Explanation The equilibrium reaction is given below. C 5 H 5 N ( a q ) + H 2 O ( l ) ⇌ C 5 H 5 N H + ( a q ) + O H − ( a q ) The base ionization constant can be written as given below. K b = [ C 5 H 5 N H + ] [ O H − ] [ C 5 H 5 N ] The value of K b can be calculated as given below. K b = K w K a = 1.00 × 10 − 14 6.3 × 10 − 6 = 1.6 × 10 − 9 . The p H of the solution is 8.5. So the concentration of hydroxide ion can be calculated as shown below. [ O H − ] = 10 − p O H = 10 − ( 14.00 − p H ) = 10 − ( 14.00 − 8.5 ) = 10 − 5.5 = 3.0 × 10 − 6 M . A table can be set up as shown below. C 5 H 5 N ( a q ) O H − ( a q ) C 5 H 5 N H + ( a q ) I n i t i a l c o n c . ( M ) 0.2 u n k n o w n 0 C h a n g e i n c o n c

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Author: John W. Moore, Conrad L. Stanitski

Publisher: Cengage Learning

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Chapter 14, Problem 63QRT

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The concentration of unreacted pyridine in the solution has to be calculated.

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Chapter 14, Problem 62QRT

Chapter 14, Problem 64QRT

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Chapter 14 Solutions

Chemistry: The Molecular Science

Ch. 14.1 - Identify each molecule or ion as a Brnsted-Lowry...Ch. 14.1 - Using Le Chatelier’s Principle Use Le Chatelier’s...Ch. 14.1 - Prob. 14.3E Ch. 14.1 - Complete the table. Ch. 14.1 - Prob. 14.4E Ch. 14.1 - Prob. 14.5E Ch. 14.2 - Prob. 14.6CE Ch. 14.2 - Prob. 14.7E Ch. 14.3 - Prob. 14.2PSP Ch. 14.3 - Prob. 14.3PSP

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The concentration of unreacted pyridine in the solution has to be calculated.

Solution Summary: The author explains that the concentration of unreacted pyridine in the solution is 0.2.

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Interpretation: The concentration of unreacted pyridine in the solution has to be calculated.

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Chapter 14 Solutions

Interpretation:

The concentration of unreacted pyridine in the solution has to be calculated.