Problem 17 - Chemistry: The Central Science

Question

(a) Calculate the percent ionization of \(0.0075 M\) butanoic acid \(\left(K_{a}=1.5 \times 10^{-5}\right) .\) (b) Calculate the percent ionization of \(0.0075 \mathrm{M}\) butanoic acid in a solution containing \(0.085 \mathrm{M}\) sodium butanoate.

Step-by-Step Solution

Answer

In conclusion, the percent ionization of 0.0075 M butanoic acid is (a) approximately 1.47% and (b) approximately 0.017% in a buffer solution containing 0.085 M sodium butanoate.

Step 1: Understand the Ionization of Butanoic Acid

Butanoic acid is a weak monoprotic acid with the chemical formula C4H8O2, denoted as HA in its protonated form. When it ionizes in a solution, it releases a hydrogen ion (H+) and a butanoate ion (A-): HA ↔ H+ + A- The equilibrium constant (\(K_a\)) for this reaction is given as \(K_a = 1.5 \times 10^{-5}\).

Step 2: Calculate the Percent Ionization of 0.0075 M Butanoic Acid (a)

First, we need to set up the ice table for this reaction as follows: \[ \begin{array}{c|c|c|c} & [\mathrm{HA}] & [\mathrm{H}^{+}] & [\mathrm{A}^{-}] \\ \hline \textbf{Initial} & 0.0075 & 0 & 0 \\ \textbf{Change} & -x & +x & +x \\ \textbf{Equilibrium} & 0.0075-x & x & x \\ \end{array} \] Now we use the equilibrium expression: \(K_a = \frac{[\mathrm{H}^{+}][\mathrm{A}^{-}]}{[\mathrm{HA}]}\) Substitute the equilibrium concentrations and given \(K_a\) value, then solve for x: \( 1.5 \times 10^{-5} = \frac{x^2}{0.0075-x} \) Since \(K_a\) is very small, we can assume that x is also small, and thus 0.0075-x ≈ 0.0075: \( x^2 = 1.5 \times 10^{-5} \times 0.0075 \) Solve for x: \( x = \sqrt{1.5 \times 10^{-5} \times 0.0075} \approx 1.1 \times 10^{-4} \) Now, we can calculate the percent ionization: Percent Ionization = \(\frac{[\text{H}^+]}{[\text{HA}_{initial}]} \times 100\%\) Percent Ionization = \(\frac{1.1 \times 10^{-4}}{0.0075} \times 100 \% \approx 1.47\%\)

Step 3: Calculate the Percent Ionization of 0.0075 M Butanoic Acid with 0.085 M Sodium Butanoate (b)

In this case, butanoic acid and sodium butanoate form a buffer solution. Sodium butanoate dissociates completely in the solution, so its initial concentration is the same as the [A-] concentration at equilibrium: \[ \begin{array}{c|c|c|c} & [\mathrm{HA}] & [\mathrm{H}^{+}] & [\mathrm{A}^{-}] \\ \hline \textbf{Initial} & 0.0075 & 0 & \boldsymbol{0.085} \\ \textbf{Change} & -x & +x & 0 \\ \textbf{Equilibrium} & 0.0075-x & x & 0.085 \\ \end{array} \] Now, substitute the equilibrium concentrations into the \(K_a\) expression: \( 1.5 \times 10^{-5} = \frac{x \times 0.085}{0.0075-x} \) Again, we can assume that x is very small, so 0.0075-x ≈ 0.0075: \( x = 1.5 \times 10^{-5} \times \frac{0.0075}{0.085} \approx 1.3 \times 10^{-6} \) Now, we can calculate the percent ionization again: Percent Ionization = \(\frac{1.3 \times 10^{-6}}{0.0075} \times 100 \% \approx 0.017\%\) In conclusion, the percent ionization of butanoic acid is (a) approximately 1.47% in a 0.0075 M solution and (b) approximately 0.017% in a 0.0075 M solution containing 0.085 M sodium butanoate.

Key Concepts

Understanding Butanoic Acid

Butanoic acid, also known as butyric acid, belongs to the class of carboxylic acids bearing the chemical formula C4H8O2. As a weak monoprotic acid, it can donate only one hydrogen ion (H+) per molecule during dissociation. This is crucial in understanding its behavior in aqueous solutions.

Found naturally in butter and plant oils, butanoic acid has a distinctive rancid smell. When it ionizes in water, it forms butanoate ions (A-) and hydrogen ions, a process that is reversible and establishes an equilibrium between the reactants and products of ionization.

In essence, butanoic acid acts as a proton donor, which is a fundamental characteristic of acids according to the Bronsted-Lowry acid-base theory. Its ionization in water is minimal, reflecting its weak acid status compared to strong acids like hydrochloric acid (HCl), which ionize completely.

Equilibrium Constant (Ka) Explained

The equilibrium constant (Ka) is a value used to express the extent of ionization of an acid in solution. Specifically, it measures the strength of an acid by quantifying the ratio of the concentration of the ionized form (the products) to the undissociated form (the reactants) at equilibrium.

For butanoic acid's ionization represented as HA ↔ H+ + A-, the equilibrium expression can be written as:
\[ K_a = \frac{[\mathrm{H}^{+}][\mathrm{A}^{-}]}{[\mathrm{HA}]} \]
Here, [H+] is the concentration of hydrogen ions, [A-] is the concentration of butanoate ions, and [HA] is the concentration of undissociated butanoic acid.

The smaller the Ka value, the weaker the acid, and vice versa. A very small Ka implies that the reactants (undissociated form) are favored and that the acid does not ionize significantly in solution.

Buffer Solution Characteristics

A buffer solution consists of a mixture of a weak acid and its conjugate base, which can also be provided by the salt of the weak acid, like sodium butanoate (the sodium salt of butanoic acid). It is designed to maintain a stable pH upon the addition of small amounts of acid or base.

When sodium butanoate is dissolved in water, it separates completely into sodium (Na+) and butanoate (A-) ions. These butanoate ions then interact with the butanoic acid present in the solution, establishing an equilibrium that resists changes in pH.

The ability of a buffer to regulate pH is derived from the presence of both the weak acid (HA) that can donate H+ to neutralize added base, and the conjugate base (A-) that can absorb H+ to neutralize added acid. The Henderson-Hasselbalch equation relates pH, pKa (the negative log of Ka), and the concentrations of the acid and its conjugate base, which is essential in buffer calculations. In this exercise, the presence of sodium butanoate significantly decreases the percent ionization of butanoic acid, showcasing the buffering effect.