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Science Chemistry Chemistry: An Atoms-Focused Approach

The amount of 6.0 M N a O H must be added to 0.500 L of a buffer, which is 0.0200 M acetic acid and 0.0250 M sodium acetate, to raise its pH to 5.75 a t 25 ℃ . Solution: The amount of 6.0 M N a O H to be added to the given buffer solution is 1.3 m L . Explanation: Concept: To find out the amount of N a O H to be added to the buffer solution, first we need to determine the ratio of the moles of acetic acid and sodium acetate required in the solution to raise the pH. This can be calculated from the Henderson-Hasselbalch Equation for buffer solutions. When a basic solution is added to the acetate buffer solution, acetic acid reacts with O H - ions and forms acetate ions. Let us assume that x m o l of N a O H is added to raise the pH to 5.75 of the solution, and, from the RICE table, we can determine the moles of acetate and acetic acid at equilibrium. The ratio of their moles at equilibrium will be the same as that of the value calculated before, which gives us the value of x . Formula: M o l a r i t y = m o l e s o f s o l u t e V o l u m e o f s o l u t i o n i n L p H = p K a + log ⁡ b a s e a c i d Given: pH of the buffer solution = 5.75 Initial concentration of acetic acid = 0.0200 M Initial concentration of sodium acetate = 0.0250 M Molarity of N a O H solution = 6.0 M Volume of the buffer solution = 0.500 L For acetic acid, p K a = 4.75 Calculation: The reaction for the formation of the acetate ion buffer is written as C H 3 C O O H a q + H 2 O l ↔ H 3 O + a q + C H 3 C O O - a q p K a = 4.75 The ratio of the moles of base and acid can be determined by the Henderson-Hasselbalch equation as follows: p H = p K a + log ⁡ b a s e a c i d log ⁡ b a s e a c i d = p H - p K a = 5.75 - 4.75 = 1.0 b a s e a c i d = 10 1 = 10 This is the desired value of the ratio of moles of base and acid required to increase the pH up to 5.75 . Further, when 6.0 M N a O H is added to the buffer solution, the acid part ( C H 3 C O O H ) will react with the added base. The reaction can be written as C H 3 C O O H a q + O H - a q ↔ C H 3 C O O - ( a q ) + H 2 O l The initial moles of acetate ions and acetic acid present in the buffer solution can be calculated as M o l e s o f C H 3 C O O - = M o l a r i t y × V o l u m e = 0.250 M × 0.500 L = 0.0125 m o l M o l e s o f C H 3 C O O H = M o l a r i t y × V o l u m e = 0.0200 M × 0.500 L = 0.0100 m o l Let us assume that x m o l of N a O H is to be added to raise the pH of the solution. The moles of the acetate ions and acetic acid in the solution at equilibrium after the addition of N a O H can be determined using the RICE table. Reaction C H 3 C O O H a q + O H - a q ↔ C H 3 C O O - ( a q ) + H 2 O l C H 3 C O O H ( m o l ) O H - ( m o l ) C H 3 C O O - ( m o l ) Initial (I) 0.0100 x 0.0125 Change (C) - x - x + x Equilibrium (E) ( 0.0100 - x ) 0 ( 0.0125 + x ) The molar ratio of the acetate ions and the acetic acid is ( 0.0125 + x ) ( 0.0100 - x ) = 10 10 ( 0.0100 - x ) = ( 0.0125 + x ) 0.1 - 10 x = 0.0125 + x 11 x = 0.0875 ∴ x = 0.0875 11 = 0.007955 m o l Thus, 0.007955 of N a O H is to be added to the buffer solution to maintain a pH of 5.75 . Molarity of N a O H solution i s 6.0 M . Therefore, the volume of this solution, which needs to be added, is V o l u m e o f N a O H t o b e a d d e d = m o l o f N a O H t o b e a d d e d M o l a r i t y o f N a O H s o l u t i o n = 0.007955 m o l 6.0 M = 0.00132 L = 1.3 m L Thus, to raise the pH of the buffer solution, we need to add 1.3 m L of 6.0 M N a O H solution to the buffer. Conclusion: Using Henderson-Hasselbalch equation for buffer solutions, the amount of base that needs to be added to the given buffer solution is calculated.

The amount of 6.0 M N a O H must be added to 0.500 L of a buffer, which is 0.0200 M acetic acid and 0.0250 M sodium acetate, to raise its pH to 5.75 a t 25 ℃ . Solution: The amount of 6.0 M N a O H to be added to the given buffer solution is 1.3 m L . Explanation: Concept: To find out the amount of N a O H to be added to the buffer solution, first we need to determine the ratio of the moles of acetic acid and sodium acetate required in the solution to raise the pH. This can be calculated from the Henderson-Hasselbalch Equation for buffer solutions. When a basic solution is added to the acetate buffer solution, acetic acid reacts with O H - ions and forms acetate ions. Let us assume that x m o l of N a O H is added to raise the pH to 5.75 of the solution, and, from the RICE table, we can determine the moles of acetate and acetic acid at equilibrium. The ratio of their moles at equilibrium will be the same as that of the value calculated before, which gives us the value of x . Formula: M o l a r i t y = m o l e s o f s o l u t e V o l u m e o f s o l u t i o n i n L p H = p K a + log ⁡ b a s e a c i d Given: pH of the buffer solution = 5.75 Initial concentration of acetic acid = 0.0200 M Initial concentration of sodium acetate = 0.0250 M Molarity of N a O H solution = 6.0 M Volume of the buffer solution = 0.500 L For acetic acid, p K a = 4.75 Calculation: The reaction for the formation of the acetate ion buffer is written as C H 3 C O O H a q + H 2 O l ↔ H 3 O + a q + C H 3 C O O - a q p K a = 4.75 The ratio of the moles of base and acid can be determined by the Henderson-Hasselbalch equation as follows: p H = p K a + log ⁡ b a s e a c i d log ⁡ b a s e a c i d = p H - p K a = 5.75 - 4.75 = 1.0 b a s e a c i d = 10 1 = 10 This is the desired value of the ratio of moles of base and acid required to increase the pH up to 5.75 . Further, when 6.0 M N a O H is added to the buffer solution, the acid part ( C H 3 C O O H ) will react with the added base. The reaction can be written as C H 3 C O O H a q + O H - a q ↔ C H 3 C O O - ( a q ) + H 2 O l The initial moles of acetate ions and acetic acid present in the buffer solution can be calculated as M o l e s o f C H 3 C O O - = M o l a r i t y × V o l u m e = 0.250 M × 0.500 L = 0.0125 m o l M o l e s o f C H 3 C O O H = M o l a r i t y × V o l u m e = 0.0200 M × 0.500 L = 0.0100 m o l Let us assume that x m o l of N a O H is to be added to raise the pH of the solution. The moles of the acetate ions and acetic acid in the solution at equilibrium after the addition of N a O H can be determined using the RICE table. Reaction C H 3 C O O H a q + O H - a q ↔ C H 3 C O O - ( a q ) + H 2 O l C H 3 C O O H ( m o l ) O H - ( m o l ) C H 3 C O O - ( m o l ) Initial (I) 0.0100 x 0.0125 Change (C) - x - x + x Equilibrium (E) ( 0.0100 - x ) 0 ( 0.0125 + x ) The molar ratio of the acetate ions and the acetic acid is ( 0.0125 + x ) ( 0.0100 - x ) = 10 10 ( 0.0100 - x ) = ( 0.0125 + x ) 0.1 - 10 x = 0.0125 + x 11 x = 0.0875 ∴ x = 0.0875 11 = 0.007955 m o l Thus, 0.007955 of N a O H is to be added to the buffer solution to maintain a pH of 5.75 . Molarity of N a O H solution i s 6.0 M . Therefore, the volume of this solution, which needs to be added, is V o l u m e o f N a O H t o b e a d d e d = m o l o f N a O H t o b e a d d e d M o l a r i t y o f N a O H s o l u t i o n = 0.007955 m o l 6.0 M = 0.00132 L = 1.3 m L Thus, to raise the pH of the buffer solution, we need to add 1.3 m L of 6.0 M N a O H solution to the buffer. Conclusion: Using Henderson-Hasselbalch equation for buffer solutions, the amount of base that needs to be added to the given buffer solution is calculated.

Solution Summary: The author calculates the amount of 6.0MNaOH to be added to a buffer solution, based on the Henderson-Hasselbalch Equation.

Chemistry: An Atoms-Focused Approach

Chemistry: An Atoms-Focused Approach

14th Edition

ISBN: 9780393600681

Author: Gilbert

Publisher: W. W. Norton & Company

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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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Chemistry: An Atoms-Focused Approach

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