### Buffer Solution Calculation**Problem Statement:**A volume of 500.0 mL of 0.170 M NaOH is added to 615 mL of 0.200 M weak acid (\( K_a = 8.41 \times 10^{-5} \)). What is the pH of the resulting buffer?\[\text{HA(aq) + OH}^-(\text{aq}) \longrightarrow \text{H}_2\text{O}(\text{l}) + \text{A}^-(\text{aq})\]**Steps to Solve:**1. **Calculate the moles of NaOH and the weak acid (HA):** - Moles of NaOH: \[ \text{Moles of NaOH} = \text{Volume} \times \text{Concentration} = 0.500 \, \text{L} \times 0.170 \, \text{M} = 0.085 \, \text{moles} \] - Moles of HA: \[ \text{Moles of HA} = \text{Volume} \times \text{Concentration} = 0.615 \, \text{L} \times 0.200 \, \text{M} = 0.123 \, \text{moles} \]2. **Identify the limiting reagent and the amount of weak acid (HA) reacted:** - Since 0.085 moles of NaOH will react with 0.085 moles of HA, the limiting reagent here is NaOH. - Number of moles of HA after reaction: \[ \text{Moles of HA remaining} = 0.123 \, \text{moles} - 0.085 \, \text{moles} = 0.038 \, \text{moles} \]3. **Calculate the moles of \( \text{A}^- \) formed:** - Moles of \( \text{A}^- \): \[ \text{Moles of A}^- \text{formed} = \text{moles of NaOH used} = 0.085 \, \text{moles} \]4. **Determine the concentrations of

Ka= 8.41 × 10-5 pKa = -log(Ka) = 4.08 Molarity of NaOH solution = 0.170 M Volume of NaOH solution…

Answered: A volume of 500.0 mL of 0.170 M NaOH is…

A volume of 500.0 mL of 0.170 M NaOH is added to 615 mL of 0.200 M weak acid (Ka = 8.41 × 10-5). What is the pH of the resulting buffer? НА(аq) + ОН (аq) H,O(1) + A¯(aq) pH

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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