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. \(1.0 M \mathrm{KC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)

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.

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.

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