hydroxide ions, use the ion product of water equation: \(K_w = \left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]\) Solve for \(\left[\mathrm{OH}^{-}\right]\): \(\left[\mathrm{OH}^{-}\right] = \frac{K_w}{\left[\mathrm{H}^{+}\right]} = \frac{1.0 \times 10^{-14}}{8.3 \times 10^{-16}} = 1.20 \times 10^{-1} \mathrm{M}\)

Step 4: Identify the solution

The solution is basic because \(\left[\mathrm{H}^{+}\right] < \left[\mathrm{OH}^{-}\right]\). c. \(\left[\mathrm{H}^{+}\right]=12 \mathrm{M}\)

Step 5: Calculate the concentration of hydroxide ions

To calculate the concentration of hydroxide ions, use the ion product of water equation: \(K_w = \left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]\) Solve for \(\left[\mathrm{OH}^{-}\right]\): \(\left[\mathrm{OH}^{-}\right] = \frac{K_w}{\left[\mathrm{H}^{+}\right]} = \frac{1.0 \times 10^{-14}}{12} = 8.33 \times 10^{-16} \mathrm{M}\)

Step 6: Identify the solution

The solution is acidic because \(\left[\mathrm{H}^{+}\right] > \left[\mathrm{OH}^{-}\right]\). d. \(\left[\mathrm{H}^{+}\right]=5.4 \times 10^{-5} \mathrm{M}\)

Step 7: Calculate the concentration of hydroxide ions

To calculate the concentration of hydroxide ions, use the ion product of water equation: \(K_w = \left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]\) Solve for \(\left[\mathrm{OH}^{-}\right]\): \(\left[\mathrm{OH}^{-}\right] = \frac{K_w}{\left[\mathrm{H}^{+}\right]} = \frac{1.0 \times 10^{-14}}{5.4 \times 10^{-5}} = 1.85 \times 10^{-10} \mathrm{M}\)

Step 8: Identify the solution

The solution is acidic because \(\left[\mathrm{H}^{+}\right] > \left[\mathrm{OH}^{-}\right]\).

Key Concepts

Ion Product of Water

Understanding the ion product of water is crucial for grasping the fundamental concepts of hydroxide ion concentration in various solutions. At 25°C, this constant value for water's ion product, denoted as \(K_w\), is always \(1.0 \times 10^{-14} M^2\), which is determined by the product of the molar concentrations of hydrogen \(\left[\mathrm{H}^{+}\right]\) and hydroxide \(\left[\mathrm{OH}^{-}\right]\) ions in water.

The equation \(K_w = \left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]\) highlights a critical balance – if the hydrogen ion concentration increases, the hydroxide ion concentration decreases to maintain the constant value of \(K_w\), and vice versa. This balance is key to determining the acidity or basicity of a solution, as shown in the textbook exercise.

pH and pOH Calculations

The relationship between pH, pOH, and the ion product of water is fundamental to analyzing the acidity or basicity of solutions. The pH is the negative logarithm of the hydrogen ion concentration, and pOH is the negative logarithm of the hydroxide ion concentration. Mathematically, they are represented as \(pH = -\log[\mathrm{H}^{+}]\) and \(pOH = -\log[\mathrm{OH}^{-}]\). Since \(pH + pOH = 14\) at 25°C, knowing one value allows us to easily calculate the other.

For the textbook exercise, after finding the hydroxide ion concentration using the ion product of water, the pOH could be calculated, offering another layer of understanding the solution's properties. A pH less than 7 indicates an acidic solution, a pH of 7 denotes a neutral solution, and a pH greater than 7 suggests a basic solution.

Acidic and Basic Solutions

An acidic solution has a higher concentration of hydrogen ions \(\left[\mathrm{H}^{+}\right]\) compared to hydroxide ions \(\left[\mathrm{OH}^{-}\right]\), whereas a basic solution contains a higher concentration of hydroxide ions. In the textbook exercise, we can infer the nature of the solution by comparing the concentrations of \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) once we've calculated them using the ion product of water.

In practical terms, for part 'a' of the exercise, the concentrations of \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) are equal, making the solution neutral. For parts 'b', with a low \(\left[\mathrm{H}^{+}\right]\) and high \(\left[\mathrm{OH}^{-}\right]\), the solution is basic, and for parts 'c' and 'd', where \(\left[\mathrm{H}^{+}\right]\) exceeds \(\left[\mathrm{OH}^{-}\right]\), the solutions are acidic.