Calculate the pH of a 50.00 mL of 0.1000 M ammonia (Ka = 5.7x 10) solution after addifion of 50.00 mtof 0.1000 M HCl in the titration.1. 05.772. 04.883. 05.034. 05.275. 05.12PreviousNexteshp

given, volume of ammonia = 50mL = 0.050L concentration of ammonia = 0.1M volume of HCl = 50mL =…

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Calculate the pH of a 50.00 mL of 0.1000 M ammonia (Ka = 5.7x 10 solution after addition of 50.00 mt of 0.1000 M HCl in the titration. 1. 05.77 2 04.88 3. 05.03 4. 05.27 5. 05.12 Previous Next

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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Calculate the pH of a 50.00 mL of 0.1000 M ammonia (Ka = 5.7x 10) solution after addifion of 50.00 mt
of 0.1000 M HCl in the titration.
1. 05.77
2. 04.88
3. 05.03
4. 05.27
5. 05.12
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Learning Goal: To calculate the pH at the equivalence point for various types of titrations. The equivalence point in an acid-base titration is the point at which stoichiometrically equivalent quantities of acid and base have been mixed together. At this point the reaction is complete because all analyte has been consumed by titrant. On a titration curve, the equivalence point is represented by the point of inflection (where the curve changes concavity). The figure (Figure 1) shows the titration of 40.0 mL of 0.100 M HCl with 0.100 M NaOH. When 40.0 mL of the NaOH solution is added, the acid-base neutralization reaction is complete. When analyzing titrations involving weak acids or bases, consider how the neutralization reaction will impact the pH of the system. For example, when titrating a weak acid with a strong base, at the equivalence point all the base has been used to neutralize the acid forming its conjugate weak base Figure E 14 12 10 8 6 4 2 0 0 Equivalence point 20.0 40.0 mL…
Answer using this data for the remaining questions on this page: For 1.0L of an acetic acid solution: Ka = 1.8 x 10-5 and [HA] = 0.250M 2. Calculate the pH of the solution. 3. What is the pka? 4. Calculate the pH of the solution if you make it a buffer by adding enough solid sodium acetate to make the concentration of acetate ion equal to 0.35M. The volume of the solution does not change. 5. What is the new concentration of [HA] if 2.0mL of 5.00M of HCI is added to 1.000L of the solution above? 6. What is the new pH?