Explanation Equilibrium for titration reaction is given as follows: C 6 H 5 NH 3 + + OH − ⇌ C 6 H 5 NH 2 + H 2 O Dilution formula is given as follows: M 1 V e = M 2 V 2 (1) Substitute 0.100 M for M 1 , 0 .100 M for M 2 , 100 mL for V 2 , in equation (1). ( 0.100 M ) V e = ( 0 .100 M ) ( 100 mL ) (2) Rearrange equation (2) to calculate value of V e . V e = ( 0 .100 M ) ( 100 mL ) ( 0 .100 M ) = 100 mL Thus, value of V e is 5 mL . Equilibrium for titration reaction is given as follows: BH + + H 2 O ⇌ B + H 3 O + Corresponding expression of K a is written as follows: K a = [ B ] [ H 3 O + ] [ BH + ] (3) When no acid is added, ICE table is drawn as follows: BH + ⇌ B + H 3 O + Initial 0.01 0 0 Change − x + x + x Equilibrium 0.01 − x x x Substitute x to [ B ] and [ H 3 O + ] , 2.51 × 10 − 5 for K a , 0.01 − x for [ BH + ] in equation (3). 2.51 × 10 − 5 = x 2 0.01 − x Simplify to obtain the value of x as 1.584 × 10 − 3 M . Formula to calculate pH is given as follows: pH = − log [ H + ] (4) Substitute 1.584 × 10 − 3 M for [ H + ] in equation (4). pH = − log ( 1.584 × 10 − 3 ) = 3.64 pH is calculated by Henderson-HasselBalch equation given as follows: pH = p K BH + + log ( [ B ] [ BH + ] ) (5) When volume of base added is 0 .1 ( V e ) mL , initial and final moles are tabulated as follows: BH + + OH − ⇌ B + H 2 O Initial 0.01 0.001 0 0 Final 0.009 0 0.001 0.001 Substitute 4.601 for p K BH + , 0.001 M for [ B ] , 0.009 M for [ BH + ] in equation (5). pH = 4.601 + log ( 0.001 0.009 ) = 3.648 ≃ 3.65 When volume of base added is 0 .5 ( V e ) mL , initial and final moles are tabulated as follows: BH + + OH − ⇌ B + H 2 O Initial 0.01 0.005 0 0 Final 0.005 0 0.005 0.005 Substitute 4.601 for p K BH + , 0.005 M for [ B ] , 0.005 M for [ BH + ] in equation (5). pH = 4.601 + log ( 0.005 0.005 ) = 4

Titration reaction has to be written and plot of pH against V b has to be made. Concept Introduction: Dilution formula is given as follows: M 1 V 1 = M 2 V 2 Here, M 1 denotes molarity of acid solution. V 1 denotes volume of acid. M 2 denotes molarity of base. V 2 denotes volume of base.

Exploring Chemical Analysis

5th Edition

ISBN: 9781429275033

Author: Daniel C. Harris

Publisher: Macmillan Higher Education

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Question

Chapter 10, Problem 10.10P

Interpretation Introduction

Interpretation:

Titration reaction has to be written and plot of pH against Vb has to be made.

Concept Introduction:

Dilution formula is given as follows:

M1V1=M2V2

Here,

M1 denotes molarity of acid solution.

V1 denotes volume of acid.

M2 denotes molarity of base.

V2 denotes volume of base.

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Chapter 10, Problem 10.9P

Chapter 10, Problem 10.11P

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