The p H of 0.10 M L i C l should be determined Concept introduction: p H is defined as the negative logarithm of the hydrogen ion of the solution p H = − log [ H 3 O + ] Strong acid dissociates completely in solution and therefore, the molar concentration of H + ion in the solution is same as that of the acid itself.But the weak acid is not completely dissociated, and the relative strength of the weak acid can be compared in terms of their dissociation or ionization constant.The dissociation equilibrium of an acid HA, may be represented as H A ( a q ) + H 2 O ( l ) ⇌ H 3 O + ( a q ) + A − ( a q ) Applying law of equilibrium K = [ H 3 O + ] [ A − ] [ H A ] [ H 2 O ] Since water is present in a large quantity its concentration remains practically constant, the concentration of water may be combined with K to give another constant K a . K [ H 2 O ] = [ H 3 O + ] [ A − ] [ H A ] K a = [ H 3 O + ] [ A − ] [ H A ] Where K a is the dissociation constant

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Chapter 15, Problem 15.155MP

Interpretation Introduction

Interpretation:

The $pH$ of $0.10 M$

$LiCl$ should be determined

Concept introduction:

$pH$ is defined as the negative logarithm of the hydrogen ion of the solution
$pH=−log[H3O+]$
Strong acid dissociates completely in solution and therefore, the molar concentration of $H+$ ion in the solution is same as that of the acid itself.But the weak acid is not completely dissociated, and the relative strength of the weak acid can be compared in terms of their dissociation or ionization constant.The dissociation equilibrium of an acid HA, may be represented as
$HA(aq)+H2O(l)⇌H3O+(aq)+A−(aq)$

Applying law of equilibrium

$K=[H3O+][A−][HA][H2O]$

Since water is present in a large quantity its concentration remains practically constant, the concentration of water may be combined with K to give another constant $Ka$ .

$K[H2O]=[ H 3 O +][ A −][HA]Ka=[ H 3 O +][ A −][HA]$

Where $Ka$ is the dissociation constant

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