Explanation The first protonation equilibrium is given below, C O 3 2 − ( a q ) + H 2 O ( l ) ⇌ O H − ( a q ) + H C O 3 − ( a q ) The expression for equilibrium constant for the first protonation reaction can be written as shown below, K b 1 = [ O H − ] [ H C O 3 − ] [ C O 3 2 − ] = K w K a 2 = 1.0 × 10 − 14 5.6 × 10 − 11 = 1 . 7 8 × 1 0 - 4 An equilibrium table can be set up as given below. C o m p o s i t i o n ( M ) C O 3 2 − H 2 O O H − H C O 3 − I n i t i a l c o n c e n t r a t i o n 0.0456 − 0 0 C h a n g e i n c o n c e n t r a t i o n − x − + x + x E q u i l i b r i u m c o n c e n t r a t i o n 0.0456 − x − x x Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below. K b 1 = ( x ) ( x ) ( 0.0456 − x ) Now, the above expression can be solved for x. K b 1 = ( x ) ( x ) ( 0.0456 − x ) 1.78 × 10 − 4 = ( x ) ( x ) ( 0.0456 − x ) ( 1.78 × 10 − 4 ) ( 0.0456 − x ) = x 2 x 2 + ( 1.78 × 10 − 4 ) x − 8.143 × 10 − 6 = 0 x = ( − 1.78 × 10 − 4 ) ± ( 1.78 × 10 − 4 ) 2 − ( 4 ) × ( 1 ) × ( − 8.143 × 10 − 6 ) 2 x = 0.0028 [ O H − ] = x = 0 . 0 0 2 8 M [ H C O 3 − ] = x = 0 . 0 0 2 8 M [ C O 3 2 − ] = 0.0456 − x = 0.0456 − 0.0028 = 0 . 0 4 2 8 M The second protonation reaction is given below, H C O 3 − ( a q ) + H 2 O ( l ) ⇌ O H − ( a q ) + H 2 C O 3 ( a q ) The expression for equilibrium constant for the second protonation reaction can be written as shown below, K b 2 = [ O H − ] [ H C O 3 ] [ H C O 3 − ] = K w K a 1 = 1

Chemical Principles: The Quest for Insight

7th Edition

ISBN: 9781464183959

Author: Peter Atkins, Loretta Jones, Leroy Laverman

Publisher: W. H. Freeman

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Chapter 6, Problem 6E.11E

Interpretation Introduction

Interpretation:

The molar concentrations of H2CO3, HCO3−, CO32−, H3O+, and OH− in 0.0456M Na2CO3(aq) have to be calculated.

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Chapter 6, Problem 6E.10E

Chapter 6, Problem 6E.12E

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Chapter 6 Solutions

Chemical Principles: The Quest for Insight

Ch. 6 - Prob. 6A.1AST Ch. 6 - Prob. 6A.1BST Ch. 6 - Prob. 6A.2AST Ch. 6 - Prob. 6A.2BST Ch. 6 - Prob. 6A.3AST Ch. 6 - Prob. 6A.3BST Ch. 6 - Prob. 6A.1E Ch. 6 - Prob. 6A.2E Ch. 6 - Prob. 6A.3E Ch. 6 - Prob. 6A.4E

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The molar concentrations of H 2 C O 3 , H C O 3 − , C O 3 2 − , H 3 O + , a n d O H − in 0.0456 M N a 2 C O 3 ( a q ) have to be calculated.

Solution Summary: The author explains how the molar concentrations for the first protonation reaction can be calculated.

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Interpretation:

The molar concentrations of H2CO3, HCO3−, CO32−, H3O+, and OH− in 0.0456M Na2CO3(aq) have to be calculated.

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Chapter 6 Solutions

The molar concentrations of H 2 C O 3 , H C O 3 − , C O 3 2 − , H 3 O + , a n d O H − in 0.0456 M N a 2 C O 3 ( a q ) have to be calculated.

Solution Summary: The author explains how the molar concentrations for the first protonation reaction can be calculated.

Concept explainers

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Interpretation: The molar concentrations of H2CO3, HCO3−, CO32−, H3O+, and OH− in 0.0456M Na2CO3(aq) have to be calculated.

Want to see the full answer?

Chapter 6 Solutions

Interpretation:

The molar concentrations of H2CO3, HCO3−, CO32−, H3O+, and OH− in 0.0456M Na2CO3(aq) have to be calculated.