Consider the titration of 100.0 mL of 0.100 M hydrazine (H2NNH2, K, = 3.0×10-6) by 0.200 MHNO3. Despite the symmetric nature of the hydrazine's molecular structure, it accepts only asingle proton from HNO3 in aqueous solution. (Kw = 1.00×10-14)%3D%3D(a) Calculate the pH of the solution after addition of 0.0, 20.0, 25.0, 40.0, 50.0, and 100.0 mL ofHNO3 to the solution. (Note: you need to consider the range of applicability of the Henderson-Hasselbalch equation and need to check if water autoionization can be neglected. See Figure15.18 in the textbook.)(b) Based on your answer to part (a), sketch the titration curve, marking the calculated pointsand labeling the equivalence and half-equivalence points.(c) Although still weak, ammonia (NH3, Kb = 1.8×10-5) is a stronger base than hydrazine. Sketchthe titration curve for 100.0 mL of 0.100 M ammonia by 0.200 M HNO3 on the same axes asyour answer to part (b); be sure to label the curve.%3DNote: a qualitative sketch is sufficient; calculations analogous to part (a) are not necessary.

Answered: Consider the titration of 100.0 mL of…

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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Consider the titration of 100.0 mL of 0.100 M hydrazine (H2NNH2, K, = 3.0×10-6) by 0.200 M
HNO3. Despite the symmetric nature of the hydrazine's molecular structure, it accepts only a
single proton from HNO3 in aqueous solution. (Kw = 1.00×10-14)
%3D
%3D
(a) Calculate the pH of the solution after addition of 0.0, 20.0, 25.0, 40.0, 50.0, and 100.0 mL of
HNO3 to the solution. (Note: you need to consider the range of applicability of the Henderson-
Hasselbalch equation and need to check if water autoionization can be neglected. See Figure
15.18 in the textbook.)
(b) Based on your answer to part (a), sketch the titration curve, marking the calculated points
and labeling the equivalence and half-equivalence points.
(c) Although still weak, ammonia (NH3, Kb = 1.8×10-5) is a stronger base than hydrazine. Sketch
the titration curve for 100.0 mL of 0.100 M ammonia by 0.200 M HNO3 on the same axes as
your answer to part (b); be sure to label the curve.
%3D
Note: a qualitative sketch is sufficient; calculations analogous to part (a) are not necessary.

Expert Solution

Step 1

Acids and bases are classified as weak or strong depending on the their ability to dissociate and the amount of dissociation. When an acid or a base undergoes a complete dissociation, then they are considered as strong acid or strong base, respectively. While those that undergo partial dissociation are weak acid or weak bases.

In both acids and bases, there exists an equilibrium between the dissociated species and undissociated species. In the case of acids, an equilibrium exists between conjugate base and hydronium ion and undissociated acid that is governed by equilibrium constant known as acid dissociation constant $(K_{a})$ . Similarly, in the case of weak base, an equilibrium exists between conjugate acid, hydroxide ion and undissociated base that is governed by base dissociation constant $(K_{b})$ .

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