Anilinium chloride (C6H5NH³Cl) is a weak acid (Ka = 2.4 x 10-5). A sample of anilinium chloride weighing 0.648 g is placedin 50.0 mL of water. Your goal is to titrate the sample with 0.185 M NaOH at 25 °Ca) (Circle one) NaOH is a strong or weak base? StrongWeakb) Provide the following:moles of anilinium chloride:moles of NaOH to reach equivalence:volume of 0.185 M NaOH to reach equivalence (L):concentration of anilinium chloride solution (M):pKa (anilinium chloride):Kw:Kb of the conjugate base:c) What is the pH of the anilinium chloride solution before any NaOH has been added?d) What is the pH after adding one small squirt (1.00 mL) of NaOH to the solution? e) What is the pH at the half equivalence point of the titration?f) What is the pH at the equivalence point?g) What is the pH after 30.0 mL of 0.185 M NaOH have been added?h) Sketch a rough titration curve using the information gathered in parts a-g, above. Use actual values as calculated above.

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Anilinium chloride (C6H5NH3Cl) is a weak acid (Ka = 2.4 x 10-5). A sample of anilinium chloride weighing 0.648 g is placed in 50.0 mL of water. Your goal is to titrate the sample with 0.185 M NaOH at 25 °C a) (Circle one) NaOH is a strong or weak base? Strong Weak b) Provide the following: moles of anilinium chloride: moles of NaOH to reach equivalence:

Anilinium chloride (C6H5NH3Cl) is a weak acid (Ka = 2.4 x 10-5). A sample of anilinium chloride weighing 0.648 g is placed in 50.0 mL of water. Your goal is to titrate the sample with 0.185 M NaOH at 25 °C a) (Circle one) NaOH is a strong or weak base? Strong Weak b) Provide the following: moles of anilinium chloride: moles of NaOH to reach equivalence:

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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