**Question:**What is the pH of a buffered system made by dissolving 17.42 g of KH₂PO₄ (MM 136.1 g/mol) and 20.41 g of K₂HPO₄ (MM 174.2 g/mol) in water to give a volume of 200.0 mL? The \( K_{a2} \) for dihydrogen phosphate is \( 6.2 \times 10^{-8} \) and the equilibrium reaction of interest is:\[ \text{H}_2\text{PO}_4^-(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{HPO}_4^{2-}(aq) \].Enter your answer in decimal format with two decimal places (value ± 0.09). **Answer Box:** [ ]---### Explanation:This question involves calculating the pH of a buffered solution using the concentrations of the buffering agents and the equilibrium constant (\( K_{a2} \)). **Steps to Solve:**1. **Calculate moles of each compound:** - Moles of KH₂PO₄ = 17.42 g / 136.1 g/mol - Moles of K₂HPO₄ = 20.41 g / 174.2 g/mol2. **Determine concentrations:** - Divide moles by the volume of the solution (0.200 L) to find molarity of each.3. **Use the Henderson-Hasselbalch equation to find pH:** - \[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{HPO}_4^{2-}]}{[\text{H}_2\text{PO}_4^-]} \right) \] - \(\text{pK}_a = -\log(K_{a2})\)**Important Considerations:**- Ensure correct significant figures in measurements.- Henderson-Hasselbalch equation applies as it is a buffer system. No graphs or diagrams are present in this image.

Given : Mass of KH2PO4 = 17.42 g Mass of K2HPO4 = 20.41 g And Ka of KH2PO4 = 6.2 X 10-8 Molar mass…

What is the pH of a buffered system made by dissolving 17.42 g of KH2PO4 (MM 136.1 g/mol) and 20.41 g of K>HPOA (MM 174.2 g/mol) in water to give a volume of 200.0 mL? The Ka2 for dihydrogen phosphate is 6.2 x 10 8 and the equilibrium reaction of interest is H2PO4 (aq) + H2O) = H30*(aq) + HPO4 (aq). Enter your answer in decimal format with two decimal places (value + 0.09).

What is the pH of a buffered system made by dissolving 17.42 g of KH2PO4 (MM 136.1 g/mol) and 20.41 g of K>HPOA (MM 174.2 g/mol) in water to give a volume of 200.0 mL? The Ka2 for dihydrogen phosphate is 6.2 x 10 8 and the equilibrium reaction of interest is H2PO4 (aq) + H2O) = H30*(aq) + HPO4 (aq). Enter your answer in decimal format with two decimal places (value + 0.09).

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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