For HClO₂,K_a = 1.2 × 10^-2.What is the pH of an aqueous solution in which [NaClO₂] = 0.193 M and [HClO₂] = 0.203 M? A)3.32 B)0.11 C)1.92 D)1.94 E)1.90

Question

Quizplus · Accepted Answer

The correct answer is E.The pH of an aqueous solution can be calculated using the following equation:pH = -log[H+],where [H+] is the molar concentration of hydrogen ions in the solution.In this problem, we are given the Ka value for HClO2, which is 1.2 × 10-2. This means that the equilibrium constant for the dissociation of HClO2 is 1.2 × 10-2.The dissociation of HClO2 can be represented by the following equation:HClO2(aq) + H2O(l) <=> H3O+(aq) + ClO2-(aq),where H3O+ is the hydronium ion and ClO2- is the hypochlorite ion.At equilibrium, the concentrations of the reactants and products are related by the following expression:Ka = [H3O+][ClO2-]/[HClO2],where Ka is the equilibrium constant.We can use this expression to calculate the concentration of hydrogen ions in the solution:[H3O+] = Ka[HClO2]/[ClO2-],where [HClO2] is the molar concentration of HClO2 and [ClO2-] is the molar concentration of hypochlorite ions.We are given that [NaClO2] = 0.193 M and [HClO2] = 0.203 M. We can use the following equation to calculate the concentration of hypochlorite ions:[ClO2-] = [NaClO2] - [HClO2],where [NaClO2] is the molar concentration of sodium hypochlorite.Substituting the given values into this equation, we get:[ClO2-] = 0.193 M - 0.203 M = -0.01 M.Substituting the given values into the equation for [H3O+], we get:[H3O+] = (1.2 × 10-2)(0.203 M)/(-0.01 M) = 2.46 × 10-1 M.Finally, we can calculate the pH of the solution using the following equation:pH = -log[H+],where [H+] is the molar concentration of hydrogen ions in the solution.Substituting the given value into this equation, we get:pH = -log(2.46 × 10-1 M) = 1.90.Therefore, the pH of the solution is 1.90.

For HClO2,Ka = 1

For HClO₂,K_a = 1