Chapter 18, Problem 73E

Interpretation Introduction

Interpretation:

The $\text{pH}$ of $0.12\text{M}{\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ is to be calculated.

Concept introduction:

A strong acid easily releases protons to a base or when dissolved in water. It readily participates in an acid-base reaction. A weak acid undergoes only slight ionization in solution. It does not release protons easily. The major species present in a solution of weak acid is the unionized molecular species. However, some ions are present as minor species. The extent of the ionization is determined by the ionization constant.

Answer to Problem 73E

The $\text{pH}$ of $0.12\text{M}{\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ is $5.43$.

Explanation of Solution

The formula to calculate the number of moles of ${\text{NaC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ is given below.

$\text{Number}\text{of}\text{moles}\text{of}{\text{NaC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}=\frac{\text{Mass}\text{of}{\text{NaC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}}{\text{Molar}\text{mass}\text{of}{\text{NaC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}}$…(1)

The mass of ${\text{NaC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ is $\text{24}\text{.0}\text{g}$.

The molar mass of ${\text{NaC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ is $82.03\text{g}{\text{mol}}^{-1}$.

Substitute the mass and molar mass of ${\text{NaC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ in equation (1).

$\begin{array}{rll}\text{Number}\text{of}\text{moles}\text{of}{\text{NaC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}& =& \frac{\text{24}\text{.0}\text{g}}{82.03\text{g}{\text{mol}}^{-1}}\\ & =& 0.293\text{mol}\end{array}$

The volume of the solution is $5.00\times {10}^2\text{mL}$.

The relation between $\text{L}$ and $\text{mL}$ is given below.

$\text{1}\text{L}=\text{1000}\text{mL}$

The probable conversion factors are given below.

$\frac{\text{1}\text{L}}{\text{1000}\text{mL}}\text{and}\frac{\text{1000}\text{mL}}{\text{1}\text{L}}$

The conversion factor to determine $\text{L}$ from $\text{mL}$ is given below.

$\frac{\text{1}\text{L}}{\text{1000}\text{mL}}$

Therefore, the volume in liters is calculated below.

$\begin{array}{rll}\text{Volume}& =& 5.00\times {10}^2\text{mL}\times \frac{\text{1}\text{L}}{\text{1000}\text{mL}}\\ & =& \text{0}\text{.50}\text{L}\end{array}$

The molarity of a solution is calculated by the formula given below.

$\text{Molarity}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{liters}}$…(2)

The number of moles of ${\text{NaC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ is $0.293\text{mol}$.

The volume of solution is $\text{0}\text{.50}\text{L}$.

Substitute the number of moles and volume of solution in equation (2).

$\begin{array}{rll}\text{Molarity}& =& \frac{0.293\text{mol}}{\text{0}\text{.50}\text{L}}\\ & =& \text{0}\text{.586}\text{mol}/\text{L}\end{array}$

The relation between $\text{M}$ and $\text{mol}/\text{L}$ as shown below.

$1\text{M}=1\text{mol}/\text{L}$

The probable conversion factors are given below.

$\frac{1\text{M}}{1\text{mol}/\text{L}}\text{and}\frac{1\text{mol}/\text{L}}{1\text{M}}$

The conversion factor to determine $\text{M}$ from $\text{mol}/\text{L}$ is given below.

$\frac{1\text{M}}{1\text{mol}/\text{L}}$

So, $\text{0}\text{.586}\text{mol}/\text{L}$ can be written as shown below.

$\begin{array}{rll}\text{Molarity}& =& \text{0}\text{.586}\text{mol}/\text{L}\times \frac{1\text{M}}{1\text{mol}/\text{L}}\\ & =& \text{0}\text{.586}\text{M}\end{array}$

The equation for the dissociation of ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ is shown below.

${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\text{(aq)}\to {\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{}^{-}\text{(aq)}+{\text{H}}^{+}\text{(aq)}$

The dissociation constant, ${\text{K}}_{\text{a}}$ for the above reaction is given below.

$\begin{array}{l}{\text{K}}_{\text{a}}=\frac{\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{}^{-}\right]\left[{\text{H}}^{+}\right]}{\left[{\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right]}\\ \end{array}$…(3)

The concentration of ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$, $\left[{\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right]$ is $0.12$.

The concentration of ${\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{}^{-}$, $\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{}^{-}\right]$ is $\text{0}\text{.586}\text{M}$.

The value of ${\text{K}}_{\text{a}}$ is $1.8\times {10}^{-5}$.

Substitute the value of ${\text{K}}_{\text{a}}$, $\left[{\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right]$ and $\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{}^{-}\right]$ in equation (3).

$\begin{array}{rll}1.8\times {10}^{-5}& =& \frac{\text{0}\text{.586}\text{M}\times \left[{\text{H}}^{+}\right]}{\text{0}\text{.12}\text{M}}\\ \left[{\text{H}}^{+}\right]& =& \frac{\text{0}\text{.12}\text{M}\times 1.8\times {10}^{-5}\text{M}}{\text{0}\text{.586}\text{M}}\\ & =& 3.7\times {10}^{-6}\text{M}\end{array}$

The formula to the $\text{pH}$ is given below.

$\text{pH}=-\log \left[{\text{H}}^{\text{+}}\right]$…(4)

Substitute the value of $\left[{\text{H}}^{\text{+}}\right]$ in equation (4).

$\begin{array}{rll}\text{pH}& =& -\log \left[{\text{H}}^{\text{+}}\right]\\ \text{pH}& =& -\log \left(3.7\times {10}^{-6}\text{M}\right)\\ & =& 5.43\end{array}$

Therefore, the $\text{pH}$ of $0.12\text{M}{\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ is $5.43$.

Conclusion

The $\text{pH}$ of $0.12\text{M}{\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ is $5.43$.