Calculate the molar equilibrium concentration ofeach chemical species of acetic acid in bothpreparations. The dissociated form of acetic acidrepresents 1.3% and 4.22% for 0.1 M and 0.01 Msolutions respectively, and the dissociationequation for acetic acid is:СНЗСООН — СHЗСОО"+ Н*Concentration ofMeasured pHacetic acid0.1M3.080.01M3.43

Answered: Calculate the molar equilibrium…

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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Determine the concentration of CH NH¸* in a buffer solution by constructing an ICE table, writing the equilibrium constant expression, and use this information to determine the concentration of CH NH+. The Kb for CH NH₂ is 3.8 × 10-10. Complete Parts 1-3 before submitting your answer. 2 1 2 3 NEXT > 2 A buffer solution contains dissolved CH NH and CH NH Cl. The initial concentration of CH NH2 is 0.50 M. The pH at equilibrium of the buffer is 4.20. Let x represent the original concentration of CH NH+ in the water. Fill in the ICE table with the appropriate value for each involved species to determine concentrations of all reactants and products. Initial (M) Change (M) Equilibrium (M) CH NH2(aq) + H₂O(I) = OH(aq) + C HẠNH, (aq) RESET 0 0.50 x 4.20 -4.20 6.3 x 10-5 -6.3 x 10-5 1.6 × 10-10 -1.6 × 10-10 3.8 × 10-10 -3.8 × 10-10 x +4.20 x-4.20 x + 6.3 x 10-5 x - 6.3 × 10-5 x + 1.6 × 10-10 x-1.6 × 10-10 x + 3.8 × 10-10 x - 3.8 × 10-10
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Fill in the blanks to show how you would solve for the pH of a 0.10 M aqueous solution of the weak acid HA. For the table and the equilibrium expression, do not ignore x. 0.10 POH Initial Concentration (M) Change 0.10 - X Equilibrium 2x² K₂ = 0.10 - 2x 4x² HA(aq) X (0.10 - x)² + 2x (0.10-2x)² H₂O(l) + +
A chemist titrates 100.0 mL of a 0.3992 M ethylamine (C,H;NH,) solution with 0.6385 M HCl solution at 25 °C. Calculate the pH at equivalence. The p K, of ethylamine is 3.19. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HCl solution added. pH =
A chemist titrates 90.0 mL of a 0.6900M ethylamine (C,H,NH,) solution with 0.0547M HNO, solution at 25 °C. Calculate the pH at equivalence. The p K of ethylamine is 3.19. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HNO, solution added. pH = da Explanation Check O 2021 McGraw-H Education All Rights Reserved Terms of Use Pvacy Accessibity 會 APR 23 MacBook Air 20 888 FS esc & delete $ 8 5 2 E R T Y Q W F G
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Determine the pH of a solution of HF by constructing an ICE table, writing the equilibrium constant expression, and using this information to determine the pH. The Ka of HF is 6.8 x 10-4. Complete Parts 1-3 before submitting your answer. NEXT > A 0.30 M aqueous solution of HF is prepared. Fill in the ICE table with the appropriate value for each involved species to determine concentrations of all reactants and products. Initial (M) Change (M) Equilibrium (M) -2x 6.8 x 10-4-2x 0 0.30 + x 1 HF(aq) 0.30 0.30 - x + H₂O(l) 6.8 x 10-4 2 0.30 + 2x 0.15 0.30 - 2x H3O+(aq) + +X 3 6.8 x 104 + x -X 6.8 x 10-4-x F-(aq) ✓ RESET +2x 6.8 x 10-4 + 2x
In the laboratory, a general chemistry student measured the pH of a 0.470 M aqueous solution of methylamine, CH3NH2 to be 12.163. Use the information she obtained to determine the Kh for this base. Ку(еxperiment) = 6.53x10^_4 An error has been detected in your answer. Check for typos, miscalculations etc. before submitting your answer.
A chemist titrates 110.0 mL of a 0.7452M aniline (C,H,NH,) solution with 0.5088M HCl solution at 25 °C. Calculate the pH at equivalence. The p K, of aniline is 4.87. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HCl solution added. pH = ||
A chemist titrates 100.0 mL of a 0.7953 M pyridine (C,H,N) solution with 0.5525M HCl solution at 25 °C. Calculate the pH at equivalence. The p K, of pyridine is 8.77. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HCl solution added. pH =
A chemist titrates 230.0 mL of a 0.4852M hypochlorous acid (HC1O) solution with 0.5571M NaOH solution at 25 °C. Calculate the pH at equivalence. The p K, of hypochlorous acid is 7.50. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of NaOH solution added. pH = |
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