Calculate the pH of a solution that contains 2.4 × 10^-⁵ M H₃O⁺ at 25°C. A) 2.40 B) 9.38 C) 4.62 D) 8.38 E) 5.62

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The balanced equation for the dissociation of H₃O⁺ is: H₃O⁺ ⇌ H⁺ + ₃ The equilibrium constant expression for this dissociation is: Ka = [H⁺][₃] / [H₃O⁺] We are given that the initial concentration of H₃O⁺ is 2.4 × 10^-⁵ M. At equilibrium, let's assume that x M of H₃O⁺ has dissociated, and therefore the concentration of [H⁺] and [₃] are also x M. Using the equilibrium constant expression, we can set up the following equation: Ka = x² / (2.4 × 10^-⁵ - x) Since Ka for H₃O⁺ is 1.2 × 10S1U1P6-S1S1P5S1U1P-14, we can plug in the values: 1.2 × 10S1U1P6-S1S1P5S1U1P-14 = x² / (2.4 × 10^-⁵ - x) Solving for x, we get: x = 1.25 × 10^- Therefore, [H⁺] = [₃] = 1.25 × 10^- M Taking the negative logarithm of [H⁺], we get: pH = -log[H⁺] = -log(1.25 × 10^-) = 4.62 Therefore, the answer is C) 4.62.

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