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

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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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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