Explanation Addition of 0 m L o f 0 . 1 0 0 M H C l to 5 0 . 0 0 m L o f 0 . 1 0 0 M ammonia: Ammonia is a weak base. The initial point in the titration when the acid is not added to the system, the system is a weak base system. The iCe table can be set up to calculate the p H of the solution. B + H 2 O ⇌ H B + O H − i ( M ) 0.200 0 0 C ( M ) − y + y + y e ( M ) 0.200 − y y y Where, B represents ammonia and H B represents the conjugate acid of ammonia. The K b value for ammonia is 1.8 × 10 − 5 . The expression for K b can be written as given below, K b = [ H B ] [ O H − ] [ B ] = 1.8 × 10 − 5 y 2 ( 0.100 − y ) = 1.8 × 10 − 5 This can be solved by approximation. If y ≪ 0.100 , then y 2 ≈ 1.8 × 10 − 5 × 0.100 = 1.8 × 10 − 6 y = 1.8 × 10 − 6 = 1.34 × 10 − 3 = [ O H − ] The p H of the solution can be calculated as given below. p O H = − log [ O H − ] = − log ( 1.34 × 10 − 3 ) = 2.87 p H = 14 − p O H = 14 − 2.87 = 1 1 . 1 3 Addition of 2 5 m L o f 0 . 1 0 0 M H C l to 5 0 . 0 0 m L o f 0 . 1 0 0 M ammonia: The neutralization reaction can be written as given, B ( a q ) + H 3 O + ( a q ) → H B ( a q ) + H 2 O ( l ) Calculation of milimoles of acid and base: A m o u n t o f H C l = 25.00 m L × ( 0.100 milimol H 3 O + m L ) = 2 . 5 m i lim o l A m o u n t o f B = 50.00 m L × ( 0.100 milimol m L ) = 5 m i lim o l Making of sRfc table: The table can be formed as shown below. The molar ratio of acid and base is 1 : 1. So the limiting reactant is H 3 O + ion. B + H 3 O + → H B + H 2 O s ( m m o l ) 5 2.5 0 E x c e s s R ( m m o l ) − 2.5 − 2.5 + 2.5 + 2.5 f ( m m o l ) 2.5 0 2.5 E x c e s s All the strong acid is consumed. There are weak base ( B ) and its conjugate acid ( H B ) in the solution. So this solution is a basic buffer. By using Henderson-Hasselbalch equation, it’s p H can be calculated. p O H = p K b + log n a n b = p K b + log 2.5 2.5 = p K b + 0 = p K b = − log ( K b ) = − log ( 1.8 × 10 − 5 ) = 4.745 p H = 14 − p O H = 14 − 4.745 = 9 . 2 5 5 Addition of 5 0 m L o f 0 . 1 0 0 M H C l to 5 0 . 0 0 m L o f 0 . 1 0 0 M ammonia: Calculation of milimoles of acid and base: A m o u n t o f H C l = 50.00 m L × ( 0.100 milimol H 3 O + m L ) = 5 m i lim o l A m o u n t o f B = 50

3rd Edition

ISBN: 9780534420123

Author: Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Publisher: Cengage Learning

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Acid Base Titration

An acid-base titration is the method of quantitative analysis for determining the concentration of an acid and base by exactly neutralizing it with a standard solution of base or acid having known concentration.

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Chapter 16, Problem 16.56QE

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Interpretation:

The pH during the titration of 50.00 mL of 0.100 M ammonia with 0.100 M HCl after the addition of 0, 25.00, 50.00,and 75.00 mL of the acid. The titration curve has to be drawn and four regions of importance have to be labeled.

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Chemistry: Principles and Practice

Ch. 16 - Prob. 16.1QE Ch. 16 - Sketch a titration curve for the titration of...Ch. 16 - Prob. 16.4QE Ch. 16 - Prob. 16.5QE Ch. 16 - Explain why the HendersonHasselbalch equation...Ch. 16 - Prob. 16.7QE Ch. 16 - Prob. 16.8QE Ch. 16 - Prob. 16.9QE Ch. 16 - Prob. 16.11QE Ch. 16 - Prob. 16.13QE

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Acid-Base Titration | Acids, Bases & Alkalis | Chemistry | FuseSchool; Author: FuseSchool - Global Education;https://www.youtube.com/watch?v=yFqx6_Y6c2M;License: Standard YouTube License, CC-BY

The p H during the titration of 50.00 m L o f 0.100 M ammonia with 0.100 M H C l after the addition of 0 , 25.00 , 50.00 , a n d 75.00 m L of the acid. The titration curve has to be drawn and four regions of importance have to be labeled.

Solution Summary: The author explains the iCe table to calculate the pH of the solution.

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Interpretation:

The pH during the titration of 50.00 mL of 0.100 M ammonia with 0.100 M HCl after the addition of 0, 25.00, 50.00,and 75.00 mL of the acid. The titration curve has to be drawn and four regions of importance have to be labeled.

Chapter 16 Solutions

The p H during the titration of 50.00 m L o f 0.100 M ammonia with 0.100 M H C l after the addition of 0 , 25.00 , 50.00 , a n d 75.00 m L of the acid. The titration curve has to be drawn and four regions of importance have to be labeled.

Solution Summary: The author explains the iCe table to calculate the pH of the solution.

Concept explainers

Videos

Interpretation: The pH during the titration of 50.00 mL of 0.100 M ammonia with 0.100 M HCl after the addition of 0, 25.00, 50.00,and 75.00 mL of the acid. The titration curve has to be drawn and four regions of importance have to be labeled.

Chapter 16 Solutions

Interpretation:

The pH during the titration of 50.00 mL of 0.100 M ammonia with 0.100 M HCl after the addition of 0, 25.00, 50.00,and 75.00 mL of the acid. The titration curve has to be drawn and four regions of importance have to be labeled.