A 25.0 mL sample of 0.723 M HClO₄is titrated with a 0.273 M KOH solution. What is the [H⁺] (molarity) before any base is added? A) 0.439 B) 1.00 × 10^-⁷ C) 0.723 D) 0.273 E) 2.81 × 10^-¹³

Question

Quizplus · Accepted Answer

The initial [H+] concentration can be found using the concentration and volume of the HClO solution: [H+] = 0.723 M x (25.0 mL / 1000 mL) = 0.0181 M Since HClO is a weak acid (Ka = 3.0 x 10^-8), we can assume that it dissociates into H+ and ClO- ions to a small extent. Using an ICE table, we can set up the equilibrium expression for the dissociation of HClO: HClO + H2O ⇌ H3O+ + ClO- [I] [HClO] [H2O] [H3O+] [ClO-] [C] -x x x x [E] (0.723-x) x (25.0/1000) x x At equilibrium, the concentration of H3O+ will be equal to the concentration of OH- added by the KOH solution. The balanced chemical equation for the neutralization reaction is: H3O+ + OH- → 2H2O The moles of KOH used can be calculated using its concentration and volume: n(KOH) = 0.273 M x (V(KOH) / 1000 mL) Since the stoichiometry of KOH and HClO is 1:1, the moles of HClO that react will also be equal to n(KOH). Therefore: n(HClO) = 0.273 M x (V(KOH) / 1000 mL) = 0.0181 mol The new concentration of HClO can be found using the remaining volume of HClO (25.0 mL - V(KOH)): [HClO] = n(HClO) / (V(HClO) / 1000 mL) = 0.0181 mol / (25.0 mL - V(KOH)) The concentration of H3O+ at equilibrium is equal to the concentration of OH- added by KOH: [H3O+] = [OH-] = n(KOH) / (V(HClO) / 1000 mL) Substituting the expressions for n(KOH) and [HClO] into [H3O+], we get: [H3O+] = 0.273 M x (V(KOH) / 1000 mL) / (25.0 mL - V(KOH)) Now we can use the equilibrium constant expression for the dissociation of HClO to solve for x: Ka = [H3O+] [ClO-] / [HClO] 3.0 x 10^-8 = x^2 / (0.723 - x) The small magnitude of x compared to 0.723 allows us to make the approximation that 0.723 - x ≈ 0.723: 3.0 x 10^-8 = x^2 / 0.723 x = 1.36 x 10^-4 M The concentration of H+ and ClO- ions, which add up to [H⁺], is equal to x: [H⁺] = x = 1.36 x 10^-4 M Therefore, the correct answer is C, 0.723.

A 250 ML Sample of 0