Explanation H 2 A is a polyprotic acid. It is strong acid in its first deprotonation. Therefore, to estimate its p H , the second deprotonation needs to be taken in account. The first and second deprotonation reactions are given below. H 2 A ( a q ) + H 2 O ( l ) → H 3 O + ( a q ) + H A − ( a q ) H A − ( a q ) + H 2 O ( l ) ⇌ H 3 O + ( a q ) + A 2 − ( a q ) , K a 2 The expression for equilibrium constant for the above reaction can be written as shown below, K a 2 = [ H 3 O + ] [ A 2 − ] [ H A − ] The conjugate base H A − is amphiprotic and it contributes protons to the solution which has been shown in the second deprotonation reaction above. An equilibrium table can be set up as given below. C o m p o s i t i o n ( M ) H A − H 2 O H 3 O + A 2 − I n i t i a l c o n c e n t r a t i o n 0.015 − 0.015 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 0.015 − x − 0.015 + x x Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below

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.8E

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The value of Ka2 for the acid H2A has to be calculated.

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Chapter 6, Problem 6E.7E

Chapter 6, Problem 6E.9E

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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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