What is the [OH⁻] for a solution at 25°C that has [H₃O⁺] = 2.35 × 10⁻³ M? A)4.26 × 10 ⁻⁵ M B)2.35 × 10 ⁻¹¹ M C)4.26 × 10 ⁻¹² M D)2.35 × 10 ⁻¹⁷ M E)None of these choices are correct.

Question

Quizplus · Accepted Answer

Using the equation for the acid dissociation constant of H₃O⁺: Ka = [H+][₃O⁺] / [H₃O⁺] We can rearrange the equation to solve for [H+]: [H+] = (Ka x [H₃O⁺]) / [₃O⁺] At 25°C, Ka for H₃O⁺ is 1.2 x 10^-2. Plugging in the given values: [H+] = (1.2 x 10^-2 x 2.35 x 10^-3) / 2.35 x 10^-3 = 1.2 x 10^-2 M Since the solution is neutral, [H+] = [OH-]. Therefore: [OH-] = 1.0 x 10^-14 / [H+] = 8.3 x 10^-13 M Now we can use the equation for the base dissociation constant of OH⁻: Kb = [OH-][OH⁻] / [OH-] Simplifying: Kb = [OH-][OH⁻] / 8.3 x 10^-13 M We can rearrange the equation to solve for [OH⁻]: [OH⁻] = (Kb x 8.3 x 10^-13 M) / [OH-] = (5.7 x 10^-6 x 8.3 x 10^-13 M) / 8.3 x 10^-13 M = 5.7 x 10^-6 M Therefore, the [OH⁻] for the solution is 5.7 x 10^-6 M, which is equivalent to 4.26 x 10^-9 mol/L. The correct answer is (C) 4.26 × 10 ⁻¹² M.

What Is the [OH−] for a Solution at 25°C That

What Is the [OH⁻] for a Solution at 25°C That