EBK EXPLORING CHEMICAL ANALYSIS
EBK EXPLORING CHEMICAL ANALYSIS
5th Edition
ISBN: 8220101443908
Author: Harris
Publisher: MAC HIGHER
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Chapter 10, Problem 10.19P
Interpretation Introduction

Interpretation:

Equivalence volume and pH at indicated volumes has to be calculated and plot of pH against Va as to be made.

Concept Introduction:

Dilution formula is given as follows:

  M1V1=M2V2  

Here,

M1 denotes molarity of  HBr solution.

V1 denotes volume of HBr solution.

M2 denotes molarity of NaOH solution.

V2 denotes volume  of NaOH solution.

Expert Solution & Answer
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Explanation of Solution

Dilution formula is given as follows:

  M1V1=M2V2        (1)

Substitute  1.00 M for M1, 0.1 M for M2 , 100 mL for V2, in equation (1).

  (1.00 M)V1=(0.1 M)(100 mL)        (2)

Rearrange equation (2) to calculate value of V1.

  V1=(0.1 M)(100 mL)(1.00 M)=10 mL

Thus, equivalence volume is 10 mL.

Formula to calculate pKb from pKa is given as follows:

  pKb=14pKa        (3)

Substitute 6.993 for pKb in equation (3).

  pKb=146.993=7.007

Formula to calculate Kb from pKb is given as follows:

  Kb=10pKb        (4)

Substitute 7.007 for pKb in equation (4).

  Kb=107.007

Equilibrium for titration reaction is given as follows:

  B+H2OHB+OH

Corresponding expression of Kb is written as follows:

  Kb=[BH+][OH][B]        (4)

When no acid is added, ICE table is drawn as follows:

BHB++OHInitial0.05000Changex+x+xEquilibrium0.050xxx

Substitute x to [BH+] and [OH] , 107.007 for Kb , 0.050x for [B] in equation (4).

  107.007=x20.050x

Simplify to obtain the value of x as 7.009×105 M.

Formula to calculate pOH is given as follows:

  pOH=log[OH]        (5)

Substitute 7.009×105 M for [OH] in equation (5).

  pOH=log(7.009×105)=4.15

Formula to calculate pH from pOH is given as follows:

  pH=14pOH        (6)

Substitute 4.15 for pOH in equation (6).

  pH=144.15=9.8459.85

pH is calculated by Henderson-HasselBalch equation given as follows:

  pH=pKb+log([Salt][Acid])        (7)

When volume added is 1 mL,  initial and final moles are tabulated as follows:

B+HNO3NO3+BH+Initial0.001250.00012500Final0.00112500.0001250.000125

Substitute 7 for pKb, 0.001125 for [Salt], 0.000125 M for [Acid] in equation (7).

  pH=7+log([Salt][Acid])=7+log(0.0090.001)=7.95

When volume added is 5 mL,  initial and final moles are tabulated as follows:

B+HNO3NO3+BH+Initial0.001250.00062500Final0.00062500.0006250.000625

Substitute 7 for pKb, 0.000625 M for [Salt], 0.000625 M for [Acid] in equation (7).

  pH=7+log([Salt][Acid])=7+log(0.0006250.000625)=7

When volume added is 9 mL,  initial and final moles are tabulated as follows:

B+HNO3NO3+BH+Initial0.001250.00112500Final0.00012500.0011250.001125

Substitute 7 for pKb, 0.000125 M for [Salt], 0.001125 M for [Acid] in equation (7).

  pH=7+log([Salt][Acid])=7+log(0.0001250.001125)=6.04

When volume added is 9.90 mL,  initial and final moles are tabulated as follows:

B+HNO3NO3+BH+Initial0.001250.001237500Final0.000012500.00123750.0012375

Substitute 7 for pKb, 0.0000125 M for [Salt], 0.0012375 M for [Acid] in equation (7).

  pH=7+log([Salt][Acid])=7+log(0.00001250.0012375)=5.00

When volume added is 10.00 mL,  initial and final moles are tabulated as follows:

B+HNO3NO3+BH+Initial0.001250.0012500Final000.001250.00125

Thus excess [BH+] is calculated as follows:

  [BH+]=0.001250.025+0.01=0.0357 M

Corresponding ICE table is as follows:

BH+B+H3O+Initial0.035700Changex+x+xEquilibrium0.0357xxx

Corresponding expression of Ka is written as follows:

  Ka=[B][H3O+][BH+]        (8)

Substitute x to [B] and [H3O+] , 1.02×107 for Ka , 0.0357x for [BH+] in equation (4).

  1.02×107=x20.0357x

Simplify to obtain the value of x as 6.029×105.

Formula to calculate pH is given as follows:

  pH=log[H+]        (9)

Substitute 6.029×105 M for [H+] in equation (9).

  pH=log(6.029×105)=4.2194.22

When volume added is 10.10 mL, excess [H+] is calculated as follows:

  [H+]=(0.0012650.00125) mol(0.025+0.0101) L=3.56×104 M

Substitute 3.56×104 M for [H+] in equation (9).

  pH=log(3.56×104)=3.4483.45

When volume added is 12.00 mL, excess [H+] is calculated as follows:

  [H+]=(0.00150.00125) mol(0.025+0.012) L=0.00675 M

Substitute 0.00675 M for [H+] in equation (9).

  pH=log(0.00675)=2.17

Corresponding plot of pH against Va is drawn as follows:

EBK EXPLORING CHEMICAL ANALYSIS, Chapter 10, Problem 10.19P

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