Chapter 6, Problem 6.43P

Interpretation Introduction

Interpretation:

The pH of the ${\text{Cl}}_{\text{3}}{\text{CCO}}_{\text{2}}\text{H}$ solution at $\text{50\%}$, $\text{90\%}$, and $\text{10\%}$ in water is to be calculated.

Concept introduction:

In an aqueous solution, an acid is protonated by water and exists in equilibrium with the protonated form, and the relative concentrations of the two species at equilibrium are determined by the pH of the solution. In an acid equilibrium reaction with water, ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ and ${\text{A}}^{\text{-}}$ are the products of the reaction. The concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ directly affects the relative amounts of the initial concentration of $\text{HA}$ at equilibrium. The change in concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ will cause an equilibrium shift to either side of the reaction. At the new equilibrium, the amount of $\text{HA}$ and ${\text{A}}^{\text{-}}$ is calculated by the percent dissociation formula. The expression of percent dissociation of $\text{HA}$ is shown below:

$\text{\% Dissociation = }\frac{{{\text{[A}}^{\text{-}}\text{]}}_{\text{eq}}}{{\text{[HA]}}_{\text{initial}}}$

The pH of the acid solution is calculated from the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in solution by using the Henderson–Hasselbalch equation shown below:

$\text{pH = pKa + log}\left(\frac{{{\text{[A}}^{\text{-}}\text{]}}_{\text{eq}}}{{\text{[HA]}}_{\text{initial}}}\right)$

Answer to Problem 6.43P

1. At $\text{50\%}$ the calculated pH of the ${\text{Cl}}_{\text{3}}{\text{CCO}}_{\text{2}}\text{H}$ solution is $\text{0}\text{.77}$.
2. At $\text{90\%}$ the calculated pH of the ${\text{Cl}}_{\text{3}}{\text{CCO}}_{\text{2}}\text{H}$ solution is $1.72$.
3. At $\text{10\%}$ the calculated pH of the ${\text{Cl}}_{\text{3}}{\text{CCO}}_{\text{2}}\text{H}$ solution is $-0.23$.

Explanation of Solution

The general reaction between acid, $\text{HA}$, with water is

${\text{HA + H}}_{\text{2}}\text{O }\rightleftharpoons {\text{ A}}^{\text{- }}\text{(aq)}\text{}{\text{ + H}}_{\text{3}}{\text{O}}^{\text{+}}\text{ (aq)}$

The pH of the acid solution is calculated from the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in solution.

$\text{pH}=\text{-}\text{}{\text{log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$

1. The pH of the ${\text{Cl}}_{\text{3}}{\text{CCO}}_{\text{2}}\text{H}$ solution at $\text{50\%}$:

The $\text{pKa}$ of the ${\text{Cl}}_{\text{3}}{\text{CCO}}_{\text{2}}\text{H}$ solution is $\text{0}\text{.77}$. At $\text{50\%}$ dissociation of acid, the equilibrium concentrations of $\text{HA}$ and ${\text{A}}^{\text{-}}$ are equal. The pH at which the two species have equal concentrations can be calculated by substituting $1$ for the ratio in the second term of Henderson-Hasselbalch equation.

$\text{pH = pKa + log}\left(\frac{{{\text{[A}}^{\text{-}}\text{]}}_{\text{eq}}}{{\text{[HA]}}_{\text{initial}}}\right)$

$\begin{array}{l}\text{pH = 0}\text{.77 + log}(1)\\ \text{pH = 0}\text{.77 + }0\\ \text{pH = 0}\text{.77}\end{array}$

So, at $\text{50\%}$ the calculated pH of the ${\text{Cl}}_{\text{3}}{\text{CCO}}_{\text{2}}\text{H}$ solution is $\text{0}\text{.77}$.

2. The pH of the ${\text{Cl}}_{\text{3}}{\text{CCO}}_{\text{2}}\text{H}$ solution at $\text{90\%}$:

Assume the initial concentration of the solution as $100$. In acid equilibrium reaction, the ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ and ${\text{A}}^{\text{-}}$ are dissociated $\text{90\%}$ from the initial concentration.

The $\text{pKa}$ of the ${\text{Cl}}_{\text{3}}{\text{CCO}}_{\text{2}}\text{H}$ solution is $\text{0}\text{.77}$. The ratio of the concentrations of the protonated $\text{HA}$ form and the deprotonated ${\text{A}}^{\text{-}}$ form are $100:90$ for the condition. Therefore, the pH is calculated using the Henderson-Hasselbalch equation.

$\text{pH = pKa + log}\left(\frac{{{\text{[A}}^{\text{-}}\text{]}}_{\text{eq}}}{{\text{[HA]}}_{\text{initial}}}\right)$

$\begin{array}{l}\text{pH = 0}\text{.77 + log}(\frac{90}{100})\\ \text{pH = 0}\text{.77 -0}\text{.04576}\\ \text{pH = 0}\text{.72}\end{array}$

So, at $\text{90\%}$ the calculated pH of the ${\text{Cl}}_{\text{3}}{\text{CCO}}_{\text{2}}\text{H}$ solution is $1.72$.

3. The pH of the ${\text{Cl}}_{\text{3}}{\text{CCO}}_{\text{2}}\text{H}$ solution at $\text{10\%}$:

Assume the initial concentration of the solution as $100$. In acid equilibrium reaction, ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ and ${\text{A}}^{\text{-}}$ are dissociated $\text{10\%}$ from the initial concentration.

The $\text{pKa}$ of the ${\text{Cl}}_{\text{3}}{\text{CCO}}_{\text{2}}\text{H}$ solution is $\text{0}\text{.77}$. The ratio of the concentrations of the protonated $\text{HA}$ form and the deprotonated ${\text{A}}^{\text{-}}$ form are $100:10$ for the condition. Therefore, the pH is calculated using the Henderson-Hasselbalch equation.

$\text{pH = pKa + log}\left(\frac{{{\text{[A}}^{\text{-}}\text{]}}_{\text{eq}}}{{\text{[HA]}}_{\text{initial}}}\right)$

$\begin{array}{l}\text{pH = 0}\text{.77 + log}(\frac{10}{100})\\ \text{pH = 0}\text{.77 -1}\\ \text{pH = -0}\text{.23}\end{array}$

So, at $\text{10\%}$ the calculated pH of the ${\text{Cl}}_{\text{3}}{\text{CCO}}_{\text{2}}\text{H}$ solution is $-0.23$.

Conclusion

The pH of the ${\text{Cl}}_{\text{3}}{\text{CCO}}_{\text{2}}\text{H}$ solution at $\text{50\%}$, $\text{90\%}$, and $\text{10\%}$ in water is calculated.