Explanation H 2 C O 3 is a diprotic protic acid. The primary transfer equilibrium is given below, H 2 C O 3 ( a q ) + H 2 O ( l ) ⇌ H 3 O + ( a q ) + H C O 3 − ( a q ) , K a 1 = 4.3 × 10 − 7 The expression for equilibrium constant for the first deprotonation reaction can be written as shown below, K a 1 = [ H 3 O + ] [ H C O 3 − ] [ H 2 C O 3 ] An equilibrium table can be set up as given below. C o m p o s i t i o n ( M ) H 2 C O 3 H 2 O H 3 O + 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 a 1 = ( x ) ( x ) ( 0.0456 − x ) Now, the above expression can be solved for x. K a 1 = ( x ) ( x ) ( 0.0456 − x ) 4.3 × 10 − 7 = ( x ) ( x ) ( 0.0456 − x ) ( 4.3 × 10 − 7 ) ( 0.0456 − x ) = x 2 x 2 + 4.3 × 10 − 7 x − 1.96 × 10 − 8 = 0 x = ( − 4.3 × 10 − 7 ) ± ( 4.3 × 10 − 7 ) 2 − ( 4 ) × ( 1 ) × ( − 1.96 × 10 − 8 ) 2 x = 1.4 × 10 − 4 [ H 3 O + ] = x = 1 . 4 × 1 0 - 4 M [ H C O 3 − ] = x = 1 . 4 × 1 0 - 4 M [ H 2 C O 3 ] = 0.0456 − x ≈ 0 . 0 4 5 6 M The second deprotonation reaction is given below, H C O 3 − ( a q ) + H 2 O ( l ) ⇌ H 3 O + ( a q ) + C O 3 2 − ( a q ) , K a 2 = 5.6 × 10 − 11 The expression for equilibrium constant for the second deprotonation reaction can be written as shown below, K a 2 = [ H 3 O + ] [ C O 3 2 − ] [ H C O 3 − ] An equilibrium table can be set up as given below

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM

7th Edition

ISBN: 9781319403959

Author: ATKINS

Publisher: MAC HIGHER

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Chapter 6, Problem 6E.9E

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The concentrations of all solute species in 0.0456 M H2CO3(aq) have to be calculated.

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Chapter 6, Problem 6E.8E

Chapter 6, Problem 6E.10E

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ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM

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 concentrations of all solute species in 0.0456 M H 2 C O 3 ( a q ) have to be calculated.

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Interpretation: The concentrations of all solute species in 0.0456 M H2CO3(aq) have to be calculated.

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Chapter 6 Solutions

Interpretation:

The concentrations of all solute species in 0.0456 M H2CO3(aq) have to be calculated.