Chapter 17, Problem 17.35QP

A 25.0-,L solution of 0n100 M CH₃COOH is titrated with a 0.200 M KOH solution. Calculate the pH after the following additions of the KOH solution: (a) 0.0 mL, (b) 5.0 mL, (c) 10.0 mL, (d) 12.5 mL, (e) 15.0 mL.

Interpretation Introduction

Interpretation:

The $\text{pH}$ of titration $\text{KOH}\text{Vs}{\text{CH}}_{\text{3}}\text{COOH}$ on adding various amount of potassium hydroxide has to be calculated.

Concept introduction:

• $\text{pH}$ is the logarithm of the reciprocal of the concentration  of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in a solution.
•   $\text{pH}$ is used to determine the acidity or basicity of an aqueous solution.
• $\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$
• The point at which amount of standard solution and analyte becomes equal and neutralisation happens in titration is called equivalence point.
• $\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}$ is Henderson-Hasselbalch equation
• Buffer solution is defined as a solution that oppose changes in $\text{pH}$ while adding little amount of either an acid or a base.

To calculate: the $\text{pH}$ when no $\text{KOH}$ is added.

Answer to Problem 17.35QP

pH = $2.87$

Explanation of Solution

Concentration of $\text{KOH}$ is $\text{0}\text{.20}0\text{M}$

Concentration of ${\text{CH}}_{\text{3}}\text{COOH}$ is $\text{0}\text{.100}\text{M}$

$\begin{array}{l}{\text{CH}}_{\text{3}}{\text{COOH}}_{\text{(aq)}}{\text{ +H}}_{\text{2}}{\text{O}}_{\text{(l)}}\to {\text{ H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{ +}{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\text{(aq)}\\ \\ \text{Initial concentration(M): }\text{0}\text{.100 }\text{0}\text{0}\\ \text{Change in concentration (M):}\text{-x}\text{-x}\text{+x}\\ \text{Equilibrium}\text{concentration (M): }\text{0}\text{.100-x}\text{x}\text{x}\\ \\ {\text{K}}_{\text{a}}\text{value for acetic acid is 1}\text{.8 ×}{\text{10}}^{\text{-5}}\\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][CH}}_{\text{3}}{\text{COO}}^{\text{-}}\text{]}}{{\text{[CH}}_{\text{3}}\text{COOH]}}\\ \text{1}\text{.8 ×}{\text{10}}^{\text{-5}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{(0}\text{.100-x)}}\\ \text{x}\text{is}\text{very}\text{small}\text{and}\text{neglect}\text{it,}\\ {\text{x = [H}}_{\text{3}}{\text{O}}^{\text{+}}\text{] = 1}\text{.3 ×}{\text{10}}^{\text{-3}}\text{M}\\ \text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\\ \text{=}\text{-log(1}\text{.3 ×}{\text{10}}^{\text{-3}}\text{)}\\ \text{pH}\text{=}\text{2}\text{.87}\end{array}$

The $\text{pH}$ of acetic acid in water can be determined from acid dissociation constant.  From the concentration of hydrogen ion, the $\text{pH}$ is calculated by taking negative log of hydrogen ion.

Interpretation Introduction

Interpretation:

The $\text{pH}$ of titration $\text{KOH}\text{Vs}{\text{CH}}_{\text{3}}\text{COOH}$ on adding various amount of potassium hydroxide has to be calculated.

Concept introduction:

• $\text{pH}$ is the logarithm of the reciprocal of the concentration  of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in a solution.
•   $\text{pH}$ is used to determine the acidity or basicity of an aqueous solution.
• $\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$
• The point at which amount of standard solution and analyte becomes equal and neutralisation happens in titration is called equivalence point.
• $\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}$ is Henderson-Hasselbalch equation
• Buffer solution is defined as a solution that oppose changes in $\text{pH}$ while adding little amount of either an acid or a base.

To calculate: the $\text{pH}$ when $\text{5}\text{mL}$ $\text{KOH}$ is added.

Answer to Problem 17.35QP

pH = $4.56$

Explanation of Solution

Calculate the number of moles of ${\text{CH}}_{\text{3}}\text{COOH}$ and $\text{KOH}$

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{of}\text{acetic}\text{acid in 25 mL}\text{=}\text{25}\text{.0 mL}\text{×}\frac{\text{0}\text{.100 mol}}{\text{1000}\text{mL}}\\ \text{=}\text{2}\text{.50}\text{×}{\text{10}}^{\text{-3}}\text{mol}\\ \text{Number}\text{of}\text{moles}\text{of}\text{KOH in 5 mL}\text{=}\text{5}\text{.0 mL}\text{×}\frac{\text{0}\text{.200 mol}}{\text{1000}\text{mL}}\\ \text{=}\text{1}\text{.0}\text{×}{\text{10}}^{\text{-3}}\text{mol}\end{array}$

In this reaction one mole of acetic acid needs to neutralise one mole of $\text{KOH}$.  The number of moles of acetic acid can be calculated using volume and given concentration of the $\text{KOH}$.

Calculate the $\text{pH}$ when $\text{5}\text{mL}$ $\text{KOH}$ is added.

$\begin{array}{l}{\text{CH}}_{\text{3}}{\text{COOH}}_{\text{(aq)}}\text{ +}{\text{KOH}}_{\text{(aq)}}\to {\text{ CH}}_{\text{3}}{\text{COOK}}_{\text{(aq)}}\text{ +}{\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}\\ \\ \text{Initial concentration(M): }\text{2}\text{.5 ×}{\text{10}}^{\text{-3}}\text{1}\text{.0 ×}{\text{10}}^{\text{-3}}\text{0}\\ \text{Change in concentration (M):}-\text{1}\text{.0 ×}{\text{10}}^{\text{-3}}\text{-1}\text{.0 ×}{\text{10}}^{\text{-3}}\text{+1}\text{.0 ×}{\text{10}}^{\text{-3}}\\ \text{Final}\text{concentration (M): }\text{1}\text{.5 ×}{\text{10}}^{\text{-3}}\text{0}\text{+1}\text{.0 ×}{\text{10}}^{\text{-3}}\\ \\ \text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}\\ \text{pH}\text{=}\text{-log(1}\text{.8 ×}{\text{10}}^{\text{-5}}\text{)}\text{+}\text{log}\frac{\text{1}\text{.0 ×}{\text{10}}^{\text{-3}}}{\text{1}\text{.5 ×}{\text{10}}^{\text{-3}}}\\ \text{pH}\text{=}\text{4}\text{.56}\end{array}$

when $\text{KOH}$ is added to acetic acid, the volume of solution rises which varies the concentration of solution but the moles remain same.  the number of moles after addition of $\text{KOH}$ changes and is given in the table.  The Henderson-Hasselbalch equation can be used to calculate the $\text{pH}$ because the solution acts as buffer system.  From the equation and using acid dissociation constant, the $\text{pH}$ is calculated.

Interpretation Introduction

Interpretation:

The $\text{pH}$ of titration $\text{KOH}\text{Vs}{\text{CH}}_{\text{3}}\text{COOH}$ on adding various amount of potassium hydroxide has to be calculated.

Concept introduction:

• $\text{pH}$ is the logarithm of the reciprocal of the concentration  of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in a solution.
•   $\text{pH}$ is used to determine the acidity or basicity of an aqueous solution.
• $\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$
• The point at which amount of standard solution and analyte becomes equal and neutralisation happens in titration is called equivalence point.
• $\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}$ is Henderson-Hasselbalch equation
• Buffer solution is defined as a solution that oppose changes in $\text{pH}$ while adding little amount of either an acid or a base.

To find: the $\text{pH}$ when $10\text{mL}$ $\text{KOH}$ is added.

Answer to Problem 17.35QP

pH = $5.34$

Explanation of Solution

Find the number of moles of ${\text{CH}}_{\text{3}}\text{COOH}$ and $\text{KOH}$

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{of}\text{acetic}\text{acid in 25 mL}\text{=}\text{25}\text{.0 mL}\text{×}\frac{\text{0}\text{.100 mol}}{\text{1000}\text{mL}}\\ \text{=}\text{2}\text{.50}\text{×}{\text{10}}^{\text{-3}}\text{mol}\\ \text{Number}\text{of}\text{moles}\text{of}\text{KOH in 10 mL}\text{=}10.\text{0 mL}\text{×}\frac{\text{0}\text{.200 mol}}{\text{1000}\text{mL}}\\ \text{=}2.\text{00}\text{×}{\text{10}}^{\text{-3}}\text{mol}\end{array}$

In this reaction one mole of acetic acid needs to neutralise one mole of $\text{KOH}$.  The number of moles of acetic acid can be calculated using volume and given concentration of the $\text{KOH}$.

Find the $\text{pH}$ when $10\text{mL}$ $\text{KOH}$ is added

$\begin{array}{l}{\text{CH}}_{\text{3}}{\text{COOH}}_{\text{(aq)}}\text{ +}{\text{KOH}}_{\text{(aq)}}\to {\text{ CH}}_{\text{3}}{\text{COOK}}_{\text{(aq)}}\text{ +}{\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}\\ \\ \text{Initial concentration(M): }\text{2}\text{.5 ×}{\text{10}}^{\text{-3}}2.\text{0 ×}{\text{10}}^{\text{-3}}\text{0}\\ \text{Change in concentration (M):}-2.\text{0 ×}{\text{10}}^{\text{-3}}\text{-2}\text{.0 ×}{\text{10}}^{\text{-3}}\text{+2}\text{.0 ×}{\text{10}}^{\text{-3}}\\ \text{Final}\text{concentration (M): }0.500\text{ ×}{\text{10}}^{\text{-3}}\text{0}\text{+2}\text{.0 ×}{\text{10}}^{\text{-3}}\\ \\ \text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}\\ \text{pH}\text{=}\text{-log(1}\text{.8 ×}{\text{10}}^{\text{-5}}\text{)}\text{+}\text{log}\frac{\text{2}\text{.0 ×}{\text{10}}^{\text{-3}}}{0.500\text{ ×}{\text{10}}^{\text{-3}}}\\ \text{pH}\text{=}\text{5}\text{.34}\end{array}$

The number of moles after addition of $\text{KOH}$ changes and is given in the table.  The Henderson-Hasselbalch equation can be used to calculate the $\text{pH}$ because the solution acts as buffer system.  From the equation and using acid dissociation constant, the $\text{pH}$ is calculated.

Interpretation Introduction

Interpretation:

The $\text{pH}$ of titration $\text{KOH}\text{Vs}{\text{CH}}_{\text{3}}\text{COOH}$ on adding various amount of potassium hydroxide has to be calculated.

Concept introduction:

• $\text{pH}$ is the logarithm of the reciprocal of the concentration  of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in a solution.
•   $\text{pH}$ is used to determine the acidity or basicity of an aqueous solution.
• $\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$
• The point at which amount of standard solution and analyte becomes equal and neutralisation happens in titration is called equivalence point.
• $\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}$ is Henderson-Hasselbalch equation
• Buffer solution is defined as a solution that oppose changes in $\text{pH}$ while adding little amount of either an acid or a base.

To find: the $\text{pH}$ when $12.5\text{mL}$ $\text{KOH}$ is added.

Answer to Problem 17.35QP

pH = $8.78$

Explanation of Solution

$\begin{array}{l}{\text{M(CH}}_{\text{3}}{\text{COO}}^{\text{-}}\text{)}\text{=}\frac{\text{2}\text{.50}\text{×}{\text{10}}^{\text{-3}}}{\text{0}\text{.0375}\text{L}}\text{=}\text{0}\text{.0667}\text{M}\\ {\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftarrows {\text{ CH}}_{\text{3}}{\text{COOH}}_{\text{(aq)}}{\text{ + OH}}^{\text{-}}{}_{\text{(aq)}}\\ \\ \text{Initial concentration(M): }\text{0}\text{.0667 }\text{0}\text{0}\\ \text{Change in concentration (M):}\text{-x}\text{+x}\text{+x}\\ \text{Equilibrium}\text{concentration (M): }\text{0}\text{.0667-x}\text{x}\text{x}\\ \\ {\text{K}}_{\text{a}}\text{ value for HCOOH is 1}\text{.8 ×}{\text{10}}^{\text{-5}}\\ {\text{K}}_{\text{b}}\text{=}\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{a}}}\text{=}\frac{\text{1}\text{.0 ×}{\text{10}}^{\text{-14}}}{\text{1}\text{.8 ×}{\text{10}}^{\text{-5}}}\text{=}\text{5}\text{.56×}{\text{10}}^{\text{-10}}\\ {\text{K}}_{\text{b}}\text{=}\frac{{\text{[CH}}_{\text{3}}{\text{COOH][OH}}^{\text{-}}\text{]}}{{\text{[CH}}_{\text{3}}{\text{COO}}^{\text{-}}\text{]}}\\ \text{5}\text{.56×}{\text{10}}^{\text{-10}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.0667-x}}\\ \text{x}\text{is}\text{very}\text{small}\text{and}\text{neglect}\text{it,}\\ {\text{x = [OH}}^{\text{-}}\text{] = 6}\text{.09 ×}{\text{10}}^{\text{-6}}\text{M}\\ \text{pOH}\text{=}{\text{-log[OH}}^{\text{-}}\text{]}\\ \text{=}\text{-log(6}\text{.09 ×}{\text{10}}^{\text{-6}}\text{)}\\ \text{pOH}\text{=}\text{5}\text{.22}\\ \text{pH = 14}\text{.00 -5}\text{.220 = 8}\text{.78}\end{array}$

By adding $\text{12}\text{.5}\text{mL}\text{KOH}$, we reached equivalence point and equal number of moles of acetic acid $\text{2}\text{.50 }\times {10}^{-3}$ reacts with $\text{2}\text{.50 }\times {10}^{-3}$ moles of $\text{KOH}$.  At equivalence point $\text{2}\text{.50 }\times {10}^{-3}$ of potassium acetate present in solution.  The concentration of hydroxide can be calculated from the final concentration of hydroxide in equilibrium table using base dissociation constant.  The $\text{pOH}$ is determined by taking negative logarithm of  hydroxide ion concentration.  From the obtained $\text{pOH}$, the $\text{pH}$ is calculated.

Interpretation Introduction

Interpretation:

The $\text{pH}$ of titration $\text{KOH}\text{Vs}{\text{CH}}_{\text{3}}\text{COOH}$ on adding various amount of potassium hydroxide has to be calculated.

Concept introduction:

• $\text{pH}$ is the logarithm of the reciprocal of the concentration  of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in a solution.
•   $\text{pH}$ is used to determine the acidity or basicity of an aqueous solution.
• $\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$
• The point at which amount of standard solution and analyte becomes equal and neutralisation happens in titration is called equivalence point.
• $\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}$ is Henderson-Hasselbalch equation
• Buffer solution is defined as a solution that oppose changes in $\text{pH}$ while adding little amount of either an acid or a base.

To find: the $\text{pH}$ when $15\text{mL}$ $\text{KOH}$ is added.

Answer to Problem 17.35QP

The $\text{pH}$ of titration $\text{KOH}\text{Vs}{\text{CH}}_{\text{3}}\text{COOH}$

pH = $12.10$

Explanation of Solution

Find the number of moles of $\text{KOH}$

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{of}\text{KOH in 15 mL}\text{=}15.\text{0 mL}\text{×}\frac{\text{0}\text{.200 mol}}{\text{1000}\text{mL}}\\ \text{=}3.\text{00}\text{×}{\text{10}}^{\text{-3}}\text{mol}\\ \end{array}$

The number of moles of $\text{KOH}$ can be calculated using volume and given concentration of the $\text{KOH}$.

Find the $\text{pH}$ when $15\text{mL}$ $\text{KOH}$ is added.

$\begin{array}{l}{\text{CH}}_{\text{3}}{\text{COOH}}_{\text{(aq)}}\text{ +}{\text{KOH}}_{\text{(aq)}}\to {\text{ CH}}_{\text{3}}{\text{COOK}}_{\text{(aq)}}\text{ +}{\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}\\ \\ \text{Initial concentration(M): }\text{2}\text{.5 ×}{\text{10}}^{\text{-3}}\text{3}\text{.0 ×}{\text{10}}^{\text{-3}}\text{0}\\ \text{Change in concentration (M):}\text{-2}\text{.50 ×}{\text{10}}^{\text{-3}}\text{-2}\text{.50 ×}{\text{10}}^{\text{-3}}\text{+2}\text{.50 ×}{\text{10}}^{\text{-3}}\\ \text{Final}\text{concentration (M): }\text{0}\text{0}\text{.500 ×}{\text{10}}^{\text{-3}}\text{+2}\text{.50 ×}{\text{10}}^{\text{-3}}\\ \\ \text{The}\text{molarity}\text{of}\text{KOH}\text{in}\text{0}\text{.0400L}\text{after}\text{reacting}\\ \text{M(KOH)}\text{=}\frac{\text{0}\text{.500}\text{×}{\text{10}}^{\text{-3}}}{\text{0}\text{.0400 L}}\text{=}\text{0}\text{.0125}\text{M}\\ \text{pOH =}\text{-log(0}\text{.0125) = 1}\text{.90}\\ \text{pH = 12}\text{.10}\end{array}$

The number of moles after addition of $\text{KOH}$ changes and is given in the table.  By calculating the strength of potassium hydroxide, the $\text{pH}$ of the solution determined.