Question

Determine \(\left[\mathrm{OH}^{-}\right],\left[\mathrm{H}^{+}\right],\) and the \(\mathrm{pH}\) of each of the following solutions. a. \(1.0 M\) KCl b. \(1.0 M \mathrm{KC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)

Short answer

For 1.0 M KCl solution: [OH⁻] = \(1 \times 10^{-7}\) M [H⁺] = \(1 \times 10^{-7}\) M pH = 7 For 1.0 M KC₂H₃O₂ solution: [OH⁻] ≈ \(2.36 \times 10^{-12}\) M [H⁺] ≈ \(4.24 \times 10^{-3}\) M pH ≈ 2.37

Step-by-step solution

Determine the nature of the solutions

We need to determine if the given solutions are acidic, basic, or neutral. a. KCl is formed from the reaction between KOH (a strong base) and HCl (a strong acid). Since both are strong, the solution is neutral. b. KC₂H₃O₂ is formed from the reaction between KOH (a strong base) and CH₃COOH (a weak acid). Since the base is strong and the acid is weak, the solution is basic.

Calculate [OH⁻], [H⁺], and pH for a neutral solution (1.0 M KCl)

Since the solution is neutral, the concentrations of H⁺ and OH⁻ ions are equal, and can be found using the ion product of water (Kw), which is equal to 1 × 10⁻¹⁴. Kw = [H⁺] × [OH⁻] 1 × 10⁻¹⁴ = [H⁺] × [OH⁻] Since [H⁺] = [OH⁻], we can write: 1 × 10⁻¹⁴ = [H⁺]² [H⁺] = √(1 × 10⁻¹⁴) = 1 × 10⁻⁷ Therefore, for 1.0 M KCl solution: [OH⁻] = 1 × 10⁻⁷ M [H⁺] = 1 × 10⁻⁷ M And the pH can be calculated using the formula: pH = -log₁₀[H⁺] pH = -log₁₀(1 × 10⁻⁷) = 7 So the pH of 1.0 M KCl solution is 7.

Calculate [OH⁻], [H⁺], and pH for a basic solution (1.0 M KC₂H₃O₂)

Since the solution is basic, we need to determine the initial concentration of the weak acid formed from the salt, which is CH₃COOH. The initial concentration of CH₃COOH is equal to the concentration of the salt KC₂H₃O₂, which is 1.0 M. The equilibrium reaction for the ionization of CH₃COOH is: CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺ (H₃O⁺ is the hydrated form of H⁺) We will assume that the ionization of CH₃COOH is small and can be approximated by x. Therefore, at equilibrium: [CH₃COO⁻] = x [H⁺] = x [CH₃COOH] = 1 - x ≈ 1 (because x << 1) Now we can use the Ka expression for CH₃COOH to relate x and the concentrations: Ka = ([CH₃COO⁻] × [H⁺]) / [CH₃COOH] 1.8 × 10⁻⁵ (given Ka for CH₃COOH) = (x × x) / 1 x² = 1.8 × 10⁻⁵ x = √(1.8 × 10⁻⁵) x = [H⁺] ≈ 4.24 × 10⁻³ M Now we can calculate the pH using the formula: pH = -log₁₀[H⁺] pH ≈ -log₁₀(4.24 × 10⁻³) ≈ 2.37 To find [OH⁻], we use the Kw expression again: Kw = [H⁺] × [OH⁻] [OH⁻] = Kw / [H⁺] [OH⁻] = (1 × 10⁻¹⁴) / (4.24 × 10⁻³) ≈ 2.36 × 10⁻¹² M So for 1.0 M KC₂H₃O₂: [OH⁻] ≈ 2.36 × 10⁻¹² M [H⁺] ≈ 4.24 × 10⁻³ M pH ≈ 2.37

Key concepts

Acid-Base Chemistry

Acid-base chemistry is a fundamental topic that explores the behavior of acids and bases in solution. Acids are substances that can donate protons (H⁺ ions), while bases are substances that can accept protons.
In the context of substance descriptions, we often categorize solutions based on their pH values:

• Acidic solutions have a pH less than 7.
• Neutral solutions have a pH exactly equal to 7.
• Basic (or alkaline) solutions have a pH greater than 7.

Understanding the nature of a solution helps predict chemical reactions and how the solution will behave under different conditions.
In the exercise, determining whether KCl and KC₂H₃O₂ are acidic, basic, or neutral sets the stage for further calculations.

Ion Product of Water

The ion product of water, or Kw, is a critical constant in acid-base chemistry that expresses the relationship between the concentration of hydrogen ions \(\text{H}^+\) and hydroxide ions \(\text{OH}^-\) in water.
Mathematically, this relationship is represented by: \[ K_w = [\text{H}^+] \times [\text{OH}^-] = 1 \times 10^{-14} \] at 25°C. This means that the product of the concentrations of these ions in pure water or a neutral solution is always \(1 \times 10^{-14}\).
This concept is pivotal when determining the pH of solutions, especially neutral ones like 1.0 M KCl, where \(\text{H}^+\) and \(\text{OH}^-\) are equal, thus both equal to \(1 \times 10^{-7} \) M.

Equilibrium Reactions

Equilibrium reactions are reactions that occur in a closed system where reactants and products are present at concentrations that do not change over time.
In the exercise, when dealing with KC₂H₃O₂, we consider the equilibrium reaction of acetic acid ionization: \[ \text{CH}_3\text{COOH} + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}_3\text{O}^+ \] Here, acetic acid (CH₃COOH) partially ionizes in water, setting up an equilibrium between the acid and its ionization products.
The ionization extent is determined by its acid dissociation constant (\(K_a\)), which gives us an insight into the strength of the weak acid and helps us calculate the equilibrium concentrations of ions.

Strong Acids and Bases

Strong acids and bases are substances that dissociate completely in water, meaning they donate or accept protons readily and leave no undissociated molecules.
When KCl is dissolved in water, it results from the combination of a strong acid (HCl) and a strong base (KOH), both of which do not remain in their undissociated forms, hence forming a neutral solution.
On the other hand, salts like KC₂H₃O₂ arise from a strong base (KOH) and a weak acid (CH₃COOH). This characteristic makes the solution basic, as the strong base dominates.

• Strong acids include HCl, H₂SO₄, and HNO₃.
• Strong bases include KOH, NaOH, and LiOH.

Understanding which components of a salt are strong or weak indicates whether the resulting solution will be acidic, basic, or neutral.