Chapter 18, Problem 71E

Interpretation Introduction

Interpretation:

The $\text{pH}$ of $\text{0}\text{.35}\text{M}{\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ 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. Some ions are present as minor species. The extent of the ionization is determined by the ionization constant.

Answer to Problem 71E

The $\text{pH}$ of $\text{0}\text{.35}\text{M}{\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ is $2.60$.

Explanation of Solution

The equation for the dissociation ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ 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 given value of ${\text{K}}_{\text{a}}$ is $1.8\times {10}^{-5}$.

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

${\text{K}}_{\text{a}}=\frac{\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{}^{-}\right]\left[{\text{H}}^{+}\right]}{\left[{\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right]}$…(1)

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 $\text{0}\text{.35}\text{M}$.

The concentration of the reactant and products before and after ionization can be tabulated as shown below.

$\begin{array}{l}{\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\text{(aq)}\to {\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{}^{-}\text{(aq)}+{\text{H}}^{+}\text{(aq)}\\ \text{Initial}\text{0}\text{.35}\text{M}\text{ 0}\text{0}\\ \text{Change }-x\text{+}x+x\\ \text{Equilibrium:}\left(\text{0}\text{.35}-x\right)\text{M}x\text{ M}x\text{ M}\end{array}$

Where,

• $x$ is the concentration of the amount of dissociated acid.

Substitute the values in equation (1).

${\text{K}}_{\text{a}}=\frac{x\times x}{\left(\text{0}\text{.35}-x\right)}$…(2)

As the acid is weak, the dissociation is less and the value of $x$ will be very less than $0.35$.

So, equation (2) can be written as shown below.

$\begin{array}{rll}{\text{K}}_{\text{a}}& =& \frac{x\times x}{\text{0}\text{.35}}\\ x^2& =& \text{0}\text{.35}\times {\text{K}}_{\text{a}}\\ x& =& \sqrt{\text{0}\text{.35}\times {\text{K}}_{\text{a}}}\end{array}$…(3)

Substitute the value of ${\text{K}}_{\text{a}}$ in equation (3).

$\begin{array}{rll}x& =& \sqrt{\text{0}\text{.35}\times 1.8\times {10}^{-5}}\\ & =& 2.5\times {10}^{-3}\text{M}\end{array}$

For the reaction given above, the concentration of hydrogen ions, $\left[{\text{H}}^{+}\right]$ is equal to that of the amount dissociated.

$\left[{\text{H}}^{+}\right]=2.5\times {10}^{-3}\text{M}$

The formula to calculate 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).

$\begin{array}{rll}\text{pH}& =& -\log \left[{\text{H}}^{\text{+}}\right]\\ \text{pH}& =& -\log \left(2.5\times {10}^{-3}\text{M}\right)\\ & =& 2.60\end{array}$

Therefore, the $\text{pH}$ of $\text{0}\text{.35}\text{M}{\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ is $2.60$.

Conclusion

The $\text{pH}$ of $\text{0}\text{.35}\text{M}{\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ is $2.60$.