Explanation A.The weak monoprotic acid is HA, having initial concentration 0.100M. HA(aq)+H 2 O(1)→H 3 O + (aq)+A − (aq) 1 (M) 0.100 0 0 C (M) X X X E (M) 0.100- X X X Ka= [ H3O ] [ A - ] HA Given as, Ka = 1.0×10 −5 By substituting the values in the above Ka equation, we get, 1 .0×10 -5 = ( x ) ( x ) ( 0 .100-x ) ( 0 .100-x ) »0 .100 x= 1 .0×10 -6 =1 .0×10 -3 = [ H3O + ] Now formula for % ionization is, [ H3O + ] [ HA ] ×100% = 1 .0×10 -3 0 .10 ×100% =1 .0% B.the weak monoprotic acid is HA, having initial concentration 0.100M. HA(aq)+H 2 O(1)→H 3 O + (aq)+A − (aq) 1 (M) 0.100 0 0 C (M) X X X E (M) 0.100- X X X Ka= [ H3O ] [ A - ] HA Given as, Ka = 1.0×10 −5 By substituting the values in the above Ka equation, we get, 1 .0×10 -3 = ( x ) ( x ) ( 0 .100-x ) ( 0 ...

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ISBN: 9780321834683

Author: Nivaldo J. Tro

Publisher: PEARSON

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Chapter 17, Problem 80E

Interpretation Introduction

To determine: Ka value.

A. Ka = 1.0×10⁻³

B. Ka = 1.0×10⁻³

C. Ka = 1.0×10⁻¹

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Chapter 17, Problem 79E

Chapter 17, Problem 81E

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Chemistry: Structure and Properties

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