Connect 2-Year Access Card for Chemistry: The Molecular Nature of Matter and Change
Connect 2-Year Access Card for Chemistry: The Molecular Nature of Matter and Change
7th Edition
ISBN: 9780078129865
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
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Concept explainers

Acid Base Titration
An acid-base titration is the method of quantitative analysis for determining the concentration of an acid and base by exactly neutralizing it with a standard solution of base or acid having known concentration.
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Question
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Chapter 19, Problem 19.54P

(a)

Interpretation Introduction

Interpretation:

The value of  pH during the titration of 40.00 ml of 0.1000MHCl with 0.00 ml of 0.1000MNaOH solution has to be calculated. 

Concept Introduction:

For the titration of a strong acid with a strong base, the  pH before the equivalence point depends on the excess concentration of acid and the  pH after the equivalence point depends on the excess concentration of base.  At the equivalence point, there is not an excess of either acid or base so pH is 7.0.  The equivalence point occurs when total volume of base has been added. 

pH=−log[H3O+]

pH = −log[OH−]

pH+pOH=14

Calculation of Moles:

1. No. of moles (n)=WM.W

Where,

W         = Weight of a given compoundM.W = Molecular weight of the same compound

2. No. of moles (n) = (Molarity)( Volume)

Molarity:

Molarity or molar concentration is the number of moles of solute dissolved per liter of solution, which can be calculated using the following equation,

Molarity = moles of solutevolume of solution in L

(a)

Expert Solution
Check Mark

Answer to Problem 19.54P

The  pH during the titration of 40.00 ml of 0.1000M HCl with 0.00 ml of 0.1000M NaOH solution has calculated as 1.0000. 

Explanation of Solution

Given Data:

Volume of  HCl                = 40.00 ml = 40.00×10−3LConcentration of HCl       = 0.1000MConcentration of NaOH  = 0.1000MVolume of NaOH added  =   0.00 ml = 15.00×10−3LThe total volume of the solution at this point = 40.00×10−3L + 0.00×10−3L                                                                                           = 40.00×10−3L 

The reaction occurring in the titration is the neutralization of H3O+ (from HCl) by OH− (from NaOH):

HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)H3O+(aq) + OH−(aq) → 2H2O(l) 

The initial number of moles of HCl= Molarity×Volume                                                                             = (0.1000 M)(40.00×10−3L)                                                                             = 4.000×10−3.

Moles of added NaOH = Molarity×Volume                                               = (0.1000 M)(0.00×10−3L)                                               =0.

No. of moles                   HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)_ Initial                                 4×10−3              0                          −                     0Change                               0                          0                          −                     0_Final                                  4×10−3               0                          −                     0       

Excess[H3O+] =  4.000×10−3 moles40.00×10−3L                                  = 0.1000 M.

pH=−log[H3O+]       = −log(0.1000)       = 1.0000.

Therefore, the  pH of the solution has been calculated to be 1.0000.

(b)

Interpretation Introduction

Interpretation:

The  pH during the titration of 40.00 ml of HCl with 0.1000M 25.00 ml of 0.1000M NaOH solution has to be calculated. 

Concept Introduction:

For the titration of a strong acid with a strong base, the  pH before the equivalence point depends on the excess concentration of acid and the  pH after the equivalence point depends on the excess concentration of base.  At the equivalence point, there is not an excess of either acid or base so pH is 7.0.  The equivalence point occurs when total volume of base has been added. 

pH=−log[H3O+]

pH = −log[OH−]

pH+pOH=14

Calculation of Moles:

1. No. of moles (n)=WM.W

Where,

W         = Weight of a given compoundM.W = Molecular weight of the same compound 

2. No. of moles (n) = (Molarity)( Volume)

Molarity:

Molarity or molar concentration is the number of moles of solute dissolved per liter of solution, which can be calculated using the following equation,

Molarity = moles of solutevolume of solution in L

(b)

Expert Solution
Check Mark

Answer to Problem 19.54P

The  pH during the titration of 40.00 ml of HCl with 0.1000M 25.00 ml of 0.1000M NaOH solution has calculated as 1.6368.

Explanation of Solution

Given Data:

Volume of  HCl                = 40.00 ml = 40.00×10−3LConcentration of HCl       = 0.1000MConcentration of NaOH  = 0.1000MVolume of NaOH added  =   25.00 ml = 25.00×10−3LThe total volume of the solution at this point = 40.00×10−3L + 25.00×10−3L                                                                                           = 65.00×10−3L 

The reaction occurring in the titration is the neutralization of H3O+ (from HCl) by OH− (from NaOH):

HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)H3O+(aq) + OH−(aq) → 2H2O(l) 

The initial number of moles of HCl= Molarity×Volume                                                                             = (0.1000 M)(40.00×10−3L)                                                                             = 4.000×10−3.

Moles of added NaOH = Molarity×Volume                                               = (0.1000 M)(25.00×10−3L)                                               = 2.5×10−3.

No. of moles                   HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)_ Initial                                 4×10−3         2.5×10−3                −                   0Change                         −2.5×10−3     − 2.5×10−3             −            +2.5×10−3_Final                                 1.5×10−3               0                         −              2.5×10−3       

Excess[H3O+] =  1.500×10−3 moles65.00×10−3L                  = 0.023077M.pH=−log[H3O+]       = −log(0.023077)       = 1.6368.

Therefore, the  pH of the solution has been calculated to be 1.6368.

(c)

Interpretation Introduction

Interpretation:

The  pH during the titration of 40.00 ml of HCl 0.1000M with 39.00 ml of 0.1000M NaOH solution has to be calculated. 

Concept Introduction:

For the titration of a strong acid with a strong base, the  pH before the equivalence point depends on the excess concentration of acid and the  pH after the equivalence point depends on the excess concentration of base.  At the equivalence point, there is not an excess of either acid or base so pH is 7.0.  The equivalence point occurs when total volume of base has been added. 

pH=−log[H3O+]

pH = −log[OH−]

pH+pOH=14

Calculation of Moles:

1. No. of moles (n)=WM.W

Where,

W         = Weight of a given compoundM.W = Molecular weight of the same compound 

2. No. of moles (n) = (Molarity)( Volume)

Molarity:

Molarity or molar concentration is the number of moles of solute dissolved per liter of solution, which can be calculated using the following equation,

Molarity = moles of solutevolume of solution in L

(c)

Expert Solution
Check Mark

Answer to Problem 19.54P

The  pH during the titration of 40.00 ml of HCl 0.1000M with 39.00 ml of 0.1000M NaOH solution has calculated as 2.898.

Explanation of Solution

Given Data:

Volume of  HCl                = 40.00 ml = 40.00×10−3LConcentration of HCl       = 0.1000MConcentration of NaOH  = 0.1000MVolume of NaOH added  =   39.00 ml = 39.00×10−3LThe total volume of the solution at this point = 40.00×10−3L + 39.00×10−3L                                                                                           = 79.00×10−3L 

The reaction occurring in the titration is the neutralization of H3O+ (from HCl) by OH− (from NaOH):

HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)H3O+(aq) + OH−(aq) → 2H2O(l) 

The initial number of moles of HCl= Molarity×Volume                                                                             = (0.1000 M)(40.00×10−3L)                                                                             = 4.000×10−3.

Moles of added NaOH = Molarity×Volume                                               = (0.1000 M)(39.00×10−3L)                                               = 3.9×10−3.

No. of moles                   HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)_ Initial                                 4×10−3         3.9×10−3                −                   0Change                         −3.9×10−3     − 3.9×10−3             −            +3.9×10−3_Final                                 0.1×10−3               0                         −              3.9×10−3       

Excess[H3O+] =  0.100×10−3 moles79.00×10−3L                                 = 0.0012658M.pH=−log[H3O+]       = −log(0.0012658)       = 2.898.

Therefore, the  pH of the solution has been calculated to be 2.898. 

(d)

Interpretation Introduction

Interpretation:

The  pH during the titration of 40.00 ml of HCl 0.1000M with 39.90 ml of 0.1000M NaOH solution has to be calculated. 

Concept Introduction:

For the titration of a strong acid with a strong base, the  pH before the equivalence point depends on the excess concentration of acid and the  pH after the equivalence point depends on the excess concentration of base.  At the equivalence point, there is not an excess of either acid or base so pH is 7.0.  The equivalence point occurs when total volume of base has been added.

pH=−log[H3O+]

pH = −log[OH−]

pH+pOH=14

Calculation of Moles:

1. No. of moles (n)=WM.W

Where,

W         = Weight of a given compoundM.W = Molecular weight of the same compound 

2. No. of moles (n) = (Molarity)( Volume)

Molarity:

Molarity or molar concentration is the number of moles of solute dissolved per liter of solution, which can be calculated using the following equation,

Molarity =  moles of solutevolume of solution in L

(d)

Expert Solution
Check Mark

Answer to Problem 19.54P

The  pH during the titration of 40.00 ml of HCl 0.1000M with 39.90 ml of 0.1000M NaOH solution has calculated as 3.903.

Explanation of Solution

Given Data:

Volume of  HCl                = 40.00 ml = 40.00×10−3LConcentration of HCl       = 0.1000MConcentration of NaOH  = 0.1000MVolume of NaOH added  =   39.90 ml = 39.90×10−3LThe total volume of the solution at this point = 40.00×10−3L + 39.90×10−3L                                                                                           = 79.90×10−3L 

The reaction occurring in the titration is the neutralization of H3O+ (from HCl) by OH− (from NaOH):

HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)H3O+(aq) + OH−(aq) → 2H2O(l) 

The initial number of moles of HCl= Molarity×Volume                                                                             = (0.1000 M)(40.00×10−3L)                                                                             = 4.000×10−3.

Moles of added NaOH = Molarity×Volume                                               = (0.1000 M)(39.90×10−3L)                                               = 3.99×10−3.

No. of moles                   HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)_ Initial                                 4×10−3         3.99×10−3                −                   0Change                      −3.99×10−3     − 3.99×10−3             −            +3.99×10−3_Final                                 0.01×10−3               0                        −              3.99×10−3       

Excess[H3O+] =  0.001×10−3 moles79.90×10−3L                  = 0.00012484M.pH=−log[H3O+]       = −log(0.00012484)       = 3.903.

Therefore, the  pH of the solution has been calculated to be 3.903. 

(e)

Interpretation Introduction

Interpretation:

The  pH during the titration of 40.00 ml of HCl 0.1000M with 40.00 ml of 0.1000M NaOH solution has to be calculated. 

Concept Introduction:

For the titration of a strong acid with a strong base, the  pH before the equivalence point depends on the excess concentration of acid and the  pH after the equivalence point depends on the excess concentration of base.  At the equivalence point, there is not an excess of either acid or base so pH is 7.0.  The equivalence point occurs when total volume of base has been added. 

pH=−log[H3O+]

pH = −log[OH−]

pH+pOH=14

Calculation of Moles:

1. No. of moles (n)=WM.W

Where,

W         = Weight of a given compoundM.W = Molecular weight of the same compound 

2. No. of moles (n) = (Molarity)( Volume)

Molarity:

Molarity or molar concentration is the number of moles of solute dissolved per liter of solution, which can be calculated using the following equation,

Molarity =  moles of solutevolume of solution in L

(e)

Expert Solution
Check Mark

Answer to Problem 19.54P

The  pH during the titration of 40.00 ml of HCl 0.1000M with 40.00 ml of 0.1000M NaOH solution has calculated as 7.00.

Explanation of Solution

Given Data:

Volume of  HCl                = 40.00 ml = 40.00×10−3LConcentration of HCl       = 0.1000MConcentration of NaOH  = 0.1000MVolume of NaOH added  =   40.00 ml = 40.00×10−3LThe total volume of the solution at this point = 40.00×10−3L + 40.00×10−3L                                                                                           = 80.00×10−3L 

The reaction occurring in the titration is the neutralization of H3O+ (from HCl) by OH− (from NaOH):

HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)H3O+(aq) + OH−(aq) → 2H2O(l) 

The initial number of moles of HCl= Molarity×Volume                                                                             = (0.1000 M)(40.00×10−3L)                                                                             = 4.000×10−3.

Moles of added NaOH = Molarity×Volume                                               = (0.1000 M)(40.00×10−3L)                                               = 4.00×10−3.

No. of moles                   HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)_ Initial                                 4×10−3         4.00×10−3                −                   0Change                      −4.00×10−3     − 4.00×10−3             −            +4.00×10−3_Final                                       0                         0                           −              4.00×10−3       

The NaOH will react with an equal amount of the HCl and 0.0 mol HCl will remain.  This is the equivalence point of a strong acid–strong base titration, thus, the pH is 7.0.  Only the neutral salt NaCl is in solution at the equivalence point. 

(f)

Interpretation Introduction

Interpretation:

The  pH during the titration of 40.00 ml of HCl 0.1000M with 40.10 ml of 0.1000M NaOH solution has to be calculated. 

Concept Introduction:

For the titration of a strong acid with a strong base, the  pH before the equivalence point depends on the excess concentration of acid and the  pH after the equivalence point depends on the excess concentration of base.  At the equivalence point, there is not an excess of either acid or base so pH is 7.0.  The equivalence point occurs when total volume of base has been added. 

pH=−log[H3O+]

pH = −log[OH−]

pH+pOH=14

Calculation of Moles:

1. No. of moles (n)=WM.W

Where,

W         = Weight of a given compoundM.W = Molecular weight of the same compound 

2. No. of moles (n) = (Molarity)( Volume)

Molarity:

Molarity or molar concentration is the number of moles of solute dissolved per liter of solution, which can be calculated using the following equation,

Molarity =  moles of solutevolume of solution in L

(f)

Expert Solution
Check Mark

Answer to Problem 19.54P

The  pH during the titration of 40.00 ml of HCl 0.1000M with 40.10 ml of 0.1000M NaOH solution has calculated as 10.10.

Explanation of Solution

Given Data:

Volume of  HCl                = 40.00 ml = 40.00×10−3LConcentration of HCl       = 0.1000MConcentration of NaOH  = 0.1000MVolume of NaOH added  =   40.10 ml = 40.10×10−3LThe total volume of the solution at this point = 40.00×10−3L + 40.10×10−3L                                                                                           = 80.10×10−3L 

The reaction occurring in the titration is the neutralization of H3O+ (from HCl) by OH− (from NaOH):

HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)H3O+(aq) + OH−(aq) → 2H2O(l) 

The initial number of moles of HCl= Molarity×Volume                                                                             = (0.1000 M)(40.00×10−3L)                                                                             = 4.000×10−3.

Moles of added NaOH = Molarity×Volume                                               = (0.1000 M)(40.10×10−3L)                                               = 4.01×10−3.

The HCl will react with an equal amount of the base. 

No. of moles                   HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)_ Initial                                 4×10−3         4.01×10−3                −                   0Change                      −4.00×10−3     − 4.00×10−3             −            +4.00×10−3_Final                                       0                   0.01×10−3               −              4.00×10−3       

The NaOH is now in excess. 

Excess[OH−] =  0.01×10−3 moles80.10×10−3L                  = 0.00012484M.pOH=−log[OH−]       = −log(0.00012484)       = 3.9036.pH = 14.00−3.9036       = 10.09637       = 10.10.

Therefore, the  pH of the solution has been calculated to be 10.10. 

(g)

Interpretation Introduction

Interpretation:

The  pH during the titration of 40.00 ml of HCl 0.1000M with 50.00 ml of 0.1000M NaOH solution has to be calculated. 

Concept Introduction:

For the titration of a strong acid with a strong base, the  pH before the equivalence point depends on the excess concentration of acid and the  pH after the equivalence point depends on the excess concentration of base.  At the equivalence point, there is not an excess of either acid or base so pH is 7.0.  The equivalence point occurs when total volume of base has been added. 

pH=−log[H3O+]

pH = −log[OH−]

pH+pOH=14

Calculation of Moles:

1. No. of moles (n)=WM.W

Where,

W         = Weight of a given compoundM.W = Molecular weight of the same compound 

2. No. of moles (n) = (Molarity)( Volume)

Molarity:

Molarity or molar concentration is the number of moles of solute dissolved per liter of solution, which can be calculated using the following equation,

Molarity =  moles of solutevolume of solution in L

(g)

Expert Solution
Check Mark

Answer to Problem 19.54P

The  pH during the titration of 40.00 ml of HCl 0.1000M with 50.00 ml of 0.1000M NaOH solution has calculated as 12.05.

Explanation of Solution

Given Data:

Volume of  HCl                = 40.00 ml = 40.00×10−3LConcentration of HCl       = 0.1000MConcentration of NaOH  = 0.1000MVolume of NaOH added  =   50.00 ml = 50.00×10−3LThe total volume of the solution at this point = 40.00×10−3L + 50.00×10−3L                                                                                           = 90.00×10−3L 

The reaction occurring in the titration is the neutralization of H3O+ (from HCl) by OH− (from NaOH):

HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)H3O+(aq) + OH−(aq) → 2H2O(l) 

The initial number of moles of HCl= Molarity×Volume                                                                             = (0.1000 M)(40.00×10−3L)                                                                             = 4.000×10−3.

Moles of added NaOH = Molarity×Volume                                               = (0.1000 M)(50.00×10−3L)                                               = 5.00×10−3.

The HCl will react with an equal amount of the base. 

No. of moles                   HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)_ Initial                                 4×10−3         5.00×10−3                −                   0Change                      −4.00×10−3     − 4.00×10−3             −            +4.00×10−3_Final                                       0                  1.00×10−3               −              4.00×10−3       

The NaOH is now in excess. 

Excess[OH−] =  1.00×10−3 moles90.00×10−3L                  = 0.011111M.pOH=−log[OH−]       = −log(0.011111)       = 1.95424.pH = 14.00−1.95424       = 12.04576       = 12.05.

Therefore, the  pH of the solution has been calculated to be 12.05. 

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Chapter 19 Solutions

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