Question

Calculate \(\left[\mathrm{OH}^{-}\right.\)] for each of the following solutions, and indicate whether the solution is acidic, basic, or neutral: (a) \(\left[\mathrm{H}^{+}\right]=0.0505 \mathrm{M}\); (b) \(\left[\mathrm{H}^{+}\right]=2.5 \times 10^{-10} \mathrm{M}\); (c) a solution in which \(\left[\mathrm{H}^{+}\right]\)is 1000 times greater than \([\mathrm{OH}]\).

Short answer

a) [OH⁻] = \(1.98 \times 10^{-13}\) M, acidic b) [OH⁻] = \(4 \times 10^{-5}\) M, basic c) [OH⁻] = \(3.16 \times 10^{-6}\) M, acidic

Step-by-step solution

Use the ion product of water formula

Since we have [H⁺] = 0.0505 M, we can find [OH⁻] using the ion product of water formula, \(K_w = [H^+][OH^-]\), and solving for [OH⁻]: \[[OH^-] = \frac{K_w}{[H^+]}\]

Insert the given values

Insert the given values of \(K_w=1 \times 10^{-14}\) and [H⁺] = 0.0505 M: \[[OH^-] = \frac{1 \times 10^{-14}}{0.0505}\]

Calculate [OH⁻] and identify the solution type

Calculate the value of [OH⁻] and identify if the solution is acidic, basic, or neutral: \[[OH^-] = 1.98 \times 10^{-13}\] Since [OH⁻] < [H⁺], the solution is acidic. #b. Calculate [OH⁻] for [H⁺] = 2.5 × 10⁻¹⁰ M#

Use the ion product of water formula

Using the ion product of water formula, \(K_w = [H^+][OH^-]\), and solving for [OH⁻] with [H⁺] = 2.5 × 10⁻¹⁰ M: \[[OH^-] = \frac{K_w}{[H^+]}\]

Insert the given values

Insert the given values of \(K_w=1 \times 10^{-14}\) and [H⁺] = 2.5 × 10⁻¹⁰ M: \[[OH^-] = \frac{1 \times 10^{-14}}{2.5 \times 10^{-10}}\]

Calculate [OH⁻] and identify the solution type

Calculate the value of [OH⁻] and identify if the solution is acidic, basic, or neutral. \[ [OH^-] = 4 \times 10^{-5} \] Since [OH⁻] > [H⁺], the solution is basic. #c. Calculate [OH⁻] for [H⁺] being 1000 times greater than [OH⁻]#

Set up the relationship between [H⁺] and [OH⁻]

We are given that [H⁺] is 1000 times greater than [OH⁻]. Let's represent this relationship: \[[H^+] = 1000[OH^-]\]

Use the ion product of water formula

Using the ion product of water formula, \(K_w = [H^+][OH^-]\), with [H⁺] = 1000 [OH⁻]: \[K_w = 1000[OH^-]^2\]

Solve for [OH⁻]

Solve for [OH⁻] by dividing both sides by 1000 and taking the square root: \[[OH^-] = \sqrt{\frac{K_w}{1000}}\]

Insert the given value of Kw and calculate [OH⁻]

Insert the given value of \(K_w=1 \times 10^{-14}\) and calculate the value of [OH⁻]: \[[OH^-] = \sqrt{\frac{1 \times 10^{-14}}{1000}} = 3.16 \times 10^{-6}\]

Identify the solution type

Since [H⁺] = 1000 [OH⁻], the solution is acidic.

Key concepts

pH Calculation

Understanding the pH scale is crucial for determining whether a solution is acidic, basic, or neutral. The pH is a logarithmic measure of the hydrogen ion concentration \( [H^+] \) in a solution. It is calculated using the formula \( pH = -\log[H^+] \). When \( [H^+] \) is known, as provided in our textbook exercise, you can easily compute the pH. For instance, if the hydrogen ion concentration is 0.0505 M, the pH is calculated as \( pH = -\log(0.0505) \), resulting in a pH less than 7, which indicates an acidic solution.

It's crucial for students to understand that a pH less than 7 is acidic, around 7 is neutral, and greater than 7 is basic. This corresponds inversely to the concentration of hydroxide ions \( [OH^-] \) in the solution, which can also be calculated using the relationship to the ion product of water. By mastering pH calculation, students can better comprehend the chemical properties of solutions and predict their behavior in various reactions.

Ion Product of Water

The ion product of water (\( K_w \) is a constant at a given temperature (25°C value of \( 1 \times 10^{-14} \) that represents the product of the molar concentrations of hydrogen \( [H^+] \) and hydroxide \( [OH^-] \) ions in water. The formula is stated as \( K_w = [H^+][OH^-] \).

This constant value allows us to solve for either ion's concentration when the other is known. In the exercise, students are shown how to manipulate this expression to find the hydroxide ion concentration given the hydrogen ion concentration. For example, if \( [H^+] = 2.5 \times 10^{-10} M \) we solve for \( [OH^-] \) by rearranging the equation to \( [OH^-] = \frac{K_w}{[H^+]} \), inserting the known values, and solving for \( [OH^-] \). This is a fundamental concept in acid-base chemistry and is essential for students to be able to shift between \( [H^+] \) and \( [OH^-] \) calculations.

Acidic and Basic Solutions

The nature of a solution, whether it is acidic or basic, is determined by its pH, which is in turn dependent on the relative concentrations of hydrogen \( [H^+] \) and hydroxide \( [OH^-] \) ions.

An acidic solution has a higher concentration of hydrogen ions than hydroxide ions \( [H^+] > [OH^-] \), while a basic solution has a lower concentration of hydrogen ions than hydroxide ions \( [H^+] < [OH^-] \). When the concentrations of \( [H^+] \) and \( [OH^-] \) are equal, the solution is neutral. In our textbook problem, the nature of the solution is deduced after calculating the \( [OH^-] \) concentration. For instance, in part a, after finding \( [OH^-] = 1.98 \times 10^{-13} \), since this value is less than the given \( [H^+] \), the solution is acidic. Conversely, in part b, where \( [OH^-] > [H^+] \) the solution is basic.

It is important for students to connect the pH and \( [OH^-] \)-concentration calculations with the concept of acidity and basicity to develop a holistic understanding of chemical solutions.