EBK EXPLORING CHEMICAL ANALYSIS
EBK EXPLORING CHEMICAL ANALYSIS
5th Edition
ISBN: 9781319416942
Author: Harris
Publisher: VST
Question
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Chapter 10, Problem 10.8P
Interpretation Introduction

Interpretation:

Titration reaction has to be written along with equivalence volume and pH at indicated volumes has to be calculated and plot of pH against Vb has to be drawn.

Concept Introduction:

Dilution formula is given as follows:

  M1V1=M2V2

Here,

M1 denotes molarity of  HA solution.

V1 denotes volume of HA solution.

M2 denotes molarity of NaOH solution.

V2 denotes volume  of NaOH solution.

Expert Solution & Answer
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Explanation of Solution

The dilution formula is given as follows:

  M1V1=M2V2        (1)

Substitute 0.05 M for M1, 0.500 M for M2 , 50 mL for V1 in equation (1).

  (0.050 M)(50.0 mL)=(0.500 M)V2        (2)

Rearrange equation (2) to calculate value of V2.

  V2=(0.050 M)(50.0 mL)(0.5 M)=5.00 mL

So, equivalence volume is 5.00 mL.

Equilibrium for titration reaction is given as follows:

  HAH++A

Formula to calculate pKb from pKb is given as follows:

  pKa=14pKb        (3)

Substitute 9.00 for pKa in equation (3).

  pKa=145=9

Formula to calculate pKa is given as follows:

  Ka=10pKa        (3)

Substitute 4.00 for pKa in equation (3).

  Ka=104

Corresponding expression of Ka is written as follows:

  Ka=[H+][A][HA]        (4)

ICE table is drawn as follows:

HAH++AInitial0.0500Changex+x+xEquilibrium0.05xxx

Substitute x to [H+] and [A] , 104 for Ka 0.05x for [HA] in equation (4)

  104=x20.05x

Simplify to obtain the value of x as 0.00218 M. Thus [HA] is calculated as follows:

  [HA]=(0.050.00218) M=0.0478 M

Until just before equivalence point there is a mixture of unionized HA and A. pH is calculated by Henderson-HasselBalch equation given as follows:

  pH=pKa+log[A][HA]        (5)

Substitute 4.00 for pKa, 0.00218 M for [A], 0.0478 M for [HA] in equation (5).

  pH=4.00+log(0.002180.0478)=2.6592.66

When volume of base added is 1 mL,  new ICE table is drawn as follows:

HA+OHA2.500.50Change0.50.5+xEquilibrium2.000.5

Substitute 4.00 for pKa, 0.5 M for [A], 2.0 M for [HA] in equation (5).

  pH=4.00+log(0.52.0)=3.3973.40

Similarly when volume of base added is 2.5 mL, new ICE table is drawn as follows:

HA+OHAInitial2.501.250Change1.251.25+1.25Equilibrium1.2501.25

Substitute 4.00 for pKa, 1.25 M for [A] and [HA] in equation (5).

  pH=4.00+log(1.251.25)=4.00

Similarly when volume of base added is 4.00 mL, new ICE table is drawn as follows:

HA+OHAInitial2.502.000Change2.002.00+2.00Equilibrium0.5002.00

Substitute 4.00 for pKa, 0.50 M for [A], 2.00 M and [HA] in equation (5).

  pH=4.00+log(2.000.50)=4.60

Similarly when volume of base added is 4.90 mL, new ICE table is drawn as follows:

HA+OHAInitial2.502.450Change2.452.45+2.45Equilibrium0.0502.45

Substitute 4.00 for pKa, 2.45 M for [A], 0.05 M and [HA] in equation (5).

  pH=4.00+log(2.450.05)=5.69

When volume of base added is 5.00 mL, value of [A] is calculated as follows:

  [A]=(50.00 mL)(0.050 M)(50.00+5.00)mL=0.0454 M

New ICE table can be drawn as follows:

  AHA+OHInitial0.045400Changey+y+yEquilibrium0.0454yyy

Expression of equilibrium constant for this reaction is given as follows:

  Kb=KwKa        (6)

Substitute 1014 for Kw, and 104 for Ka in equation (6).

  Kb=1014104=1010

Therefore expression of equilibrium constant can be written as follows:

  1010=y20.0454y

Simplify to obtain the value of y as 2.13×106. Expression to calculate pH is then given as follows:

  pH=log(Kwy)        (7)

Substitute 1014 for Kw, and 2.13×106 for y in equation (7).

  1010=y20.04537y=2.129×106

When 5.10 mL of base is added it exceeds acid by 0.05 M, and total volume is 55.10 mL, thus excess [OH] is calculated as follows:

  [OH]=0.05 M55.10 mL=9.0744×104

Formula to calculate pOH is given as follows:

  pOH=log[OH]        (3)

Substitute 9.0744×104 M for [OH] in equation (3).

  pOH=log(9.0744×104)=3.04

Formula to calculate pH from pOH is given as follows:

  pH=14pOH        (4)

Substitute 3.04 for pOH in equation (4).

  pH=143.04=10.95

Similarly when 6.00 mL of base is added it exceeds acid by 0.5 M, and total volume is 56.00 mL, thus excess [OH] is calculated as follows:

  [OH]=0.5 M56.00 mL=8.92×103

Formula to calculate pOH is given as follows:

  pOH=log[OH]        (3)

Substitute 8.92×103 M for [OH] in equation (3).

  pOH=log(8.92×103)=2.04

Formula to calculate pH from pOH is given as follows:

  pH=14pOH        (4)

Substitute 2.04 for pOH in equation (4).

  pH=143.04=11.95

Corresponding plot of pH against Va is drawn as follows:

EBK EXPLORING CHEMICAL ANALYSIS, Chapter 10, Problem 10.8P

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