Question

Distribution curves show how the fractions of a nonionized acid and its conjugate base vary as a function of the $\mathrm{pH}$ of the medium. Plot distribution curves for ${\mathrm{CH}}_3\mathrm{COOH}$ and its conjugate base ${\mathrm{CH}}_3{\mathrm{COO}}^{-}$ in solution. Your graph should show fraction as the $y$ axis and $\mathrm{pH}$ as the $x$ axis. What are the fractions and $\mathrm{pH}$ at the point where these two curves intersect?

Short answer

The intersection occurs at pH 4.76, with both fractions equal to 0.5.

Step-by-step solution

Understand the Chemical Equilibrium

The fraction of non-ionized acid and its conjugate base in solution depends on the pH. The equation for chemical equilibrium in this context is $${\text{CH}}_3\text{COOH}\rightleftharpoons {\text{CH}}_3{\text{COO}}^{-}+{\text{H}}^{+}$$The acid dissociation constant $K_a$ is used to describe the equilibrium.

Use the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is used to relate pH, pKa, and the ratio of concentrations: $$\text{pH}=\text{pKa}+\log \left(\frac{[{\text{A}}^{-}]}{[\text{HA}]}\right)$$For acetic acid ${\text{CH}}_3\text{COOH}$, $\text{pKa}\approx 4.76$.

Calculate Fraction of Non-ionized and Ionized Forms

Let $[\text{HA}]$ be the concentration of ${\text{CH}}_3\text{COOH}$, and $[{\text{A}}^{-}]$ be the concentration of ${\text{CH}}_3{\text{COO}}^{-}$.The fraction of ${\text{CH}}_3\text{COOH}$ is: $$\text{Fraction of }{\text{CH}}_3\text{COOH}=\frac{[\text{HA}]}{[\text{HA}]+[{\text{A}}^{-}]}$$The fraction of ${\text{CH}}_3{\text{COO}}^{-}$ is:$$\text{Fraction of }{\text{CH}}_3{\text{COO}}^{-}=\frac{[{\text{A}}^{-}]}{[\text{HA}]+[{\text{A}}^{-}]}$$

Simplify Using Henderson-Hasselbalch

Using the Henderson-Hasselbalch equation, express the fraction equations in terms of pH and pKa:$$[{\text{A}}^{-}]/[\text{HA}]={10}^{(\text{pH}-\text{pKa})}$$Thus,$$\text{Fraction of }{\text{CH}}_3\text{COOH}=\frac{1}{1+{10}^{(\text{pH}-\text{pKa})}}$$$$\text{Fraction of }{\text{CH}}_3{\text{COO}}^{-}=\frac{{10}^{(\text{pH}-\text{pKa})}}{1+{10}^{(\text{pH}-\text{pKa})}}$$

Plot the Distribution Curves

Create a plot with pH on the x-axis ranging from about 2 to 12. Calculate fractions for each pH value using the formulas from the previous step. Plot the fraction of ${\text{CH}}_3\text{COOH}$ and ${\text{CH}}_3{\text{COO}}^{-}$ on the y-axis.

Find the Intersection Point

The intersection of the curves occurs where both fractions are equal, i.e.,$$\frac{1}{1+{10}^{(\text{pH}-\text{pKa})}}=\frac{{10}^{(\text{pH}-\text{pKa})}}{1+{10}^{(\text{pH}-\text{pKa})}}$$Solving, $$1={10}^{(\text{pH}-\text{pKa})}$$Thus, $\text{pH}=\text{pKa}=4.76$. At this point, both fractions are 0.5.

Key concepts

Chemical Equilibrium

Chemical equilibrium is a foundational concept in chemistry. It describes a state where the concentrations of reactants and products remain constant over time. This occurs when a chemical reaction proceeds at the same rate in both forward and reverse directions.
In the context of the given acetic acid system, the reaction is $${\text{CH}}_3\text{COOH}\rightleftharpoons {\text{CH}}_3{\text{COO}}^{-}+{\text{H}}^{+}$$
This means acetic acid, ${\text{CH}}_3\text{COOH}$, can dissociate into acetate ions ${\text{CH}}_3{\text{COO}}^{-}$ and hydrogen ions ${\text{H}}^{+}$. At equilibrium, the formation of these products happens at the same rate as they recombine to form acetic acid.

• The importance of equilibrium lies in understanding how acids and bases behave in solutions and how changes in conditions (like pH) affect these systems.
• This equilibrium concept is crucial for predicting the proportion between the non-ionized acid and its conjugate base.

Understanding this balance helps in visualizing how molecules distribute in a solution, and ties directly back to the fascinating distribution curves you may plot in the exercise.

Acid Dissociation Constant

The Acid Dissociation Constant, or $K_a$, is a specific equilibrium constant for the dissociation of an acid in aqueous solution. It provides a quantitative measure of the strength of an acid in a solution. For acetic acid:$$K_a=\frac{[{\text{CH}}_3{\text{COO}}^{-}][{\text{H}}^{+}]}{[{\text{CH}}_3\text{COOH}]}$$
Let's break it down:

• A larger $K_a$ value indicates a stronger acid, which ionizes more in solution.
• For acetic acid, $K_a\approx 1.8\times {10}^{-5}$, is relatively small, meaning it's a weak acid.

Another way to represent $K_a$ is through $\text{pKa}$, calculated as:
$$\text{pKa}=-\log (K_a)\approx 4.76$$
This equation helps to relate the strength of an acid to the pH of the solution using the Henderson-Hasselbalch equation. Understanding $K_a$ and $\text{pKa}$ not only lets you determine how much of an acid is ionized in solution but also guides you in predicting changes in chemical behavior as the pH varies.

Distribution Curves

Distribution curves are graphical representations showing how fractions of a weak acid and its conjugate base change with pH. When plotting these for acetic acid and its conjugate base, acetate, you will use fractions determined by the Henderson-Hasselbalch equation.
These curves typically have pH on the x-axis and fraction on the y-axis, showcasing how the concentration of ${\text{CH}}_3\text{COOH}$ and ${\text{CH}}_3{\text{COO}}^{-}$ shift as the medium’s pH changes.
Here's the insight:

• As pH increases (becoming more basic), the fraction of ${\text{CH}}_3{\text{COO}}^{-}$ increases, while the ${\text{CH}}_3\text{COOH}$ fraction decreases.
• At $\text{pH}=\text{pKa}=4.76$, the fractions are equal, intersecting at 0.5.

Why are distribution curves important?

• They provide a visual interpretation of how buffers work and how effective an acid or base can be in a specific pH range.
• They are instrumental in various applications, from pharmaceuticals to environmental science, wherever pH balance is a factor.

Conjugate Base

In acid-base chemistry, the conjugate base is what remains after an acid donates a proton (hydrogen ion). Understanding the role of a conjugate base helps explain how bases form and balance reactions in solution.
For acetic acid:$${\text{CH}}_3\text{COOH}\to {\text{CH}}_3{\text{COO}}^{-}+{\text{H}}^{+}$$
Upon losing a ${\text{H}}^{+}$, acetic acid forms its conjugate base, acetate ${\text{CH}}_3{\text{COO}}^{-}$.

• The conjugate base has the ability to accept a proton, reversing the reaction and reforming the acid.
• In equilibrium, the presence of the conjugate base influences how the solution maintains its pH.

Understanding how conjugate acids and bases work provides insight into:

• Their reactivity and interaction under different conditions.
• Stability of the solutions, influencing their use as buffers at certain pH levels.

This foundational knowledge is critical when evaluating acid-base reactions and their results at various pH levels, a key aspect exemplified in the distribution curves for acetic acid in your exercise.