Bundle: Chemistry: The Molecular Science, 5th, Loose-Leaf + OWLv2 with Quick Prep 24-Months Printed Access Card
Bundle: Chemistry: The Molecular Science, 5th, Loose-Leaf + OWLv2 with Quick Prep 24-Months Printed Access Card
5th Edition
ISBN: 9781305367487
Author: John W. Moore, Conrad L. Stanitski
Publisher: Cengage Learning
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Chapter 15, Problem 30QRT

(a)

Interpretation Introduction

Interpretation:

The value of pH change when 1.0 M of HCl is mixed with 0.10 M of CH3COOH and 0.10 M of CH3COONa has to be determined.

Concept Introduction:

The Henderson-Hasselbalch equation can be used to determine the pH of the buffer which is formed by the known concentrations of conjugate base and conjugate acid.  This equation can also be applied to calculate the ratio of conjugate base to conjugate acid concentration.

The Henderson-Hasselbalch equation can be represented as:

  pH=pKa+log[Conjugatebase][Conjugateacid]

(a)

Expert Solution
Check Mark

Answer to Problem 30QRT

The change in pH value is -0.087_.

Explanation of Solution

The value of pH of the buffer solution before NaOH is added to the buffer solution is calculated using the Henderson-Hasselbalch equation shown below.

  pH=pKa+log[Conjugate base][Conjugateacid]        (1)

The concentration of conjugate acid is 1.0M.

The concentration of conjugate base is 1.0M.

Substitute the value of concentration of conjugate acid and the concentration of conjugate base in the equation (1).

  pH=4.74+log1.0M1.0M=4.74+log1=4.74+0.0=4.74

The number of moles of HCl is calculated by the formula shown below.

  n=M×V        (2)

Where,

  • M is the molarity.
  • n is the number of moles.
  • V is the volume.

The value of M for HCl is 1.0 M.

The value of V for HCl is 1.0mL.

Substitute the values of  M and  V for HCl in the equation (2).

  nHCl=(1.0 M)(1.0 mL×1 L1000mL)=0.001mol

Therefore, the number of moles of HCl is 0.001 mol.

The value of M for CH3COOH is 0.1 M.

The value of V for CH3COOH is 0.1 L.

Substitute the values of M and V for CH3COOH in the equation (2).

  nCH3COOH=(0.1 M)(0.1 L)=0.01mol

The concentration of CH3COOH and CH3COONa are same.  Thus, the mole of CH3COONa is equal to the mole of CH3COOH.

Therefore, the number of mole of CH3COONa is 0.01mol.

The total volume is calculated is as shown below:

  Vtotal =(1.0 L×1000mL1.0 L)+0.1 mL=1000.1 mL

The chemical reaction between CH3COONa with HCl is shown below.

  CH3COONa+HClCH3COOH+H2O

The summarized data for the number of moles of all species after NaOH is added to the buffer solution is shown below.

  CH3COONa+HClCH3COOH+H2OInitial0.0100.01Final0.010.0010.0010.0010.01+0.001

Therefore, the new number of moles of CH3COOH and CH3COONa is 0.011mol and 0.009mol respectively.

The concentration of CH3COONa is calculated by the formula shown below.

  M=nV        (3)

Where,

  • M is the molarity.
  • n is the number of moles.
  • V is the volume.

The value of n for CH3COONa is 0.009mol.

The value of V for CH3COONa is 1000.1 mL.

Substitute the values of  n and  V for CH3COONa in the equation (3).

  [CH3COONa]=(0.009 mol)(1000.1 mL×1 L1000mL)=0.0089M

The value of n for CH3COOH is 0.011mol.

The value of V for CH3COOH is 1000.1 mL.

Substitute the values of n and V for CH3COOH in the equation (3).

  [CH3COOH]=(0.011 mol)(1000.1 mL×1 L1000mL)=0.0109M

The standard value of pKa for CH3COOH is 4.74.

The concentration of conjugate acid CH3COOH is 0.0109M.

The concentration of conjugate base CH3COONa is 0.0089M.

Substitute the value of pKa, concentration of conjugate acid and concentration of conjugate base in the equation (1).

  pH=4.74+log0.0089M0.0109M=4.74+log0.8165=4.740.080=4.66

Therefore, the pH of the buffer solution is 4.66

The change in pH that is ΔpH calculated as shown below.

  ΔpH=4.664.74=0.08

Thus, the pH change of the buffer solution is -0.08_.

(b)

Interpretation Introduction

Interpretation:

The value of pH change when 1.0M of HCl is mixed with 0.010 M of CH3COOH and 0.010 M of CH3COONa has to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 30QRT

The change in pH value is -1.00_.

Explanation of Solution

The value of pH of the buffer solution before HCl is added to the buffer solution is calculated using equation (1).

The concentration of conjugate acid is 1.0M.

The concentration of conjugate base is 1.0M.

Substitute the value of concentration of conjugate acid and the concentration of conjugate base in the equation (1).

  pH=4.74+log1.0M1.0M=4.74+log1=4.74+0.0=4.74

The value of M for HCl is  1.0 M.

The value of V for HCl is 1.0mL.

Substitute the values of M and V for HCl in the equation (2).

  nHCl=(1.0 M)(1.0 mL×1 L1000mL)=0.001mol

The value of M for CH3COOH is 0.01 M.

The value of V for CH3COOH is 0.1 L.

Substitute the values of M and V for CH3COOH in the equation (2).

  nCH3COOH=(0.01 M)(0.1 L)=0.001mol

The concentration of CH3COOH and CH3COONa are equal. Thus, the number of moles of CH3COONa is equal to that of CH3COOH.

Therefore, the number of moles of CH3COONa is 0.001 M.

The total volume is calculated is as shown below:

  Vsolution=(1.0 L×1000mL1.0 L)+0.1 mL=1000.1 mL

The chemical reaction between CH3COONa and HCl is shown below.

  CH3COONa+HClCH3COOH+H2O

The summarized data of number of moles of all species after HCl is added to the buffer solution is shown below.

  CH3COONa+HClCH3COOH+H2OInitial0.0010.0010.001Final0.0010.0010.0010.0010.001+0.001

Therefore, the new number of moles of CH3COOH and CH3COONa is 0.002mol and 0mol respectively.

The value of n for CH3COOH is 0.002mol.

The value of V for CH3COOH is 1000.1 mL.

Substitute the values of n and V for CH3COOH in the equation (3).

  [CH3COOH]=(0.002 mol)(1000.1 mL×1 L1000mL)=0.0019M

The equation for the ionization of CH3COOH is shown below.

  CH3COOHCH3COO+H+

In the above equation, one mole of CH3COOH is dissociated to formed one mole of CH3COO and one mole of H+.  Thus, concentration of CH3COO and H+ are same.

The ionization constantexpression for the above equation is shown below.

  Ka=[CH3COO][H+][CH3COOH]        (4)

Since concentration of CH3COO and H+ are same.  Rearrange equation (4) to calculate [H+].

  [H+]=(Ka)([CH3COOH])        (5)

The value for the concentration of CH3COOH is 0.0019M.

The value of Ka is 1.8×105.

Substitute the value of Ka and [CH3COOH] in the equation (5).

  [H+]=(1.8×105)(0.0019M)=1.8×104 M

The relation between pH and [H+] is shown below.

  pH=log[H+]        (6)

Substitute the value of [H+] in the equation (6).

  pH=log(1.8×104)=3.74

Therefore, the pH of the buffer solution is 3.74.

The change in pH that is ΔpH is calculated as shown below.

  ΔpH=3.744.74=-1.00_

Thus, the pH change of the buffer solution is -1.00_.

(c)

Interpretation Introduction

Interpretation:

The value of pH change when 1.0 M of HCl is mixed with 0.0010 M of CH3COOH and 0.0010 M of CH3COONa has to be determined.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 30QRT

The change in pH value is -2.7_.

Explanation of Solution

The value of pH of the buffer solution before HCl is added to the buffer solution is calculated using equation (1).

The concentration of conjugate acid is 1.0 M.

The concentration of conjugate base is 1.0 M.

Substitute the concentration of conjugate acid and the concentration of conjugate base in the equation (1).

  pH=4.74+log1.0M1.0M=4.74+log1=4.74+0.0=4.74

The value of M for HCl is 1.0 M.

The value of V for HCl is 1.0mL.

Substitute the values of M and V for HCl in the equation (2).

  nNaOH=(1.0 M)(1.0 mL×1 L1000mL)=0.001mol

The value of M for CH3COOH is 0.0010 M.

The value of V for CH3COOH is 0.1 L.

Substitute the values of M and V for CH3COOH in the equation (2).

  nCH3COOH=(0.0010 M)(0.1 L)=0.0001mol

The concentration of CH3COOH and CH3COONa are same. Thus, the mole of CH3COONa is equal to the mole of CH3COOH.

Therefore, the number of moles of CH3COONa is 0.0001 M.

The total volume is calculated as shown below.

  Vsolution=(1.0 L×1000mL1.0 L)+0.1 mL=1000.1 mL

The chemical reaction between CH3COONa with HCl is shown below.

  CH3COONa+HClCH3COOH+H2O

The summarized data of the number of moles of all species after HCl is added to the buffer solution is shown below.

  CH3COONa+HClCH3COOH+H2OInitial0.00010.0010.0001Final0.00010.00010.0010.00010.0001+0.0001

There are greater number of moles of NaOH as compared to CH3COONa and CH3COOH. Thus, total number of moles of CH3COOH will be consumed in the reaction while HCl will remains unreacted.

Therefore, the new number of moles of CH3COONa, CH3COOH and HCl is 0.0mol, 0.0002mol and 0.0009mol respectively.

The value of n for HCl is 0.0009mol.

The value of V for HCl is 0.1 L.

Substitute the values of n and V for HCl in the equation (3).

  [HCl]=(0.0009 mol)(0.1 L)=0.009M

As HCl is the strong acid so, it will be dissociated completely into aqueous solution.

Thus, the concentration of [H+] is 0.009M.

Substitute the value of [H+] in the equation (6).

  pH=log(0.009)=2.04

Therefore, the pH of the buffer solution is 2.04.

The change in pH that is ΔpH is calculated as shown below.

  ΔpH=2.044.74=-2.7_

Thus, the pH change of the buffer solution is -2.7_.

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Chapter 15 Solutions

Bundle: Chemistry: The Molecular Science, 5th, Loose-Leaf + OWLv2 with Quick Prep 24-Months Printed Access Card

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