Problem 29 - Chemistry: The Central Science in SI Units, Global Edition

Question

Calculate \(\left[\mathrm{H}^{+}\right]\) for each of the following solutions, and indicate whether the solution is acidic, basic, or neutral: (a) \(\left[\mathrm{OH}^{-}\right]=7.3 \times 10^{-10} \mathrm{M}(\mathbf{b})\left[\mathrm{OH}^{-}\right]=0.015 \mathrm{M} ;\) (c) a solution in which \(\left[\mathrm{H}^{+}\right]\) is 10 times greater than \(\left[\mathrm{OH}^{-}\right]\).

Step-by-Step Solution

Answer

Short Answer: (a) For \(\left[\mathrm{OH}^{-}\right]=7.3 \times 10^{-10} \mathrm{M}\), we find \(\left[\mathrm{H}^{+}\right]=1.37 \times 10^{-6} \mathrm{M}\). The solution is acidic. (b) For \(\left[\mathrm{OH}^{-}\right]=0.015 \mathrm{M}\), we find \(\left[\mathrm{H}^{+}\right]=6.67 \times 10^{-13} \mathrm{M}\). The solution is basic. (c) For a solution in which \(\left[\mathrm{H}^{+}\right]\) is 10 times greater than \(\left[\mathrm{OH}^{-}\right]\), we find \(\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-6.5} \mathrm{M}\). The solution is acidic.

Step 1: Part (a) Calculate \(\left[\mathrm{H}^{+}\right]\) for \(\left[\mathrm{OH}^{-}\right]=7.3\times10^{-10}\) M

To find the \(\left[\mathrm{H}^{+}\right]\), we will apply the ion product of water: \[K_w = \left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]= 1.0 \times 10^{-14}\] Now, substitute the given \(\left[\mathrm{OH}^{-}\right]\) value: \[1.0 \times 10^{-14} = \left[\mathrm{H}^{+}\right]\times(7.3\times10^{-10})\] Next, solve for \(\left[\mathrm{H}^{+}\right]\): \[\left[\mathrm{H}^{+}\right] = \frac{1.0 \times 10^{-14}}{7.3\times10^{-10}}\] \[\left[\mathrm{H}^{+}\right] = 1.37\times10^{-6}\] Since \(\left[\mathrm{H}^{+}\right] > \left[\mathrm{OH}^{-}\right]\), the solution is acidic.

Step 2: Part (b) Calculate \(\left[\mathrm{H}^{+}\right]\) for \(\left[\mathrm{OH}^{-}\right]=0.015\) M

Apply the ion product of water again and substitute the given \(\left[\mathrm{OH}^{-}\right]\) value: \[1.0 \times 10^{-14} = \left[\mathrm{H}^{+}\right]\times 0.015\] Solve for \(\left[\mathrm{H}^{+}\right]\): \[\left[\mathrm{H}^{+}\right] = \frac{1.0 \times 10^{-14}}{0.015}\] \[\left[\mathrm{H}^{+}\right] = 6.67\times10^{-13}\] Since \(\left[\mathrm{H}^{+}\right] < \left[\mathrm{OH}^{-}\right]\), the solution is basic.

Step 3: Part (c) Calculate \(\left[\mathrm{H}^{+}\right]\) when it is 10 times greater than \(\left[\mathrm{OH}^{-}\right]\)

We know that \(\left[\mathrm{H}^{+}\right]\) is 10 times greater than \(\left[\mathrm{OH}^{-}\right]\), so let's denote this relationship as: \[\left[\mathrm{H}^{+}\right] = 10\left[\mathrm{OH}^{-}\right]\] Now apply the ion product of water again: \[1.0 \times 10^{-14} = (10\left[\mathrm{OH}^{-}\right])\left[\mathrm{OH}^{-}\right]\] Solve for \(\left[\mathrm{OH}^{-}\right]\): \[\left[\mathrm{OH}^{-}\right]^2 = \frac{1.0 \times 10^{-14}}{10}\] \[\left[\mathrm{OH}^{-}\right] = \sqrt{1.0 \times 10^{-15}}\] \[\left[\mathrm{OH}^{-}\right] = 1.0\times10^{-7.5}\] Now we can find the \(\left[\mathrm{H}^{+}\right]\): \[\left[\mathrm{H}^{+}\right] = 10\left[\mathrm{OH}^{-}\right] = 10\times(1.0\times10^{-7.5}) = 1.0\times10^{-6.5}\] Since \(\left[\mathrm{H}^{+}\right] > \left[\mathrm{OH}^{-}\right]\), the solution is acidic.

Key Concepts

Ion Product of Water

Water, a simple molecule, has an interesting ability to partially ionize into hydrogen ions ( \(\mathrm{H}^{+} \)) and hydroxide ions ( \(\mathrm{OH}^{-} \)). At a temperature of 25°C, the concentration of these ions is defined by the equilibrium constant known as the ion product of water, ( \(K_w\)). The value of \(K_w\) is \(1.0 \times 10^{-14}\). This relationship can be written as:

• \[\mathrm{K_w} = [\mathrm{H}^{+}][\mathrm{OH}^{-}] = 1.0 \times 10^{-14} \]

To maintain the state of equilibrium, the product of the concentrations of hydrogen and hydroxide ions will always equal \(K_w\), assuming standard conditions. This constant helps in calculating the concentration of one ion if the other is known, thus determining the nature of the solution based on its acidity or basicity.

Acidic and Basic Solutions

Solutions can be categorized based on the relative concentration of hydrogen ions ( \(\mathrm{H}^{+} \)) and hydroxide ions ( \(\mathrm{OH}^{-} \)). The pH scale is a measure used to express the acidity or basicity of a solution. Here's how different solutions are described:

• **Acidic**: When the concentration of hydrogen ions is higher than hydroxide ions, the solution is acidic. Typically, this occurs at a pH less than 7.
• **Neutral**: If both ion concentrations are equal, as in pure water, the solution is neutral with a pH of 7.
• **Basic**: This is when hydroxide ions surpass hydrogen ions in concentration, resulting in a pH greater than 7.

By understanding whether a solution is acidic, neutral, or basic, we can infer its properties and potential reactions with various substances.

Hydrogen Ion Concentration

Hydrogen ion concentration ( \([\mathrm{H}^{+}]\)) is crucial in determining the pH of a solution. The concentration directly influences the solution’s acidity. The pH is calculated using the following relationship:

• \[\text{pH} = -\log{[\mathrm{H}^{+}]}\]

The lower the concentration of \(\mathrm{H}^{+}\), the higher the pH and more basic the solution. Conversely, a higher concentration of \(\mathrm{H}^{+}\) indicates a lower pH and a more acidic solution.In practice, if you know the \( [\mathrm{OH}^{-}] \), you can use the \(K_w\) to find \([\mathrm{H}^{+}]\) and thus calculate the pH. This makes understanding hydrogen ion concentration valuable for many scientific and real-world applications, such as chemical manufacturing and environmental science.