**Problem Statement:**Calculate the equilibrium concentration for \( \text{H}_3\text{O}^+ \) for part a from question 1 based on the fact that \( K_a \) is \( 2.5 \times 10^{-9} \) and that the initial concentration of HBrO is 0.15 M.---**Solution Explanation:**1. **Understand the Given Data:** - \( K_a \) (acid dissociation constant) = \( 2.5 \times 10^{-9} \) - Initial concentration of HBrO (weak acid) = 0.15 M2. **Set Up the Equilibrium Expression:** The dissociation of HBrO in water can be represented as: \[ \text{HBrO} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{BrO}^- \]3. **Use the \( K_a \) Expression:** The expression for the acid dissociation constant (\( K_a \)) is: \[ K_a = \frac{[\text{H}_3\text{O}^+][\text{BrO}^-]}{[\text{HBrO}]} \]4. **Assume Initial Concentrations and Changes:** Let \( x \) be the concentration of \( \text{H}_3\text{O}^+ \) at equilibrium. Therefore: - Initial concentration of \( \text{HBrO} \) = 0.15 M - Change in \( \text{H}_3\text{O}^+ \) and \( \text{BrO}^- \) = \( +x \) - Change in \( \text{HBrO} \) = \( -x \) At equilibrium: \[ [\text{H}_3\text{O}^+] = x \] \[ [\text{BrO}^-] = x \] \[ [\text{HBrO}] = 0.15 - x \]5. **Apply the Initial Assumptions to the \( K_a \) Expression:** \[ K_a = \frac{x \cdot

The dissociation constant (Ka): This constant measures the strength of an acid in the solution…

Answered: Calculate the equilibrium concentration…

Calculate the equilibrium concentration for H30* for part a from question 1 based on the fact that K, is 2.5 x 10° and that the initial concentration of HBrO is 0.15 M.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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