Answered: 25.00 mL sample of 0.08000 M NH3 (Kb =…

Chemistry: An Atoms First Approach

2nd Edition

ISBN:9781305079243

Author:Steven S. Zumdahl, Susan A. Zumdahl

Publisher:Steven S. Zumdahl, Susan A. Zumdahl

Chapter14: Acid- Base Equilibria

Section: Chapter Questions

Problem 106CWP: Consider the titration of 100.0 mL of 0.200 M HONH2 by 0.100 M HCI. (Kb for HONH2 = 1.1 108.) a....

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Question

25.00 mL sample of 0.08000 M NH3 (Kb = 1.8 × 10-5) is being titrated with 0.1000 M HCl.

a. Calculate the initial pH of the base.

b. Calculate the pH at 5.00 mL before the equivalence point.

c. Calculate the pH at the equivalence point.

d. Calculate the pH at 5.00 mL after the equivalence point.

Expert Solution

Step 1

$(a) . m o l a r i t y o f N a O H [O H^{-}] = 0.080000 M p O H = l o g [O H^{-}] t h e r e f o r e, p O H = - \log 0.080000 M p O H = 1.097 p H = 14 - 1.097 p H = 12.903 (b) . A t e q u i v a l e n t p o i n t m o l e s o f n u m b e r i n a c i d (H C l) = m o l e s o f n u m b e r b a s e (N a O H) s o, a t e q u i v a l e n t p o i n t v o l u m e o f H C l = m o l a r i t y o f N a O H \times \frac{v o l u m e o f N a O H}{m o l a r i t y o f H C l} 0.080000 M \times \frac{25 m L}{0.1000 M} = 20 m L b e f o r e A t e q u i v a l e n c e p o i n t v o l u m e o f H C l = 20 m L - 5 m L = 15 m L n u m b e r o f m o l e s o f N a O H = m o l a r i t y (N a O H) \times v o l u m e (L) = 0.080000 M \times 0.025 m L = 0.002 m o l s i m i l a r l y, f o r H C l, n u m b e r o f m o l e s o f H C l = m o l a r i t y (H C l) \times v o l u m e (L) = 0.1000 M \times 0.015 L = 0.0015 m o l t o t a l v o l u m e = 25 m L + 15 m L = 40 m L = 0.040 m L [O H^{-}] = \frac{m o l e s o f N a O H - m o l e s o f H C l}{t o t a l v o l u m e i n L} = \frac{0.002 m o l - 0.0015 m o l}{0.040} = 0.0125 M t h e r e f o r e t h e p O H = - \log 0.0125 M p O H = 1.90 p H = 14 - 1.90 12.10$

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