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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First, let's write out the chemical equilibrium: H₃O⁺ ⇌ OH^- The equilibrium constant expression is: K = [OH^-]/[H₃O⁺] We are given that [H₃O⁺] = 2.35 × 10^-3 M at 25°C, so we can substitute into the equilibrium constant expression: K = [OH^-]/(2.35 × 10^-3) We want to solve for [OH^-], which is the same as solving for K times [H₃O⁺]. We can do this by multiplying K by the given concentration: [OH^-] = K × [H₃O⁺] [OH^-] = (K) × (2.35 × 10^-3) Now we need to calculate the equilibrium constant, K. We can do this using the acid dissociation constant (Ka) for hydrogen sulfide (H2S) and the equation: Ka = [H+][H₃O]/[H2S] At 25°C, the value of Ka for H2S is 1.3 × 10^-7. We can use this value to calculate the equilibrium constant K: K = [H+][SO4S1U1B20S1U1B2-]/[H₃O⁺] At neutral pH, we can assume that [H+] and [SO4S1U1B20S1U1B2-] are both equal to 10^-7 M (i.e. the concentration of H+ in pure water at this temperature). Therefore: K = (1.3 × 10^-7) × (10^-7)/(2.35 × 10^-3) K = 5.52 × 10^-12 Now we can substitute this value into the equation we derived earlier: [OH^-] = (5.52 × 10^-12) × (2.35 × 10^-3) [OH^-] = 12.993 × 10^-15 [OH^-] = 1.2993 × 10^-14 Therefore, [OH^-] = 4.26 × 10^-12 M. The correct answer is (C), 4.26 × 10^-12 M.

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

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