What is [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) 2.35 ×10^-3^M

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Quizplus · Accepted Answer

We need to use the equation for the dissociation of water: H2O ⇌ H+ + OH- At 25°C, Kw (the ion product constant for water) is 1.0 × 10^-14. Kw = [H+][OH-] We are given that [H₃O⁺] = 2.35 × 10^-3^M, which represents the concentration of [H+] and [OH-]. Therefore, [H+] = [OH-] = 2.35 × 10^-7 M Now we need to calculate the pH using the equation: pH = -log[H+] pH = -log(2.35 × 10^-7) pH = 6.63 Finally, we use the equation: pOH = 14 - pH pOH = 14 - 6.63 pOH = 7.37 Therefore, the [OH-] is: [OH-] = 10^-pOH [OH-] = 2.29 × 10^-8 M Now we have all the concentrations needed to calculate the equilibrium constant for the reaction: OH- + HSO4- ⇌ SO42- + H2O Kc = [SO42-][H2O] / [OH-][HSO4-] We are given that [HSO4-] = [SO42-] = 2.35 × 10^-3 M and [OH-] = 2.29 × 10^-8 M [H2O] is usually considered constant, but its concentration is so much greater than the concentrations of the other species that we can assume it is essentially unchanged during the reaction. Therefore, we can simplify the equation to: Kc = [SO42-] / [OH-][HSO4-] Kc = (2.35 × 10^-3)^2 / (2.29 × 10^-8)(2.35 × 10^-3) Kc = 4.26 × 10^-12 Therefore, the answer is (C) 4.26 × 10^-12.

What Is [OH-] for a Solution at 25°C That Has

What Is [OH^-] for a Solution at 25°C That Has