The initial pH of 0.200 M acetic acid should be calculated. Concept Introduction : The pH of the solution can be calculated by, pH = − log [ H + ]

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Answer 1

Question

Chapter 8, Problem 65E

(a)

Interpretation Introduction

Interpretation:

The initial pH of 0.200 M acetic acid should be calculated.

Concept Introduction :

The pH of the solution can be calculated by,

$pH = −log[H+]$

(a)

Expert Solution

Answer to Problem 65E

$pH = 2.72$

Explanation of Solution

$CH3COOH ⇌ H+ + CH3COO−initial(M) 0.200 - -eqm (0.200−x) x x$

$Ka=[ H +][ CH 3 COO −][ CH 3COOH]1.8×10−5=x2( 0.200−x)≈x20.200x=[H+]=1.9×10−3 M$

$pH = −log[H+]pH = −log(1.9×10−3)pH = 2.72$

(b)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 50.0 mL of 0.100 M KOH to 100.0 mL of 0.200 M acetic acid should be calculated.

Concept Introduction :

Henderson-Hasselbalch equation is represented as follows:

$pH = pKa+log[conjugate base][acid]$

This is used to calculate the pH of a buffer solution containing an acid or its conjugate base.

(b)

Expert Solution

Answer to Problem 65E

$pH = 4.26$

Explanation of Solution

The added OH- reacts completely with acetic acid.

$CH3COOH + OH−→CH3COO− + H2O$

Number of moles of acetic acid present = $0.200 mol1000 mL×100.0 mL = 0.02 mol$

Number of moles of OH- added = $0.100 mol1000 mL×50.0 mL = 0.005 mol$

Number of moles of acetic acid left = $0.02−0.005=0.015 mol$

Number of moles of acetate ions formed = $0.005 mol$

If the final volume is V,

$pH = pKa+log[ CH 3 COO −][ CH 3COOH]pH = −log(1.8× 10 −5)+log( 0.005/V 0.015/V)pH = 4.74 + (−0.477)pH = 4.26$

(c)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 100.0 mL of 0.100 M KOH to 100.0 mL of 0.200 M acetic acid should be calculated.

Concept Introduction :

Henderson-Hasselbalch equation is represented as follows:

$pH = pKa+log[conjugate base][acid]$

This is used to calculate the pH of a buffer solution containing an acid or its conjugate base.

(c)

Expert Solution

Answer to Problem 65E

$pH = 4.74$

Explanation of Solution

$CH3COOH + OH−→CH3COO− + H2O$

Number of moles of acetic acid present = $0.200 mol1000 mL×100.0 mL = 0.02 mol$

Number of moles of OH- added = $0.100 mol1000 mL×100.0 mL = 0.01 mol$

Number of moles of acetic acid left = $0.02−0.01=0.01 mol$

Number of moles of acetate ions formed = $0.01 mol$

If the final volume is V,

$pH = pKa+log[ CH 3 COO −][ CH 3COOH]pH = −log(1.8× 10 −5)+log( 0.01/V 0.01/V)pH = 4.74$

(d)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 150.0 mL of 0.100 M KOH to 100.0 mL of 0.200 M acetic acid should be calculated.

Concept Introduction :

Henderson-Hasselbalch equation is represented as follows:

$pH = pKa+log[conjugate base][acid]$

This is used to calculate the pH of a buffer solution containing an acid or its conjugate base.

(d)

Expert Solution

Answer to Problem 65E

$pH = 5.22$

Explanation of Solution

$CH3COOH + OH−→CH3COO− + H2O$

Number of moles of acetic acid present = $0.200 mol1000 mL×100.0 mL = 0.02 mol$

Number of moles of OH- added = $0.100 mol1000 mL×150.0 mL = 0.015 mol$

Number of moles of acetic acid left = $0.02−0.015=0.005 mol$

Number of moles of acetate ions formed = $0.015 mol$

If the final volume is V,

$pH = pKa+log[ CH 3 COO −][ CH 3COOH]pH = −log(1.8× 10 −5)+log( 0.015/V 0.005/V)pH = 4.74 + 0.48pH = 5.22$

(e)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 200.0 mL of 0.100 M KOH to 100.0 mL of 0.200 M acetic acid should be calculated.

Concept Introduction :

Henderson-Hasselbalch equation is represented as follows:

$pH = pKa+log[conjugate base][acid]$

This is used to calculate the pH of a buffer solution containing an acid or its conjugate base.

(e)

Expert Solution

Answer to Problem 65E

$pH = 8.79$

Explanation of Solution

$CH3COOH + OH−→CH3COO− + H2O$

Number of moles of acetic acid present = $0.200 mol1000 mL×100.0 mL = 0.02 mol$

Number of moles of OH- added = $0.100 mol1000 mL×200.0 mL = 0.02 mol$

Number of moles of acetic acid left = $0.02−0.02=0.0 mol$

Number of moles of acetate ions formed = $0.02 mol$

This is the equivalence point of the reaction. At the equivalence point pH is affected by the conjugate base, acetate ions.

$CH3COO− + H2O ⇌ CH3COOH + OH−initial(M) 0.02mol/0.3 L - -change -x +x +xeqm (0.0667−x) x x$

$Kb=KwKa=1.0×10−141.8×10−5=5.6×10−10$

$Kb=[ CH 3COOH][ OH −][ CH 3 COO −]5.6×10−10=x20.0667−x≈x20.0667x=[OH−]=6.1×10−6 M$

$pOH = −log(6.1× 10 −6)pOH = 5.21pH = 14 - 5.21pH = 8.79$

(f)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 250.0 mL of 0.100 M KOH to 100.0 mL of 0.200 M acetic acid should be calculated.

Concept Introduction :

Henderson-Hasselbalch equation is represented as follows:

$pH = pKa+log[conjugate base][acid]$

This is used to calculate the pH of a buffer solution containing an acid or its conjugate base.

(f)

Expert Solution

Answer to Problem 65E

$pH = 12.14$

Explanation of Solution

Number of moles of OH- added = $0.100 mol1000 mL×250.0 mL = 0.025 mol$

Number of moles of acetate ions formed = $0.02 mol$

Number of excess OH- moles from KOH = $0.025−0.020=0.005 mol$

When a strong base and a weak base both are present in a solution, the amount of OH- added from weak base can be neglected as its dissociation constant is very small. The pH past the equivalence point is determined by the amount of excess base added.

$[OH−]excess=0.005 mol(100.0+250.0) mL=0.014 M$

$pOH = −log(0.014)pOH = 1.86pH = 14 - 1.86pH = 12.14$

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Answer 2

Question

Chapter 8, Problem 65E

(a)

Interpretation Introduction

Interpretation:

The initial pH of 0.200 M acetic acid should be calculated.

Concept Introduction :

The pH of the solution can be calculated by,

$pH = −log[H+]$

(a)

Expert Solution

Answer to Problem 65E

$pH = 2.72$

Explanation of Solution

$CH3COOH ⇌ H+ + CH3COO−initial(M) 0.200 - -eqm (0.200−x) x x$

$Ka=[ H +][ CH 3 COO −][ CH 3COOH]1.8×10−5=x2( 0.200−x)≈x20.200x=[H+]=1.9×10−3 M$

$pH = −log[H+]pH = −log(1.9×10−3)pH = 2.72$

(b)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 50.0 mL of 0.100 M KOH to 100.0 mL of 0.200 M acetic acid should be calculated.

Concept Introduction :

Henderson-Hasselbalch equation is represented as follows:

$pH = pKa+log[conjugate base][acid]$

This is used to calculate the pH of a buffer solution containing an acid or its conjugate base.

(b)

Expert Solution

Answer to Problem 65E

$pH = 4.26$

Explanation of Solution

The added OH- reacts completely with acetic acid.

$CH3COOH + OH−→CH3COO− + H2O$

Number of moles of acetic acid present = $0.200 mol1000 mL×100.0 mL = 0.02 mol$

Number of moles of OH- added = $0.100 mol1000 mL×50.0 mL = 0.005 mol$

Number of moles of acetic acid left = $0.02−0.005=0.015 mol$

Number of moles of acetate ions formed = $0.005 mol$

If the final volume is V,

$pH = pKa+log[ CH 3 COO −][ CH 3COOH]pH = −log(1.8× 10 −5)+log( 0.005/V 0.015/V)pH = 4.74 + (−0.477)pH = 4.26$

(c)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 100.0 mL of 0.100 M KOH to 100.0 mL of 0.200 M acetic acid should be calculated.

Concept Introduction :

Henderson-Hasselbalch equation is represented as follows:

$pH = pKa+log[conjugate base][acid]$

This is used to calculate the pH of a buffer solution containing an acid or its conjugate base.

(c)

Expert Solution

Answer to Problem 65E

$pH = 4.74$

Explanation of Solution

$CH3COOH + OH−→CH3COO− + H2O$

Number of moles of acetic acid present = $0.200 mol1000 mL×100.0 mL = 0.02 mol$

Number of moles of OH- added = $0.100 mol1000 mL×100.0 mL = 0.01 mol$

Number of moles of acetic acid left = $0.02−0.01=0.01 mol$

Number of moles of acetate ions formed = $0.01 mol$

If the final volume is V,

$pH = pKa+log[ CH 3 COO −][ CH 3COOH]pH = −log(1.8× 10 −5)+log( 0.01/V 0.01/V)pH = 4.74$

(d)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 150.0 mL of 0.100 M KOH to 100.0 mL of 0.200 M acetic acid should be calculated.

Concept Introduction :

Henderson-Hasselbalch equation is represented as follows:

$pH = pKa+log[conjugate base][acid]$

This is used to calculate the pH of a buffer solution containing an acid or its conjugate base.

(d)

Expert Solution

Answer to Problem 65E

$pH = 5.22$

Explanation of Solution

$CH3COOH + OH−→CH3COO− + H2O$

Number of moles of acetic acid present = $0.200 mol1000 mL×100.0 mL = 0.02 mol$

Number of moles of OH- added = $0.100 mol1000 mL×150.0 mL = 0.015 mol$

Number of moles of acetic acid left = $0.02−0.015=0.005 mol$

Number of moles of acetate ions formed = $0.015 mol$

If the final volume is V,

$pH = pKa+log[ CH 3 COO −][ CH 3COOH]pH = −log(1.8× 10 −5)+log( 0.015/V 0.005/V)pH = 4.74 + 0.48pH = 5.22$

(e)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 200.0 mL of 0.100 M KOH to 100.0 mL of 0.200 M acetic acid should be calculated.

Concept Introduction :

Henderson-Hasselbalch equation is represented as follows:

$pH = pKa+log[conjugate base][acid]$

This is used to calculate the pH of a buffer solution containing an acid or its conjugate base.

(e)

Expert Solution

Answer to Problem 65E

$pH = 8.79$

Explanation of Solution

$CH3COOH + OH−→CH3COO− + H2O$

Number of moles of acetic acid present = $0.200 mol1000 mL×100.0 mL = 0.02 mol$

Number of moles of OH- added = $0.100 mol1000 mL×200.0 mL = 0.02 mol$

Number of moles of acetic acid left = $0.02−0.02=0.0 mol$

Number of moles of acetate ions formed = $0.02 mol$

This is the equivalence point of the reaction. At the equivalence point pH is affected by the conjugate base, acetate ions.

$CH3COO− + H2O ⇌ CH3COOH + OH−initial(M) 0.02mol/0.3 L - -change -x +x +xeqm (0.0667−x) x x$

$Kb=KwKa=1.0×10−141.8×10−5=5.6×10−10$

$Kb=[ CH 3COOH][ OH −][ CH 3 COO −]5.6×10−10=x20.0667−x≈x20.0667x=[OH−]=6.1×10−6 M$

$pOH = −log(6.1× 10 −6)pOH = 5.21pH = 14 - 5.21pH = 8.79$

(f)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 250.0 mL of 0.100 M KOH to 100.0 mL of 0.200 M acetic acid should be calculated.

Concept Introduction :

Henderson-Hasselbalch equation is represented as follows:

$pH = pKa+log[conjugate base][acid]$

This is used to calculate the pH of a buffer solution containing an acid or its conjugate base.

(f)

Expert Solution

Answer to Problem 65E

$pH = 12.14$

Explanation of Solution

Number of moles of OH- added = $0.100 mol1000 mL×250.0 mL = 0.025 mol$

Number of moles of acetate ions formed = $0.02 mol$

Number of excess OH- moles from KOH = $0.025−0.020=0.005 mol$

When a strong base and a weak base both are present in a solution, the amount of OH- added from weak base can be neglected as its dissociation constant is very small. The pH past the equivalence point is determined by the amount of excess base added.

$[OH−]excess=0.005 mol(100.0+250.0) mL=0.014 M$

$pOH = −log(0.014)pOH = 1.86pH = 14 - 1.86pH = 12.14$

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Answer 3

Question

Chapter 8, Problem 65E

(a)

Interpretation Introduction

Interpretation:

The initial pH of 0.200 M acetic acid should be calculated.

Concept Introduction :

The pH of the solution can be calculated by,

$pH = −log[H+]$

(a)

Expert Solution

Answer to Problem 65E

$pH = 2.72$

Explanation of Solution

$CH3COOH ⇌ H+ + CH3COO−initial(M) 0.200 - -eqm (0.200−x) x x$

$Ka=[ H +][ CH 3 COO −][ CH 3COOH]1.8×10−5=x2( 0.200−x)≈x20.200x=[H+]=1.9×10−3 M$

$pH = −log[H+]pH = −log(1.9×10−3)pH = 2.72$

(b)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 50.0 mL of 0.100 M KOH to 100.0 mL of 0.200 M acetic acid should be calculated.

Concept Introduction :

Henderson-Hasselbalch equation is represented as follows:

$pH = pKa+log[conjugate base][acid]$

This is used to calculate the pH of a buffer solution containing an acid or its conjugate base.

(b)

Expert Solution

Answer to Problem 65E

$pH = 4.26$

Explanation of Solution

The added OH- reacts completely with acetic acid.

$CH3COOH + OH−→CH3COO− + H2O$

Number of moles of acetic acid present = $0.200 mol1000 mL×100.0 mL = 0.02 mol$

Number of moles of OH- added = $0.100 mol1000 mL×50.0 mL = 0.005 mol$

Number of moles of acetic acid left = $0.02−0.005=0.015 mol$

Number of moles of acetate ions formed = $0.005 mol$

If the final volume is V,

$pH = pKa+log[ CH 3 COO −][ CH 3COOH]pH = −log(1.8× 10 −5)+log( 0.005/V 0.015/V)pH = 4.74 + (−0.477)pH = 4.26$

(c)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 100.0 mL of 0.100 M KOH to 100.0 mL of 0.200 M acetic acid should be calculated.

Concept Introduction :

Henderson-Hasselbalch equation is represented as follows:

$pH = pKa+log[conjugate base][acid]$

This is used to calculate the pH of a buffer solution containing an acid or its conjugate base.

(c)

Expert Solution

Answer to Problem 65E

$pH = 4.74$

Explanation of Solution

$CH3COOH + OH−→CH3COO− + H2O$

Number of moles of acetic acid present = $0.200 mol1000 mL×100.0 mL = 0.02 mol$

Number of moles of OH- added = $0.100 mol1000 mL×100.0 mL = 0.01 mol$

Number of moles of acetic acid left = $0.02−0.01=0.01 mol$

Number of moles of acetate ions formed = $0.01 mol$

If the final volume is V,

$pH = pKa+log[ CH 3 COO −][ CH 3COOH]pH = −log(1.8× 10 −5)+log( 0.01/V 0.01/V)pH = 4.74$

(d)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 150.0 mL of 0.100 M KOH to 100.0 mL of 0.200 M acetic acid should be calculated.

Concept Introduction :

Henderson-Hasselbalch equation is represented as follows:

$pH = pKa+log[conjugate base][acid]$

This is used to calculate the pH of a buffer solution containing an acid or its conjugate base.

(d)

Expert Solution

Answer to Problem 65E

$pH = 5.22$

Explanation of Solution

$CH3COOH + OH−→CH3COO− + H2O$

Number of moles of acetic acid present = $0.200 mol1000 mL×100.0 mL = 0.02 mol$

Number of moles of OH- added = $0.100 mol1000 mL×150.0 mL = 0.015 mol$

Number of moles of acetic acid left = $0.02−0.015=0.005 mol$

Number of moles of acetate ions formed = $0.015 mol$

If the final volume is V,

$pH = pKa+log[ CH 3 COO −][ CH 3COOH]pH = −log(1.8× 10 −5)+log( 0.015/V 0.005/V)pH = 4.74 + 0.48pH = 5.22$

(e)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 200.0 mL of 0.100 M KOH to 100.0 mL of 0.200 M acetic acid should be calculated.

Concept Introduction :

Henderson-Hasselbalch equation is represented as follows:

$pH = pKa+log[conjugate base][acid]$

This is used to calculate the pH of a buffer solution containing an acid or its conjugate base.

(e)

Expert Solution

Answer to Problem 65E

$pH = 8.79$

Explanation of Solution

$CH3COOH + OH−→CH3COO− + H2O$

Number of moles of acetic acid present = $0.200 mol1000 mL×100.0 mL = 0.02 mol$

Number of moles of OH- added = $0.100 mol1000 mL×200.0 mL = 0.02 mol$

Number of moles of acetic acid left = $0.02−0.02=0.0 mol$

Number of moles of acetate ions formed = $0.02 mol$

This is the equivalence point of the reaction. At the equivalence point pH is affected by the conjugate base, acetate ions.

$CH3COO− + H2O ⇌ CH3COOH + OH−initial(M) 0.02mol/0.3 L - -change -x +x +xeqm (0.0667−x) x x$

$Kb=KwKa=1.0×10−141.8×10−5=5.6×10−10$

$Kb=[ CH 3COOH][ OH −][ CH 3 COO −]5.6×10−10=x20.0667−x≈x20.0667x=[OH−]=6.1×10−6 M$

$pOH = −log(6.1× 10 −6)pOH = 5.21pH = 14 - 5.21pH = 8.79$

(f)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 250.0 mL of 0.100 M KOH to 100.0 mL of 0.200 M acetic acid should be calculated.

Concept Introduction :

Henderson-Hasselbalch equation is represented as follows:

$pH = pKa+log[conjugate base][acid]$

This is used to calculate the pH of a buffer solution containing an acid or its conjugate base.

(f)

Expert Solution

Answer to Problem 65E

$pH = 12.14$

Explanation of Solution

Number of moles of OH- added = $0.100 mol1000 mL×250.0 mL = 0.025 mol$

Number of moles of acetate ions formed = $0.02 mol$

Number of excess OH- moles from KOH = $0.025−0.020=0.005 mol$

When a strong base and a weak base both are present in a solution, the amount of OH- added from weak base can be neglected as its dissociation constant is very small. The pH past the equivalence point is determined by the amount of excess base added.

$[OH−]excess=0.005 mol(100.0+250.0) mL=0.014 M$

$pOH = −log(0.014)pOH = 1.86pH = 14 - 1.86pH = 12.14$

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Answer 4

Question

Chapter 8, Problem 65E

(a)

Interpretation Introduction

Interpretation:

The initial pH of 0.200 M acetic acid should be calculated.

Concept Introduction :

The pH of the solution can be calculated by,

$pH = −log[H+]$

(a)

Expert Solution

Answer to Problem 65E

$pH = 2.72$

Explanation of Solution

$CH3COOH ⇌ H+ + CH3COO−initial(M) 0.200 - -eqm (0.200−x) x x$

$Ka=[ H +][ CH 3 COO −][ CH 3COOH]1.8×10−5=x2( 0.200−x)≈x20.200x=[H+]=1.9×10−3 M$

$pH = −log[H+]pH = −log(1.9×10−3)pH = 2.72$

(b)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 50.0 mL of 0.100 M KOH to 100.0 mL of 0.200 M acetic acid should be calculated.

Concept Introduction :

Henderson-Hasselbalch equation is represented as follows:

$pH = pKa+log[conjugate base][acid]$

This is used to calculate the pH of a buffer solution containing an acid or its conjugate base.

(b)

Expert Solution

Answer to Problem 65E

$pH = 4.26$

Explanation of Solution

The added OH- reacts completely with acetic acid.

$CH3COOH + OH−→CH3COO− + H2O$

Number of moles of acetic acid present = $0.200 mol1000 mL×100.0 mL = 0.02 mol$

Number of moles of OH- added = $0.100 mol1000 mL×50.0 mL = 0.005 mol$

Number of moles of acetic acid left = $0.02−0.005=0.015 mol$

Number of moles of acetate ions formed = $0.005 mol$

If the final volume is V,

$pH = pKa+log[ CH 3 COO −][ CH 3COOH]pH = −log(1.8× 10 −5)+log( 0.005/V 0.015/V)pH = 4.74 + (−0.477)pH = 4.26$

(c)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 100.0 mL of 0.100 M KOH to 100.0 mL of 0.200 M acetic acid should be calculated.

Concept Introduction :

Henderson-Hasselbalch equation is represented as follows:

$pH = pKa+log[conjugate base][acid]$

This is used to calculate the pH of a buffer solution containing an acid or its conjugate base.

(c)

Expert Solution

Answer to Problem 65E

$pH = 4.74$

Explanation of Solution

$CH3COOH + OH−→CH3COO− + H2O$

Number of moles of acetic acid present = $0.200 mol1000 mL×100.0 mL = 0.02 mol$

Number of moles of OH- added = $0.100 mol1000 mL×100.0 mL = 0.01 mol$

Number of moles of acetic acid left = $0.02−0.01=0.01 mol$

Number of moles of acetate ions formed = $0.01 mol$

If the final volume is V,

$pH = pKa+log[ CH 3 COO −][ CH 3COOH]pH = −log(1.8× 10 −5)+log( 0.01/V 0.01/V)pH = 4.74$

(d)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 150.0 mL of 0.100 M KOH to 100.0 mL of 0.200 M acetic acid should be calculated.

Concept Introduction :

Henderson-Hasselbalch equation is represented as follows:

$pH = pKa+log[conjugate base][acid]$

This is used to calculate the pH of a buffer solution containing an acid or its conjugate base.

(d)

Expert Solution

Answer to Problem 65E

$pH = 5.22$

Explanation of Solution

$CH3COOH + OH−→CH3COO− + H2O$

Number of moles of acetic acid present = $0.200 mol1000 mL×100.0 mL = 0.02 mol$

Number of moles of OH- added = $0.100 mol1000 mL×150.0 mL = 0.015 mol$

Number of moles of acetic acid left = $0.02−0.015=0.005 mol$

Number of moles of acetate ions formed = $0.015 mol$

If the final volume is V,

$pH = pKa+log[ CH 3 COO −][ CH 3COOH]pH = −log(1.8× 10 −5)+log( 0.015/V 0.005/V)pH = 4.74 + 0.48pH = 5.22$

(e)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 200.0 mL of 0.100 M KOH to 100.0 mL of 0.200 M acetic acid should be calculated.

Concept Introduction :

Henderson-Hasselbalch equation is represented as follows:

$pH = pKa+log[conjugate base][acid]$

This is used to calculate the pH of a buffer solution containing an acid or its conjugate base.

(e)

Expert Solution

Answer to Problem 65E

$pH = 8.79$

Explanation of Solution

$CH3COOH + OH−→CH3COO− + H2O$

Number of moles of acetic acid present = $0.200 mol1000 mL×100.0 mL = 0.02 mol$

Number of moles of OH- added = $0.100 mol1000 mL×200.0 mL = 0.02 mol$

Number of moles of acetic acid left = $0.02−0.02=0.0 mol$

Number of moles of acetate ions formed = $0.02 mol$

This is the equivalence point of the reaction. At the equivalence point pH is affected by the conjugate base, acetate ions.

$CH3COO− + H2O ⇌ CH3COOH + OH−initial(M) 0.02mol/0.3 L - -change -x +x +xeqm (0.0667−x) x x$

$Kb=KwKa=1.0×10−141.8×10−5=5.6×10−10$

$Kb=[ CH 3COOH][ OH −][ CH 3 COO −]5.6×10−10=x20.0667−x≈x20.0667x=[OH−]=6.1×10−6 M$

$pOH = −log(6.1× 10 −6)pOH = 5.21pH = 14 - 5.21pH = 8.79$

(f)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 250.0 mL of 0.100 M KOH to 100.0 mL of 0.200 M acetic acid should be calculated.

Concept Introduction :

Henderson-Hasselbalch equation is represented as follows:

$pH = pKa+log[conjugate base][acid]$

This is used to calculate the pH of a buffer solution containing an acid or its conjugate base.

(f)

Expert Solution

Answer to Problem 65E

$pH = 12.14$

Explanation of Solution

Number of moles of OH- added = $0.100 mol1000 mL×250.0 mL = 0.025 mol$

Number of moles of acetate ions formed = $0.02 mol$

Number of excess OH- moles from KOH = $0.025−0.020=0.005 mol$

When a strong base and a weak base both are present in a solution, the amount of OH- added from weak base can be neglected as its dissociation constant is very small. The pH past the equivalence point is determined by the amount of excess base added.

$[OH−]excess=0.005 mol(100.0+250.0) mL=0.014 M$

$pOH = −log(0.014)pOH = 1.86pH = 14 - 1.86pH = 12.14$

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Answer 5

Chapter 8, Problem 65E

(a)

Interpretation Introduction

Interpretation:

The initial pH of 0.200 M acetic acid should be calculated.

Concept Introduction :

The pH of the solution can be calculated by,

$pH = −log[H+]$

(a)

Expert Solution

Answer to Problem 65E

$pH = 2.72$

Explanation of Solution

$CH3COOH ⇌ H+ + CH3COO−initial(M) 0.200 - -eqm (0.200−x) x x$

$Ka=[ H +][ CH 3 COO −][ CH 3COOH]1.8×10−5=x2( 0.200−x)≈x20.200x=[H+]=1.9×10−3 M$

$pH = −log[H+]pH = −log(1.9×10−3)pH = 2.72$

(b)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 50.0 mL of 0.100 M KOH to 100.0 mL of 0.200 M acetic acid should be calculated.

Concept Introduction :

Henderson-Hasselbalch equation is represented as follows:

$pH = pKa+log[conjugate base][acid]$

This is used to calculate the pH of a buffer solution containing an acid or its conjugate base.

(b)

Expert Solution

Answer to Problem 65E

$pH = 4.26$

Explanation of Solution

The added OH- reacts completely with acetic acid.

$CH3COOH + OH−→CH3COO− + H2O$

Number of moles of acetic acid present = $0.200 mol1000 mL×100.0 mL = 0.02 mol$

Number of moles of OH- added = $0.100 mol1000 mL×50.0 mL = 0.005 mol$

Number of moles of acetic acid left = $0.02−0.005=0.015 mol$

Number of moles of acetate ions formed = $0.005 mol$

If the final volume is V,

$pH = pKa+log[ CH 3 COO −][ CH 3COOH]pH = −log(1.8× 10 −5)+log( 0.005/V 0.015/V)pH = 4.74 + (−0.477)pH = 4.26$

(c)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 100.0 mL of 0.100 M KOH to 100.0 mL of 0.200 M acetic acid should be calculated.

Concept Introduction :

Henderson-Hasselbalch equation is represented as follows:

$pH = pKa+log[conjugate base][acid]$

This is used to calculate the pH of a buffer solution containing an acid or its conjugate base.

(c)

Expert Solution

Answer to Problem 65E

$pH = 4.74$

Explanation of Solution

$CH3COOH + OH−→CH3COO− + H2O$

Number of moles of acetic acid present = $0.200 mol1000 mL×100.0 mL = 0.02 mol$

Number of moles of OH- added = $0.100 mol1000 mL×100.0 mL = 0.01 mol$

Number of moles of acetic acid left = $0.02−0.01=0.01 mol$

Number of moles of acetate ions formed = $0.01 mol$

If the final volume is V,

$pH = pKa+log[ CH 3 COO −][ CH 3COOH]pH = −log(1.8× 10 −5)+log( 0.01/V 0.01/V)pH = 4.74$

(d)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 150.0 mL of 0.100 M KOH to 100.0 mL of 0.200 M acetic acid should be calculated.

Concept Introduction :

Henderson-Hasselbalch equation is represented as follows:

$pH = pKa+log[conjugate base][acid]$

This is used to calculate the pH of a buffer solution containing an acid or its conjugate base.

(d)

Expert Solution

Answer to Problem 65E

$pH = 5.22$

Explanation of Solution

$CH3COOH + OH−→CH3COO− + H2O$

Number of moles of acetic acid present = $0.200 mol1000 mL×100.0 mL = 0.02 mol$

Number of moles of OH- added = $0.100 mol1000 mL×150.0 mL = 0.015 mol$

Number of moles of acetic acid left = $0.02−0.015=0.005 mol$

Number of moles of acetate ions formed = $0.015 mol$

If the final volume is V,

$pH = pKa+log[ CH 3 COO −][ CH 3COOH]pH = −log(1.8× 10 −5)+log( 0.015/V 0.005/V)pH = 4.74 + 0.48pH = 5.22$

(e)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 200.0 mL of 0.100 M KOH to 100.0 mL of 0.200 M acetic acid should be calculated.

Concept Introduction :

Henderson-Hasselbalch equation is represented as follows:

$pH = pKa+log[conjugate base][acid]$

This is used to calculate the pH of a buffer solution containing an acid or its conjugate base.

(e)

Expert Solution

Answer to Problem 65E

$pH = 8.79$

Explanation of Solution

$CH3COOH + OH−→CH3COO− + H2O$

Number of moles of acetic acid present = $0.200 mol1000 mL×100.0 mL = 0.02 mol$

Number of moles of OH- added = $0.100 mol1000 mL×200.0 mL = 0.02 mol$

Number of moles of acetic acid left = $0.02−0.02=0.0 mol$

Number of moles of acetate ions formed = $0.02 mol$

This is the equivalence point of the reaction. At the equivalence point pH is affected by the conjugate base, acetate ions.

$CH3COO− + H2O ⇌ CH3COOH + OH−initial(M) 0.02mol/0.3 L - -change -x +x +xeqm (0.0667−x) x x$

$Kb=KwKa=1.0×10−141.8×10−5=5.6×10−10$

$Kb=[ CH 3COOH][ OH −][ CH 3 COO −]5.6×10−10=x20.0667−x≈x20.0667x=[OH−]=6.1×10−6 M$

$pOH = −log(6.1× 10 −6)pOH = 5.21pH = 14 - 5.21pH = 8.79$

(f)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 250.0 mL of 0.100 M KOH to 100.0 mL of 0.200 M acetic acid should be calculated.

Concept Introduction :

Henderson-Hasselbalch equation is represented as follows:

$pH = pKa+log[conjugate base][acid]$

This is used to calculate the pH of a buffer solution containing an acid or its conjugate base.

(f)

Expert Solution

Answer to Problem 65E

$pH = 12.14$

Explanation of Solution

Number of moles of OH- added = $0.100 mol1000 mL×250.0 mL = 0.025 mol$

Number of moles of acetate ions formed = $0.02 mol$

Number of excess OH- moles from KOH = $0.025−0.020=0.005 mol$

When a strong base and a weak base both are present in a solution, the amount of OH- added from weak base can be neglected as its dissociation constant is very small. The pH past the equivalence point is determined by the amount of excess base added.

$[OH−]excess=0.005 mol(100.0+250.0) mL=0.014 M$

$pOH = −log(0.014)pOH = 1.86pH = 14 - 1.86pH = 12.14$

Answer 6

Chapter 8, Problem 65E

(a)

Interpretation Introduction

Interpretation:

The initial pH of 0.200 M acetic acid should be calculated.

Concept Introduction :

The pH of the solution can be calculated by,

$pH = −log[H+]$

(a)

Expert Solution

Answer to Problem 65E

$pH = 2.72$

Explanation of Solution

$CH3COOH ⇌ H+ + CH3COO−initial(M) 0.200 - -eqm (0.200−x) x x$

$Ka=[ H +][ CH 3 COO −][ CH 3COOH]1.8×10−5=x2( 0.200−x)≈x20.200x=[H+]=1.9×10−3 M$

$pH = −log[H+]pH = −log(1.9×10−3)pH = 2.72$

Answer 7

Chapter 8, Problem 65E

(a)

Interpretation Introduction

Interpretation:

The initial pH of 0.200 M acetic acid should be calculated.

Concept Introduction :

The pH of the solution can be calculated by,

$pH = −log[H+]$

Answer 8

(a)

Expert Solution

Answer to Problem 65E

$pH = 2.72$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

$CH3COOH ⇌ H+ + CH3COO−initial(M) 0.200 - -eqm (0.200−x) x x$

$Ka=[ H +][ CH 3 COO −][ CH 3COOH]1.8×10−5=x2( 0.200−x)≈x20.200x=[H+]=1.9×10−3 M$

$pH = −log[H+]pH = −log(1.9×10−3)pH = 2.72$

Answer 13

(b)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 50.0 mL of 0.100 M KOH to 100.0 mL of 0.200 M acetic acid should be calculated.

Concept Introduction :

Henderson-Hasselbalch equation is represented as follows:

$pH = pKa+log[conjugate base][acid]$

This is used to calculate the pH of a buffer solution containing an acid or its conjugate base.

(b)

Expert Solution

Answer to Problem 65E

$pH = 4.26$

Explanation of Solution

The added OH- reacts completely with acetic acid.

$CH3COOH + OH−→CH3COO− + H2O$

Number of moles of acetic acid present = $0.200 mol1000 mL×100.0 mL = 0.02 mol$

Number of moles of OH- added = $0.100 mol1000 mL×50.0 mL = 0.005 mol$

Number of moles of acetic acid left = $0.02−0.005=0.015 mol$

Number of moles of acetate ions formed = $0.005 mol$

If the final volume is V,

$pH = pKa+log[ CH 3 COO −][ CH 3COOH]pH = −log(1.8× 10 −5)+log( 0.005/V 0.015/V)pH = 4.74 + (−0.477)pH = 4.26$

Answer 14

(b)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 50.0 mL of 0.100 M KOH to 100.0 mL of 0.200 M acetic acid should be calculated.

Concept Introduction :

Henderson-Hasselbalch equation is represented as follows:

$pH = pKa+log[conjugate base][acid]$

This is used to calculate the pH of a buffer solution containing an acid or its conjugate base.

Answer 15

(b)

Expert Solution

Answer to Problem 65E

$pH = 4.26$

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

The added OH- reacts completely with acetic acid.

$CH3COOH + OH−→CH3COO− + H2O$

Number of moles of acetic acid present = $0.200 mol1000 mL×100.0 mL = 0.02 mol$

Number of moles of OH- added = $0.100 mol1000 mL×50.0 mL = 0.005 mol$

Number of moles of acetic acid left = $0.02−0.005=0.015 mol$

Number of moles of acetate ions formed = $0.005 mol$

If the final volume is V,

$pH = pKa+log[ CH 3 COO −][ CH 3COOH]pH = −log(1.8× 10 −5)+log( 0.005/V 0.015/V)pH = 4.74 + (−0.477)pH = 4.26$

Answer 20

(c)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 100.0 mL of 0.100 M KOH to 100.0 mL of 0.200 M acetic acid should be calculated.

Concept Introduction :

Henderson-Hasselbalch equation is represented as follows:

$pH = pKa+log[conjugate base][acid]$

This is used to calculate the pH of a buffer solution containing an acid or its conjugate base.

(c)

Expert Solution

Answer to Problem 65E

$pH = 4.74$

Explanation of Solution

$CH3COOH + OH−→CH3COO− + H2O$

Number of moles of acetic acid present = $0.200 mol1000 mL×100.0 mL = 0.02 mol$

Number of moles of OH- added = $0.100 mol1000 mL×100.0 mL = 0.01 mol$

Number of moles of acetic acid left = $0.02−0.01=0.01 mol$

Number of moles of acetate ions formed = $0.01 mol$

If the final volume is V,

$pH = pKa+log[ CH 3 COO −][ CH 3COOH]pH = −log(1.8× 10 −5)+log( 0.01/V 0.01/V)pH = 4.74$

Answer 21

(c)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 100.0 mL of 0.100 M KOH to 100.0 mL of 0.200 M acetic acid should be calculated.

Concept Introduction :

Henderson-Hasselbalch equation is represented as follows:

$pH = pKa+log[conjugate base][acid]$

This is used to calculate the pH of a buffer solution containing an acid or its conjugate base.

Answer 22

(c)

Expert Solution

Answer to Problem 65E

$pH = 4.74$

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

$CH3COOH + OH−→CH3COO− + H2O$

Number of moles of acetic acid present = $0.200 mol1000 mL×100.0 mL = 0.02 mol$

Number of moles of OH- added = $0.100 mol1000 mL×100.0 mL = 0.01 mol$

Number of moles of acetic acid left = $0.02−0.01=0.01 mol$

Number of moles of acetate ions formed = $0.01 mol$

If the final volume is V,

$pH = pKa+log[ CH 3 COO −][ CH 3COOH]pH = −log(1.8× 10 −5)+log( 0.01/V 0.01/V)pH = 4.74$

Answer 27

(d)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 150.0 mL of 0.100 M KOH to 100.0 mL of 0.200 M acetic acid should be calculated.

Concept Introduction :

Henderson-Hasselbalch equation is represented as follows:

$pH = pKa+log[conjugate base][acid]$

This is used to calculate the pH of a buffer solution containing an acid or its conjugate base.

(d)

Expert Solution

Answer to Problem 65E

$pH = 5.22$

Explanation of Solution

$CH3COOH + OH−→CH3COO− + H2O$

Number of moles of acetic acid present = $0.200 mol1000 mL×100.0 mL = 0.02 mol$

Number of moles of OH- added = $0.100 mol1000 mL×150.0 mL = 0.015 mol$

Number of moles of acetic acid left = $0.02−0.015=0.005 mol$

Number of moles of acetate ions formed = $0.015 mol$

If the final volume is V,

$pH = pKa+log[ CH 3 COO −][ CH 3COOH]pH = −log(1.8× 10 −5)+log( 0.015/V 0.005/V)pH = 4.74 + 0.48pH = 5.22$

Answer 28

(d)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 150.0 mL of 0.100 M KOH to 100.0 mL of 0.200 M acetic acid should be calculated.

Concept Introduction :

Henderson-Hasselbalch equation is represented as follows:

$pH = pKa+log[conjugate base][acid]$

This is used to calculate the pH of a buffer solution containing an acid or its conjugate base.

Answer 29

(d)

Expert Solution

Answer to Problem 65E

$pH = 5.22$

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

$CH3COOH + OH−→CH3COO− + H2O$

Number of moles of acetic acid present = $0.200 mol1000 mL×100.0 mL = 0.02 mol$

Number of moles of OH- added = $0.100 mol1000 mL×150.0 mL = 0.015 mol$

Number of moles of acetic acid left = $0.02−0.015=0.005 mol$

Number of moles of acetate ions formed = $0.015 mol$

If the final volume is V,

$pH = pKa+log[ CH 3 COO −][ CH 3COOH]pH = −log(1.8× 10 −5)+log( 0.015/V 0.005/V)pH = 4.74 + 0.48pH = 5.22$

Answer 34

(e)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 200.0 mL of 0.100 M KOH to 100.0 mL of 0.200 M acetic acid should be calculated.

Concept Introduction :

Henderson-Hasselbalch equation is represented as follows:

$pH = pKa+log[conjugate base][acid]$

This is used to calculate the pH of a buffer solution containing an acid or its conjugate base.

(e)

Expert Solution

Answer to Problem 65E

$pH = 8.79$

Explanation of Solution

$CH3COOH + OH−→CH3COO− + H2O$

Number of moles of acetic acid present = $0.200 mol1000 mL×100.0 mL = 0.02 mol$

Number of moles of OH- added = $0.100 mol1000 mL×200.0 mL = 0.02 mol$

Number of moles of acetic acid left = $0.02−0.02=0.0 mol$

Number of moles of acetate ions formed = $0.02 mol$

This is the equivalence point of the reaction. At the equivalence point pH is affected by the conjugate base, acetate ions.

$CH3COO− + H2O ⇌ CH3COOH + OH−initial(M) 0.02mol/0.3 L - -change -x +x +xeqm (0.0667−x) x x$

$Kb=KwKa=1.0×10−141.8×10−5=5.6×10−10$

$Kb=[ CH 3COOH][ OH −][ CH 3 COO −]5.6×10−10=x20.0667−x≈x20.0667x=[OH−]=6.1×10−6 M$

$pOH = −log(6.1× 10 −6)pOH = 5.21pH = 14 - 5.21pH = 8.79$

Answer 35

(e)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 200.0 mL of 0.100 M KOH to 100.0 mL of 0.200 M acetic acid should be calculated.

Concept Introduction :

Henderson-Hasselbalch equation is represented as follows:

$pH = pKa+log[conjugate base][acid]$

This is used to calculate the pH of a buffer solution containing an acid or its conjugate base.

Answer 36

(e)

Expert Solution

Answer to Problem 65E

$pH = 8.79$

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

$CH3COOH + OH−→CH3COO− + H2O$

Number of moles of acetic acid present = $0.200 mol1000 mL×100.0 mL = 0.02 mol$

Number of moles of OH- added = $0.100 mol1000 mL×200.0 mL = 0.02 mol$

Number of moles of acetic acid left = $0.02−0.02=0.0 mol$

Number of moles of acetate ions formed = $0.02 mol$

This is the equivalence point of the reaction. At the equivalence point pH is affected by the conjugate base, acetate ions.

$CH3COO− + H2O ⇌ CH3COOH + OH−initial(M) 0.02mol/0.3 L - -change -x +x +xeqm (0.0667−x) x x$

$Kb=KwKa=1.0×10−141.8×10−5=5.6×10−10$

$Kb=[ CH 3COOH][ OH −][ CH 3 COO −]5.6×10−10=x20.0667−x≈x20.0667x=[OH−]=6.1×10−6 M$

$pOH = −log(6.1× 10 −6)pOH = 5.21pH = 14 - 5.21pH = 8.79$

Answer 41

(f)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 250.0 mL of 0.100 M KOH to 100.0 mL of 0.200 M acetic acid should be calculated.

Concept Introduction :

Henderson-Hasselbalch equation is represented as follows:

$pH = pKa+log[conjugate base][acid]$

This is used to calculate the pH of a buffer solution containing an acid or its conjugate base.

(f)

Expert Solution

Answer to Problem 65E

$pH = 12.14$

Explanation of Solution

Number of moles of OH- added = $0.100 mol1000 mL×250.0 mL = 0.025 mol$

Number of moles of acetate ions formed = $0.02 mol$

Number of excess OH- moles from KOH = $0.025−0.020=0.005 mol$

When a strong base and a weak base both are present in a solution, the amount of OH- added from weak base can be neglected as its dissociation constant is very small. The pH past the equivalence point is determined by the amount of excess base added.

$[OH−]excess=0.005 mol(100.0+250.0) mL=0.014 M$

$pOH = −log(0.014)pOH = 1.86pH = 14 - 1.86pH = 12.14$

Answer 42

(f)

Interpretation Introduction

Interpretation:

The pH of the resulting solution after adding 250.0 mL of 0.100 M KOH to 100.0 mL of 0.200 M acetic acid should be calculated.

Concept Introduction :

Henderson-Hasselbalch equation is represented as follows:

$pH = pKa+log[conjugate base][acid]$

This is used to calculate the pH of a buffer solution containing an acid or its conjugate base.

Answer 43

(f)

Expert Solution

Answer to Problem 65E

$pH = 12.14$

Answer 44

(f)

Expert Solution

Answer 45

(f)

Expert Solution

Answer 46

Expert Solution

Answer 47

Explanation of Solution

Number of moles of OH- added = $0.100 mol1000 mL×250.0 mL = 0.025 mol$

Number of moles of acetate ions formed = $0.02 mol$

Number of excess OH- moles from KOH = $0.025−0.020=0.005 mol$

When a strong base and a weak base both are present in a solution, the amount of OH- added from weak base can be neglected as its dissociation constant is very small. The pH past the equivalence point is determined by the amount of excess base added.

$[OH−]excess=0.005 mol(100.0+250.0) mL=0.014 M$