For the titration of 10.0 mL of 0.15 M acetic acid with 0.10 M sodium hydroxide, determine the pH when (a) 10.0 mL of base has been added, (b) 15 .0 mL of base has been added, and (c) 20.0 mL of base has been added.

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Answer 1

Textbook Question

Chapter 17.3, Problem 4PPA

For the titration of 10.0 mL of 0.15 M acetic acid with 0.10 M sodium hydroxide, determine the pH when (a) 10.0 mL of base has been added, (b) 15 .0 mL of base has been added, and (c) 20.0 mL of base has been added.

(a)

Expert Solution

Interpretation Introduction

Interpretation:

The $pH$ of titration $NaOH Vs CH3COOH$ on adding various amount of sodium hydroxide has to be calculated.

Concept introduction:

$pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution.
The point at which amount of standard solution and analyte becomes equal and neutralisation happens in titration is called equivalence point.
$pH = pKa + log[conjugate base][acid]$ is Henderson-Hasselbalch equation.
Buffer solution is defined as a solution that oppose changes in $pH$ while adding little amount of either an acid or a base.
Buffer solution is a combination of a weak acid and its conjugate base or a weak base and its conjugate acid.

To calculate: The $pH$ when $10mL$ $NaOH$ is added.

Answer to Problem 4PPA

$pH = 5.04$

Explanation of Solution

The $pH$ of buffer solution when $10mL$ $NaOH$ is added is calculated below.

$The number of moles of acetic acid is(0.15 mmol/mL)(10.0 mL) = 1.50 mmol of acetic acidThe number of moles of NaOH is(0.10 mmol/mL)(10.0 mL) = 1.00 mmol of NaOH$

$CH3COOH(aq) +OH-(aq) →⇄ H2O(l) + CH3COO-(aq) Initial concentration(M): 1.50mmol 0 Change in concentration (M): -1.00mmol +1.00mmol Equilibrium concentration (M): 0.50mmol 1.00mmol pH = pKa + log[conjugate base][acid] pH = 4.74 + log1.000.50 pH = 4.74 + 0.30 pH = 5.04$

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The $pH$ of titration $NaOH Vs CH3COOH$ on adding various amount of sodium hydroxide has to be calculated.

Concept introduction:

$pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution.
The point at which amount of standard solution and analyte becomes equal and neutralisation happens in titration is called equivalence point.
$pH = pKa + log[conjugate base][acid]$ is Henderson-Hasselbalch equation.
Buffer solution is defined as a solution that oppose changes in $pH$ while adding little amount of either an acid or a base.
Buffer solution is a combination of a weak acid and its conjugate base or a weak base and its conjugate acid.

To calculate: The $pH$ when $15mL$ $NaOH$ is added that is at equivalence point.

Answer to Problem 4PPA

$pH = 8.87$

Explanation of Solution

The $pH$ of buffer solution when $15mL$ $NaOH$ is added is calculated below.

$The number of moles of acetic acid is(0.15 mmol/mL)(10.0 mL) = 1.50 mmol of acetic acidThe number of moles of NaOH is(0.10 mmol/mL)(15.0 mL) = 1.50 mmol of NaOH$

At equivalence point, the total volume of solution is

$10.0mL(acetic acid) + 15.0 mL(base) = 25.0 mL$

At equivalence point, the concentration of acetate ion is

$1.50mmol acetate ion15.0mL = 0.1M$

$CH3COO-(aq) +H2O(l) →⇄ OH-(aq) + CH3COO-(aq) Initial concentration(M): 0.1 0 0 Change in concentration (M): -x -x +x Equilibrium concentration (M): 0.1-x x x Kb value for acetate ion is 5.6 × 10-10 Kb = [OH-][CH3COOH][CH3COO-] 5.6 × 10-10 = x2(0.1-x)x is very small and neglect it, x2 = 5.6 × 10-11 x = 7.48 × 10-6M x = [OH-] = 7.48 × 10-6M pOH = -log[OH-] = -log(7.48 × 10-6) pOH = 5.13 pH = 14.00 - 5.13 = 8.87$

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The $pH$ of titration $NaOH Vs CH3COOH$ on adding various amount of sodium hydroxide has to be calculated.

Concept introduction:

$pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution.
The point at which amount of standard solution and analyte becomes equal and neutralisation happens in titration is called equivalence point.
$pH = pKa + log[conjugate base][acid]$ is Henderson-Hasselbalch equation.
Buffer solution is defined as a solution that oppose changes in $pH$ while adding little amount of either an acid or a base.
Buffer solution is a combination of a weak acid and its conjugate base or a weak base and its conjugate acid.

To calculate: The $pH$ when $20mL$ $NaOH$ is added that is beyond equivalence point.

Answer to Problem 4PPA

$pH = 12.82$

Explanation of Solution

The $pH$ of buffer solution when $20mL$ $NaOH$ is added is calculated below.

$The number of moles of NaOH is(0.10 mmol/mL)(20.0 mL) = 2.0 mmol of NaOHTotal volume of solution is 10.0 + 20.0 = 35.0 mL$

The concentration of sodium hydroxide is

$[OH-] = 2.00 mmol/30.0 mL = 0.0666 MpOH = -log (0.0666) = 1.24pH = 14.000 - 1.176 = 12.82$

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Answer 2

Textbook Question

Chapter 17.3, Problem 4PPA

For the titration of 10.0 mL of 0.15 M acetic acid with 0.10 M sodium hydroxide, determine the pH when (a) 10.0 mL of base has been added, (b) 15 .0 mL of base has been added, and (c) 20.0 mL of base has been added.

(a)

Expert Solution

Interpretation Introduction

Interpretation:

The $pH$ of titration $NaOH Vs CH3COOH$ on adding various amount of sodium hydroxide has to be calculated.

Concept introduction:

$pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution.
The point at which amount of standard solution and analyte becomes equal and neutralisation happens in titration is called equivalence point.
$pH = pKa + log[conjugate base][acid]$ is Henderson-Hasselbalch equation.
Buffer solution is defined as a solution that oppose changes in $pH$ while adding little amount of either an acid or a base.
Buffer solution is a combination of a weak acid and its conjugate base or a weak base and its conjugate acid.

To calculate: The $pH$ when $10mL$ $NaOH$ is added.

Answer to Problem 4PPA

$pH = 5.04$

Explanation of Solution

The $pH$ of buffer solution when $10mL$ $NaOH$ is added is calculated below.

$The number of moles of acetic acid is(0.15 mmol/mL)(10.0 mL) = 1.50 mmol of acetic acidThe number of moles of NaOH is(0.10 mmol/mL)(10.0 mL) = 1.00 mmol of NaOH$

$CH3COOH(aq) +OH-(aq) →⇄ H2O(l) + CH3COO-(aq) Initial concentration(M): 1.50mmol 0 Change in concentration (M): -1.00mmol +1.00mmol Equilibrium concentration (M): 0.50mmol 1.00mmol pH = pKa + log[conjugate base][acid] pH = 4.74 + log1.000.50 pH = 4.74 + 0.30 pH = 5.04$

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The $pH$ of titration $NaOH Vs CH3COOH$ on adding various amount of sodium hydroxide has to be calculated.

Concept introduction:

$pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution.
The point at which amount of standard solution and analyte becomes equal and neutralisation happens in titration is called equivalence point.
$pH = pKa + log[conjugate base][acid]$ is Henderson-Hasselbalch equation.
Buffer solution is defined as a solution that oppose changes in $pH$ while adding little amount of either an acid or a base.
Buffer solution is a combination of a weak acid and its conjugate base or a weak base and its conjugate acid.

To calculate: The $pH$ when $15mL$ $NaOH$ is added that is at equivalence point.

Answer to Problem 4PPA

$pH = 8.87$

Explanation of Solution

The $pH$ of buffer solution when $15mL$ $NaOH$ is added is calculated below.

$The number of moles of acetic acid is(0.15 mmol/mL)(10.0 mL) = 1.50 mmol of acetic acidThe number of moles of NaOH is(0.10 mmol/mL)(15.0 mL) = 1.50 mmol of NaOH$

At equivalence point, the total volume of solution is

$10.0mL(acetic acid) + 15.0 mL(base) = 25.0 mL$

At equivalence point, the concentration of acetate ion is

$1.50mmol acetate ion15.0mL = 0.1M$

$CH3COO-(aq) +H2O(l) →⇄ OH-(aq) + CH3COO-(aq) Initial concentration(M): 0.1 0 0 Change in concentration (M): -x -x +x Equilibrium concentration (M): 0.1-x x x Kb value for acetate ion is 5.6 × 10-10 Kb = [OH-][CH3COOH][CH3COO-] 5.6 × 10-10 = x2(0.1-x)x is very small and neglect it, x2 = 5.6 × 10-11 x = 7.48 × 10-6M x = [OH-] = 7.48 × 10-6M pOH = -log[OH-] = -log(7.48 × 10-6) pOH = 5.13 pH = 14.00 - 5.13 = 8.87$

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The $pH$ of titration $NaOH Vs CH3COOH$ on adding various amount of sodium hydroxide has to be calculated.

Concept introduction:

$pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution.
The point at which amount of standard solution and analyte becomes equal and neutralisation happens in titration is called equivalence point.
$pH = pKa + log[conjugate base][acid]$ is Henderson-Hasselbalch equation.
Buffer solution is defined as a solution that oppose changes in $pH$ while adding little amount of either an acid or a base.
Buffer solution is a combination of a weak acid and its conjugate base or a weak base and its conjugate acid.

To calculate: The $pH$ when $20mL$ $NaOH$ is added that is beyond equivalence point.

Answer to Problem 4PPA

$pH = 12.82$

Explanation of Solution

The $pH$ of buffer solution when $20mL$ $NaOH$ is added is calculated below.

$The number of moles of NaOH is(0.10 mmol/mL)(20.0 mL) = 2.0 mmol of NaOHTotal volume of solution is 10.0 + 20.0 = 35.0 mL$

The concentration of sodium hydroxide is

$[OH-] = 2.00 mmol/30.0 mL = 0.0666 MpOH = -log (0.0666) = 1.24pH = 14.000 - 1.176 = 12.82$

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Answer 3

Textbook Question

Chapter 17.3, Problem 4PPA

For the titration of 10.0 mL of 0.15 M acetic acid with 0.10 M sodium hydroxide, determine the pH when (a) 10.0 mL of base has been added, (b) 15 .0 mL of base has been added, and (c) 20.0 mL of base has been added.

(a)

Expert Solution

Interpretation Introduction

Interpretation:

The $pH$ of titration $NaOH Vs CH3COOH$ on adding various amount of sodium hydroxide has to be calculated.

Concept introduction:

$pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution.
The point at which amount of standard solution and analyte becomes equal and neutralisation happens in titration is called equivalence point.
$pH = pKa + log[conjugate base][acid]$ is Henderson-Hasselbalch equation.
Buffer solution is defined as a solution that oppose changes in $pH$ while adding little amount of either an acid or a base.
Buffer solution is a combination of a weak acid and its conjugate base or a weak base and its conjugate acid.

To calculate: The $pH$ when $10mL$ $NaOH$ is added.

Answer to Problem 4PPA

$pH = 5.04$

Explanation of Solution

The $pH$ of buffer solution when $10mL$ $NaOH$ is added is calculated below.

$The number of moles of acetic acid is(0.15 mmol/mL)(10.0 mL) = 1.50 mmol of acetic acidThe number of moles of NaOH is(0.10 mmol/mL)(10.0 mL) = 1.00 mmol of NaOH$

$CH3COOH(aq) +OH-(aq) →⇄ H2O(l) + CH3COO-(aq) Initial concentration(M): 1.50mmol 0 Change in concentration (M): -1.00mmol +1.00mmol Equilibrium concentration (M): 0.50mmol 1.00mmol pH = pKa + log[conjugate base][acid] pH = 4.74 + log1.000.50 pH = 4.74 + 0.30 pH = 5.04$

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The $pH$ of titration $NaOH Vs CH3COOH$ on adding various amount of sodium hydroxide has to be calculated.

Concept introduction:

$pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution.
The point at which amount of standard solution and analyte becomes equal and neutralisation happens in titration is called equivalence point.
$pH = pKa + log[conjugate base][acid]$ is Henderson-Hasselbalch equation.
Buffer solution is defined as a solution that oppose changes in $pH$ while adding little amount of either an acid or a base.
Buffer solution is a combination of a weak acid and its conjugate base or a weak base and its conjugate acid.

To calculate: The $pH$ when $15mL$ $NaOH$ is added that is at equivalence point.

Answer to Problem 4PPA

$pH = 8.87$

Explanation of Solution

The $pH$ of buffer solution when $15mL$ $NaOH$ is added is calculated below.

$The number of moles of acetic acid is(0.15 mmol/mL)(10.0 mL) = 1.50 mmol of acetic acidThe number of moles of NaOH is(0.10 mmol/mL)(15.0 mL) = 1.50 mmol of NaOH$

At equivalence point, the total volume of solution is

$10.0mL(acetic acid) + 15.0 mL(base) = 25.0 mL$

At equivalence point, the concentration of acetate ion is

$1.50mmol acetate ion15.0mL = 0.1M$

$CH3COO-(aq) +H2O(l) →⇄ OH-(aq) + CH3COO-(aq) Initial concentration(M): 0.1 0 0 Change in concentration (M): -x -x +x Equilibrium concentration (M): 0.1-x x x Kb value for acetate ion is 5.6 × 10-10 Kb = [OH-][CH3COOH][CH3COO-] 5.6 × 10-10 = x2(0.1-x)x is very small and neglect it, x2 = 5.6 × 10-11 x = 7.48 × 10-6M x = [OH-] = 7.48 × 10-6M pOH = -log[OH-] = -log(7.48 × 10-6) pOH = 5.13 pH = 14.00 - 5.13 = 8.87$

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The $pH$ of titration $NaOH Vs CH3COOH$ on adding various amount of sodium hydroxide has to be calculated.

Concept introduction:

$pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution.
The point at which amount of standard solution and analyte becomes equal and neutralisation happens in titration is called equivalence point.
$pH = pKa + log[conjugate base][acid]$ is Henderson-Hasselbalch equation.
Buffer solution is defined as a solution that oppose changes in $pH$ while adding little amount of either an acid or a base.
Buffer solution is a combination of a weak acid and its conjugate base or a weak base and its conjugate acid.

To calculate: The $pH$ when $20mL$ $NaOH$ is added that is beyond equivalence point.

Answer to Problem 4PPA

$pH = 12.82$

Explanation of Solution

The $pH$ of buffer solution when $20mL$ $NaOH$ is added is calculated below.

$The number of moles of NaOH is(0.10 mmol/mL)(20.0 mL) = 2.0 mmol of NaOHTotal volume of solution is 10.0 + 20.0 = 35.0 mL$

The concentration of sodium hydroxide is

$[OH-] = 2.00 mmol/30.0 mL = 0.0666 MpOH = -log (0.0666) = 1.24pH = 14.000 - 1.176 = 12.82$

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Answer 4

Textbook Question

Chapter 17.3, Problem 4PPA

For the titration of 10.0 mL of 0.15 M acetic acid with 0.10 M sodium hydroxide, determine the pH when (a) 10.0 mL of base has been added, (b) 15 .0 mL of base has been added, and (c) 20.0 mL of base has been added.

(a)

Expert Solution

Interpretation Introduction

Interpretation:

The $pH$ of titration $NaOH Vs CH3COOH$ on adding various amount of sodium hydroxide has to be calculated.

Concept introduction:

$pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution.
The point at which amount of standard solution and analyte becomes equal and neutralisation happens in titration is called equivalence point.
$pH = pKa + log[conjugate base][acid]$ is Henderson-Hasselbalch equation.
Buffer solution is defined as a solution that oppose changes in $pH$ while adding little amount of either an acid or a base.
Buffer solution is a combination of a weak acid and its conjugate base or a weak base and its conjugate acid.

To calculate: The $pH$ when $10mL$ $NaOH$ is added.

Answer to Problem 4PPA

$pH = 5.04$

Explanation of Solution

The $pH$ of buffer solution when $10mL$ $NaOH$ is added is calculated below.

$The number of moles of acetic acid is(0.15 mmol/mL)(10.0 mL) = 1.50 mmol of acetic acidThe number of moles of NaOH is(0.10 mmol/mL)(10.0 mL) = 1.00 mmol of NaOH$

$CH3COOH(aq) +OH-(aq) →⇄ H2O(l) + CH3COO-(aq) Initial concentration(M): 1.50mmol 0 Change in concentration (M): -1.00mmol +1.00mmol Equilibrium concentration (M): 0.50mmol 1.00mmol pH = pKa + log[conjugate base][acid] pH = 4.74 + log1.000.50 pH = 4.74 + 0.30 pH = 5.04$

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The $pH$ of titration $NaOH Vs CH3COOH$ on adding various amount of sodium hydroxide has to be calculated.

Concept introduction:

$pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution.
The point at which amount of standard solution and analyte becomes equal and neutralisation happens in titration is called equivalence point.
$pH = pKa + log[conjugate base][acid]$ is Henderson-Hasselbalch equation.
Buffer solution is defined as a solution that oppose changes in $pH$ while adding little amount of either an acid or a base.
Buffer solution is a combination of a weak acid and its conjugate base or a weak base and its conjugate acid.

To calculate: The $pH$ when $15mL$ $NaOH$ is added that is at equivalence point.

Answer to Problem 4PPA

$pH = 8.87$

Explanation of Solution

The $pH$ of buffer solution when $15mL$ $NaOH$ is added is calculated below.

$The number of moles of acetic acid is(0.15 mmol/mL)(10.0 mL) = 1.50 mmol of acetic acidThe number of moles of NaOH is(0.10 mmol/mL)(15.0 mL) = 1.50 mmol of NaOH$

At equivalence point, the total volume of solution is

$10.0mL(acetic acid) + 15.0 mL(base) = 25.0 mL$

At equivalence point, the concentration of acetate ion is

$1.50mmol acetate ion15.0mL = 0.1M$

$CH3COO-(aq) +H2O(l) →⇄ OH-(aq) + CH3COO-(aq) Initial concentration(M): 0.1 0 0 Change in concentration (M): -x -x +x Equilibrium concentration (M): 0.1-x x x Kb value for acetate ion is 5.6 × 10-10 Kb = [OH-][CH3COOH][CH3COO-] 5.6 × 10-10 = x2(0.1-x)x is very small and neglect it, x2 = 5.6 × 10-11 x = 7.48 × 10-6M x = [OH-] = 7.48 × 10-6M pOH = -log[OH-] = -log(7.48 × 10-6) pOH = 5.13 pH = 14.00 - 5.13 = 8.87$

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The $pH$ of titration $NaOH Vs CH3COOH$ on adding various amount of sodium hydroxide has to be calculated.

Concept introduction:

$pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution.
The point at which amount of standard solution and analyte becomes equal and neutralisation happens in titration is called equivalence point.
$pH = pKa + log[conjugate base][acid]$ is Henderson-Hasselbalch equation.
Buffer solution is defined as a solution that oppose changes in $pH$ while adding little amount of either an acid or a base.
Buffer solution is a combination of a weak acid and its conjugate base or a weak base and its conjugate acid.

To calculate: The $pH$ when $20mL$ $NaOH$ is added that is beyond equivalence point.

Answer to Problem 4PPA

$pH = 12.82$

Explanation of Solution

The $pH$ of buffer solution when $20mL$ $NaOH$ is added is calculated below.

$The number of moles of NaOH is(0.10 mmol/mL)(20.0 mL) = 2.0 mmol of NaOHTotal volume of solution is 10.0 + 20.0 = 35.0 mL$

The concentration of sodium hydroxide is

$[OH-] = 2.00 mmol/30.0 mL = 0.0666 MpOH = -log (0.0666) = 1.24pH = 14.000 - 1.176 = 12.82$

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

Interpretation Introduction

Interpretation:

The $pH$ of titration $NaOH Vs CH3COOH$ on adding various amount of sodium hydroxide has to be calculated.

Concept introduction:

$pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution.
The point at which amount of standard solution and analyte becomes equal and neutralisation happens in titration is called equivalence point.
$pH = pKa + log[conjugate base][acid]$ is Henderson-Hasselbalch equation.
Buffer solution is defined as a solution that oppose changes in $pH$ while adding little amount of either an acid or a base.
Buffer solution is a combination of a weak acid and its conjugate base or a weak base and its conjugate acid.

To calculate: The $pH$ when $10mL$ $NaOH$ is added.

Answer to Problem 4PPA

$pH = 5.04$

Explanation of Solution

The $pH$ of buffer solution when $10mL$ $NaOH$ is added is calculated below.

$The number of moles of acetic acid is(0.15 mmol/mL)(10.0 mL) = 1.50 mmol of acetic acidThe number of moles of NaOH is(0.10 mmol/mL)(10.0 mL) = 1.00 mmol of NaOH$

$CH3COOH(aq) +OH-(aq) →⇄ H2O(l) + CH3COO-(aq) Initial concentration(M): 1.50mmol 0 Change in concentration (M): -1.00mmol +1.00mmol Equilibrium concentration (M): 0.50mmol 1.00mmol pH = pKa + log[conjugate base][acid] pH = 4.74 + log1.000.50 pH = 4.74 + 0.30 pH = 5.04$

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The $pH$ of titration $NaOH Vs CH3COOH$ on adding various amount of sodium hydroxide has to be calculated.

Concept introduction:

$pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution.
The point at which amount of standard solution and analyte becomes equal and neutralisation happens in titration is called equivalence point.
$pH = pKa + log[conjugate base][acid]$ is Henderson-Hasselbalch equation.
Buffer solution is defined as a solution that oppose changes in $pH$ while adding little amount of either an acid or a base.
Buffer solution is a combination of a weak acid and its conjugate base or a weak base and its conjugate acid.

To calculate: The $pH$ when $15mL$ $NaOH$ is added that is at equivalence point.

Answer to Problem 4PPA

$pH = 8.87$

Explanation of Solution

The $pH$ of buffer solution when $15mL$ $NaOH$ is added is calculated below.

$The number of moles of acetic acid is(0.15 mmol/mL)(10.0 mL) = 1.50 mmol of acetic acidThe number of moles of NaOH is(0.10 mmol/mL)(15.0 mL) = 1.50 mmol of NaOH$

At equivalence point, the total volume of solution is

$10.0mL(acetic acid) + 15.0 mL(base) = 25.0 mL$

At equivalence point, the concentration of acetate ion is

$1.50mmol acetate ion15.0mL = 0.1M$

$CH3COO-(aq) +H2O(l) →⇄ OH-(aq) + CH3COO-(aq) Initial concentration(M): 0.1 0 0 Change in concentration (M): -x -x +x Equilibrium concentration (M): 0.1-x x x Kb value for acetate ion is 5.6 × 10-10 Kb = [OH-][CH3COOH][CH3COO-] 5.6 × 10-10 = x2(0.1-x)x is very small and neglect it, x2 = 5.6 × 10-11 x = 7.48 × 10-6M x = [OH-] = 7.48 × 10-6M pOH = -log[OH-] = -log(7.48 × 10-6) pOH = 5.13 pH = 14.00 - 5.13 = 8.87$

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The $pH$ of titration $NaOH Vs CH3COOH$ on adding various amount of sodium hydroxide has to be calculated.

Concept introduction:

$pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution.
The point at which amount of standard solution and analyte becomes equal and neutralisation happens in titration is called equivalence point.
$pH = pKa + log[conjugate base][acid]$ is Henderson-Hasselbalch equation.
Buffer solution is defined as a solution that oppose changes in $pH$ while adding little amount of either an acid or a base.
Buffer solution is a combination of a weak acid and its conjugate base or a weak base and its conjugate acid.

To calculate: The $pH$ when $20mL$ $NaOH$ is added that is beyond equivalence point.

Answer to Problem 4PPA

$pH = 12.82$

Explanation of Solution

The $pH$ of buffer solution when $20mL$ $NaOH$ is added is calculated below.

$The number of moles of NaOH is(0.10 mmol/mL)(20.0 mL) = 2.0 mmol of NaOHTotal volume of solution is 10.0 + 20.0 = 35.0 mL$

The concentration of sodium hydroxide is

$[OH-] = 2.00 mmol/30.0 mL = 0.0666 MpOH = -log (0.0666) = 1.24pH = 14.000 - 1.176 = 12.82$

Answer 8

(a)

Expert Solution

Interpretation Introduction

Interpretation:

The $pH$ of titration $NaOH Vs CH3COOH$ on adding various amount of sodium hydroxide has to be calculated.

Concept introduction:

$pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution.
The point at which amount of standard solution and analyte becomes equal and neutralisation happens in titration is called equivalence point.
$pH = pKa + log[conjugate base][acid]$ is Henderson-Hasselbalch equation.
Buffer solution is defined as a solution that oppose changes in $pH$ while adding little amount of either an acid or a base.
Buffer solution is a combination of a weak acid and its conjugate base or a weak base and its conjugate acid.

To calculate: The $pH$ when $10mL$ $NaOH$ is added.

Answer to Problem 4PPA

$pH = 5.04$

Explanation of Solution

The $pH$ of buffer solution when $10mL$ $NaOH$ is added is calculated below.

$The number of moles of acetic acid is(0.15 mmol/mL)(10.0 mL) = 1.50 mmol of acetic acidThe number of moles of NaOH is(0.10 mmol/mL)(10.0 mL) = 1.00 mmol of NaOH$

$CH3COOH(aq) +OH-(aq) →⇄ H2O(l) + CH3COO-(aq) Initial concentration(M): 1.50mmol 0 Change in concentration (M): -1.00mmol +1.00mmol Equilibrium concentration (M): 0.50mmol 1.00mmol pH = pKa + log[conjugate base][acid] pH = 4.74 + log1.000.50 pH = 4.74 + 0.30 pH = 5.04$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Interpretation Introduction

Interpretation:

The $pH$ of titration $NaOH Vs CH3COOH$ on adding various amount of sodium hydroxide has to be calculated.

Concept introduction:

$pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution.
The point at which amount of standard solution and analyte becomes equal and neutralisation happens in titration is called equivalence point.
$pH = pKa + log[conjugate base][acid]$ is Henderson-Hasselbalch equation.
Buffer solution is defined as a solution that oppose changes in $pH$ while adding little amount of either an acid or a base.
Buffer solution is a combination of a weak acid and its conjugate base or a weak base and its conjugate acid.

To calculate: The $pH$ when $10mL$ $NaOH$ is added.

Answer 13

Answer to Problem 4PPA

$pH = 5.04$

Answer 14

Explanation of Solution

The $pH$ of buffer solution when $10mL$ $NaOH$ is added is calculated below.

$The number of moles of acetic acid is(0.15 mmol/mL)(10.0 mL) = 1.50 mmol of acetic acidThe number of moles of NaOH is(0.10 mmol/mL)(10.0 mL) = 1.00 mmol of NaOH$

$CH3COOH(aq) +OH-(aq) →⇄ H2O(l) + CH3COO-(aq) Initial concentration(M): 1.50mmol 0 Change in concentration (M): -1.00mmol +1.00mmol Equilibrium concentration (M): 0.50mmol 1.00mmol pH = pKa + log[conjugate base][acid] pH = 4.74 + log1.000.50 pH = 4.74 + 0.30 pH = 5.04$

Answer 15

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The $pH$ of titration $NaOH Vs CH3COOH$ on adding various amount of sodium hydroxide has to be calculated.

Concept introduction:

$pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution.
The point at which amount of standard solution and analyte becomes equal and neutralisation happens in titration is called equivalence point.
$pH = pKa + log[conjugate base][acid]$ is Henderson-Hasselbalch equation.
Buffer solution is defined as a solution that oppose changes in $pH$ while adding little amount of either an acid or a base.
Buffer solution is a combination of a weak acid and its conjugate base or a weak base and its conjugate acid.

To calculate: The $pH$ when $15mL$ $NaOH$ is added that is at equivalence point.

Answer to Problem 4PPA

$pH = 8.87$

Explanation of Solution

The $pH$ of buffer solution when $15mL$ $NaOH$ is added is calculated below.

$The number of moles of acetic acid is(0.15 mmol/mL)(10.0 mL) = 1.50 mmol of acetic acidThe number of moles of NaOH is(0.10 mmol/mL)(15.0 mL) = 1.50 mmol of NaOH$

At equivalence point, the total volume of solution is

$10.0mL(acetic acid) + 15.0 mL(base) = 25.0 mL$

At equivalence point, the concentration of acetate ion is

$1.50mmol acetate ion15.0mL = 0.1M$

$CH3COO-(aq) +H2O(l) →⇄ OH-(aq) + CH3COO-(aq) Initial concentration(M): 0.1 0 0 Change in concentration (M): -x -x +x Equilibrium concentration (M): 0.1-x x x Kb value for acetate ion is 5.6 × 10-10 Kb = [OH-][CH3COOH][CH3COO-] 5.6 × 10-10 = x2(0.1-x)x is very small and neglect it, x2 = 5.6 × 10-11 x = 7.48 × 10-6M x = [OH-] = 7.48 × 10-6M pOH = -log[OH-] = -log(7.48 × 10-6) pOH = 5.13 pH = 14.00 - 5.13 = 8.87$

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Interpretation Introduction

Interpretation:

The $pH$ of titration $NaOH Vs CH3COOH$ on adding various amount of sodium hydroxide has to be calculated.

Concept introduction:

$pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution.
The point at which amount of standard solution and analyte becomes equal and neutralisation happens in titration is called equivalence point.
$pH = pKa + log[conjugate base][acid]$ is Henderson-Hasselbalch equation.
Buffer solution is defined as a solution that oppose changes in $pH$ while adding little amount of either an acid or a base.
Buffer solution is a combination of a weak acid and its conjugate base or a weak base and its conjugate acid.

To calculate: The $pH$ when $15mL$ $NaOH$ is added that is at equivalence point.

Answer 20

Answer to Problem 4PPA

$pH = 8.87$

Answer 21

Explanation of Solution

The $pH$ of buffer solution when $15mL$ $NaOH$ is added is calculated below.

$The number of moles of acetic acid is(0.15 mmol/mL)(10.0 mL) = 1.50 mmol of acetic acidThe number of moles of NaOH is(0.10 mmol/mL)(15.0 mL) = 1.50 mmol of NaOH$

At equivalence point, the total volume of solution is

$10.0mL(acetic acid) + 15.0 mL(base) = 25.0 mL$

At equivalence point, the concentration of acetate ion is

$1.50mmol acetate ion15.0mL = 0.1M$

$CH3COO-(aq) +H2O(l) →⇄ OH-(aq) + CH3COO-(aq) Initial concentration(M): 0.1 0 0 Change in concentration (M): -x -x +x Equilibrium concentration (M): 0.1-x x x Kb value for acetate ion is 5.6 × 10-10 Kb = [OH-][CH3COOH][CH3COO-] 5.6 × 10-10 = x2(0.1-x)x is very small and neglect it, x2 = 5.6 × 10-11 x = 7.48 × 10-6M x = [OH-] = 7.48 × 10-6M pOH = -log[OH-] = -log(7.48 × 10-6) pOH = 5.13 pH = 14.00 - 5.13 = 8.87$

Answer 22

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The $pH$ of titration $NaOH Vs CH3COOH$ on adding various amount of sodium hydroxide has to be calculated.

Concept introduction:

$pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution.
The point at which amount of standard solution and analyte becomes equal and neutralisation happens in titration is called equivalence point.
$pH = pKa + log[conjugate base][acid]$ is Henderson-Hasselbalch equation.
Buffer solution is defined as a solution that oppose changes in $pH$ while adding little amount of either an acid or a base.
Buffer solution is a combination of a weak acid and its conjugate base or a weak base and its conjugate acid.

To calculate: The $pH$ when $20mL$ $NaOH$ is added that is beyond equivalence point.

Answer to Problem 4PPA

$pH = 12.82$

Explanation of Solution

The $pH$ of buffer solution when $20mL$ $NaOH$ is added is calculated below.

$The number of moles of NaOH is(0.10 mmol/mL)(20.0 mL) = 2.0 mmol of NaOHTotal volume of solution is 10.0 + 20.0 = 35.0 mL$

The concentration of sodium hydroxide is

$[OH-] = 2.00 mmol/30.0 mL = 0.0666 MpOH = -log (0.0666) = 1.24pH = 14.000 - 1.176 = 12.82$

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Interpretation Introduction

Interpretation:

The $pH$ of titration $NaOH Vs CH3COOH$ on adding various amount of sodium hydroxide has to be calculated.

Concept introduction:

$pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution.
The point at which amount of standard solution and analyte becomes equal and neutralisation happens in titration is called equivalence point.
$pH = pKa + log[conjugate base][acid]$ is Henderson-Hasselbalch equation.
Buffer solution is defined as a solution that oppose changes in $pH$ while adding little amount of either an acid or a base.
Buffer solution is a combination of a weak acid and its conjugate base or a weak base and its conjugate acid.