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.

Interpretation Introduction

Interpretation:

The $\text{pH}$ of titration $\text{NaOH}\text{Vs}{\text{CH}}_{\text{3}}\text{COOH}$ on adding various amount of sodium 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.
• 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.
• Buffer solution is a combination of a weak acid and its conjugate base or a weak base and its conjugate acid.

To calculate: The $\text{pH}$ when $\text{10mL}$ $\text{NaOH}$ is added.

Answer to Problem 4PPA

$\text{pH}\text{=}5.04$

Explanation of Solution

The $\text{pH}$ of buffer solution when $\text{10mL}$ $\text{NaOH}$ is added is calculated below.

$\begin{array}{l}\text{The}\text{number}\text{of}\text{moles}\text{of}\text{acetic}\text{acid}\text{is}\\ \\ \text{(0}\text{.15 mmol/mL)(10}\text{.0 mL) = 1}\text{.50 mmol of acetic acid}\\ \\ \text{The}\text{number}\text{of}\text{moles}\text{of}\text{NaOH}\text{is}\\ \\ \text{(0}\text{.10 mmol/mL)(10}\text{.0 mL) = 1}\text{.00 mmol of NaOH}\end{array}$

$\begin{array}{l}{\text{CH}}_{\text{3}}{\text{COOH}}_{\text{(aq)}}{\text{ +OH}}^{\text{-}}{}_{\text{(aq)}}\stackrel{\rightleftarrows }{\to }{\text{ H}}_2{\text{O}}_{\text{(l)}}\text{ +}{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{\text{(aq)}}\\ \\ \text{Initial concentration(M): }1.\text{50mmol }\text{0}\\ \text{Change in concentration (M):}\text{-1}\text{.00mmol}\text{+1}\text{.00mmol}\\ \text{Equilibrium}\text{concentration (M): }0.\text{50mmol}\text{1}\text{.00mmol}\\ \\ \text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}\\ \\ \text{pH}\text{=}4.74\text{+}\text{log}\frac{\text{1}\text{.00}}{0.50}\\ \\ \text{pH}\text{=}4.74+0.30\\ \\ \text{pH}\text{=}5.04\end{array}$

Interpretation Introduction

Interpretation:

The $\text{pH}$ of titration $\text{NaOH}\text{Vs}{\text{CH}}_{\text{3}}\text{COOH}$ on adding various amount of sodium 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.
• 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.
• Buffer solution is a combination of a weak acid and its conjugate base or a weak base and its conjugate acid.

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

Answer to Problem 4PPA

$\text{pH = 8}\text{.87}$

Explanation of Solution

The $\text{pH}$ of buffer solution when $\text{15mL}$ $\text{NaOH}$ is added is calculated below.

$\begin{array}{l}\text{The}\text{number}\text{of}\text{moles}\text{of}\text{acetic}\text{acid}\text{is}\\ \\ \text{(0}\text{.15 mmol/mL)(10}\text{.0 mL) = 1}\text{.50 mmol of acetic acid}\\ \\ \text{The}\text{number}\text{of}\text{moles}\text{of}\text{NaOH}\text{is}\\ \\ \text{(0}\text{.10 mmol/mL)(15}\text{.0 mL) = 1}\text{.50 mmol of NaOH}\end{array}$

At equivalence point, the total volume of solution is

$\text{10}\text{.0mL(acetic acid) + 15}\text{.0 mL(base) = 25}\text{.0 mL}$

At equivalence point, the concentration of acetate ion is

$\frac{\text{1}\text{.50mmol}\text{acetate}\text{ion}}{\text{15}\text{.0mL}}\text{=}\text{0}\text{.1M}$

$\begin{array}{l}{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{\text{(aq)}}{\text{ +H}}_{\text{2}}{\text{O}}_{\text{(l)}}\stackrel{\rightleftarrows }{\to }{\text{ OH}}^{\text{-}}{}_{\text{(aq)}}\text{ +}{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{\text{(aq)}}\\ \\ \text{Initial concentration(M): }\text{0}\text{.1 }\text{0}\text{0}\\ \text{Change in concentration (M):}\text{-x}\text{-x}\text{+x}\\ \text{Equilibrium}\text{concentration (M): }\text{0}\text{.1-x}\text{x}\text{x}\\ \\ {\text{K}}_{\text{b}}\text{value for acetate ion is 5}\text{.6 ×}{\text{10}}^{\text{-10}}\\ \\ {\text{K}}_{\text{b}}\text{=}\frac{{\text{[OH}}^{\text{-}}{\text{][CH}}_{\text{3}}\text{COOH]}}{{\text{[CH}}_{\text{3}}{\text{COO}}^{\text{-}}\text{]}}\\ \text{5}\text{.6 ×}{\text{10}}^{\text{-10}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{(0}\text{.1-x)}}\\ \text{x}\text{is}\text{very}\text{small}\text{and}\text{neglect}\text{it,}\\ {\text{x}}^{\text{2}}\text{=}\text{5}\text{.6 ×}{\text{10}}^{\text{-11}}\\ \text{x}\text{=}\text{7}\text{.48}\text{×}{\text{10}}^{\text{-6}}\text{M}\\ {\text{x = [OH}}^{\text{-}}\text{] = 7}\text{.48}\text{×}{\text{10}}^{\text{-6}}\text{M}\\ \text{pOH}\text{=}{\text{-log[OH}}^{\text{-}}\text{]}\\ \text{=}\text{-log(7}\text{.48}\text{×}{\text{10}}^{\text{-6}}\text{)}\\ \text{pOH}\text{=}\text{5}\text{.13}\\ \text{pH = 14}\text{.00 - 5}\text{.13 = 8}\text{.87}\end{array}$

Interpretation Introduction

Interpretation:

The $\text{pH}$ of titration $\text{NaOH}\text{Vs}{\text{CH}}_{\text{3}}\text{COOH}$ on adding various amount of sodium 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.
• 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.
• Buffer solution is a combination of a weak acid and its conjugate base or a weak base and its conjugate acid.

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

Answer to Problem 4PPA

$\text{pH = 12}\text{.82}$

Explanation of Solution

The $\text{pH}$ of buffer solution when $\text{20mL}$ $\text{NaOH}$ is added is calculated below.

$\begin{array}{l}\text{The}\text{number}\text{of}\text{moles}\text{of}\text{NaOH}\text{is}\\ \\ \text{(0}\text{.10 mmol/mL)(20}\text{.0 mL) = 2}\text{.0 mmol of NaOH}\\ \\ \text{Total volume of solution is }\\ \\ \text{10}\text{.0 + 20}\text{.0 = 35}\text{.0 mL}\end{array}$

The concentration of  sodium hydroxide is

$\begin{array}{l}{\text{[OH}}^{\text{-}}\text{] = 2}\text{.00 mmol/30}\text{.0 mL = 0}\text{.0666 M}\\ \\ \text{pOH = -log (0}\text{.0666) = 1}\text{.24}\\ \\ \text{pH = 14}\text{.000 - 1}\text{.176 = 12}\text{.82}\end{array}$