Chapter 17, Problem 17.32QP

Interpretation Introduction

Interpretation:

The equilibrium concentrations of ${\text{H}}^{\text{+}}{\text{, CH}}_{\text{3}}{\text{COOH, CH}}_{\text{3}}{\text{COO}}^{\text{-}}{\text{, OH}}^{\text{-}}{\text{, and Na}}^{\text{+}}$ has to be calculated.

Concept introduction:

• 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.
• Base dissociation constant ${\text{K}}_{\text{b}}$ is defined as measure of base strength in a given solution
• ${\text{K}}_{\text{b}}\text{=}\frac{{\text{[OH}}^{\text{-}}{\text{][HB}}^{\text{+}}\text{]}}{\text{[B]}}$

To calculate: the concentration of acetic acid and ${\text{H}}^{\text{+}}{\text{, CH}}_{\text{3}}{\text{COO}}^{\text{-}}{\text{, OH}}^{\text{-}}{\text{, and Na}}^{\text{+}}$

Answer to Problem 17.32QP

$\begin{array}{l}[H^{+}]is3.0\times 10^{-13}M\\ [C{H}_{3}COOH]is8.4\times 10^{-10}M\\ [C{H}_{3}CO{O}^{-}]is0.0500M\\ [O{H}^{-}]is0.0335M\\ [N{a}^{+}]is0.0835M\end{array}$

Explanation of Solution

Concentration of acetic acid is $\text{0}\text{.100}\text{M}$

Concentration of $\text{NaOH}$ is $0.167\text{M}$

Calculate the number of moles of acetic acid and $\text{NaOH}$

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{of}\text{NaOH}\text{=}\text{0}\text{.500}\text{L}\text{×}\frac{\text{0}\text{.167}\text{mol}}{\text{1}\text{L}}\\ \text{=}\text{0}\text{.0835}\text{mol}\\ \text{Number}\text{of}\text{moles}\text{of}\text{acetic}\text{acid}\text{=}\text{0}\text{.500}\text{L}\text{×}\frac{\text{0}\text{.100}\text{mol}}{\text{1}\text{L}}\\ \text{=}\text{0}\text{.0500}\text{mol}\end{array}$

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

Calculate the concentration of ${\text{H}}^{\text{+}}{\text{, CH}}_{\text{3}}{\text{COO}}^{\text{-}}{\text{, OH}}^{\text{-}}{\text{, and Na}}^{\text{+}}$

$\begin{array}{l}{\text{CH}}_{\text{3}}{\text{COOH}}_{\text{(aq)}}\text{+}{\text{NaOH}}_{\text{(aq)}}\to {\text{CH}}_{\text{3}}{\text{COONa}}_{\text{(aq)}}\text{+}{\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}\\ \\ \text{Initial concentration(M): }\text{0}\text{.0500 }\text{0}\text{.0835}\text{0}\\ \text{Change in concentration (M):}\text{-0}\text{.0500}\text{-0}\text{.0500}\text{+0}\text{.0500}\\ \text{Final}\text{concentration (M): }\text{0 }\text{0}\text{.0355}\text{0}\text{.0500}\\ \text{The volume of solution is 1}\text{.00 L}\\ {\text{[OH}}^{\text{-}}\text{]}\text{=}\frac{\text{0}\text{.0335}\text{mol}}{\text{1}\text{.00}\text{L}}\text{=}\text{0}\text{.0335}\text{M}\\ {\text{[Na}}^{\text{+}}\text{]}\text{=}\frac{\text{(0}\text{.0500+0}\text{.0335)}\text{mol}}{\text{1}\text{.00}\text{L}}\text{=}\text{0}\text{.0835}\text{M}\\ {\text{[H}}^{\text{+}}\text{]}\text{=}\frac{{\text{K}}_{\text{w}}}{{\text{[OH}}^{\text{-}}\text{]}}=\frac{\text{1}\text{.0}\text{×}{\text{10}}^{\text{-14}}}{\text{0}\text{.0335}}\text{=}\text{3}\text{.0}\text{×}{\text{10}}^{\text{-13}}\text{M}\\ {\text{[CH}}_{\text{3}}{\text{COO}}^{\text{-}}\text{]}\text{=}\frac{\text{0}\text{.0500}\text{mol}}{\text{1}\text{.00}\text{L}}\text{=}\text{0}\text{.0500}\text{M}\end{array}$

The concentrations of ${\text{H}}^{\text{+}}{\text{, CH}}_{\text{3}}{\text{COOH, CH}}_{\text{3}}{\text{COO}}^{\text{-}}{\text{, OH}}^{\text{-}}{\text{, and Na}}^{\text{+}}$ can be infer from the reaction table.  the obtained concentrations should be converted to one litre solution on dividing concentrations by one litre.

Calculate the concentration of acetic acid.

$\begin{array}{l}{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{\text{(aq)}}\text{+}{\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}\to {\text{CH}}_{\text{3}}{\text{COOH}}_{\text{(aq)}}\text{+}{\text{OH}}^{\text{-}}{}_{\text{(aq)}}\\ \\ \text{Initial concentration(M): }\text{0}\text{.0500 }\text{0}\text{0}\text{.0335}\\ \text{Change in concentration (M):}\text{-x }\text{+x}\text{+x}\\ \text{Final}\text{concentration (M): }\text{0}\text{.0500-x }\text{x}\text{0}\text{.0335+x}\\ {\text{K}}_{\text{b}}\text{=}\frac{{\text{[CH}}_{\text{3}}{\text{COOH][OH}}^{\text{-}}\text{]}}{{\text{[CH3COO}}^{\text{-}}\text{]}}\\ \text{5}\text{.6}\text{×}{\text{10}}^{\text{-10}}\text{=}\frac{\text{( x)(0}\text{.0335+x )}}{\text{(0}\text{.0500-x)}}\\ \text{x}\text{is}\text{very}\text{small}\text{and}\text{neglect}\text{it,}\\ \text{=}\frac{\text{(x)(0}\text{.0335 )}}{\text{(0}\text{.0500)}}\\ \text{x}\text{=}{\text{[CH}}_{\text{3}}\text{COOH]}\text{=}\text{8}\text{.4}\text{×}{\text{10}}^{\text{-10}}\text{M}\end{array}$

The concentration of acetic acid can be calculated by using base dissociation constant.  By substituting concentrations from equilibrium table and doing mathematical operations the strength of acetic acid can be found.

Conclusion

The equilibrium concentrations of ${\text{H}}^{\text{+}}{\text{, CH}}_{\text{3}}{\text{COOH, CH}}_{\text{3}}{\text{COO}}^{\text{-}}{\text{, OH}}^{\text{-}}{\text{, and Na}}^{\text{+}}$ was calculated

$\begin{array}{l}[H^{+}]is3.0\times 10^{-13}M\\ [C{H}_{3}COOH]is8.4\times 10^{-10}M\\ [C{H}_{3}CO{O}^{-}]is0.0500M\\ [O{H}^{-}]is0.0335M\\ [N{a}^{+}]is0.0835M\end{array}$