Problem 109 - Chemistry The Central Science In Si Units

Question

Butyric acid is responsible for the foul smell of rancid butter. The \(\mathrm{p} K_{b}\) of the butyrate ion is 9.16. (a) Calculate the \(K_{a}\) for butyric acid. (b) Calculate the pH of a \(0.075 \mathrm{M}\) solution of butyric acid. (c) Calculate the pH of a \(0.075 \mathrm{M}\) solution of sodium butyrate.

Step-by-Step Solution

Answer

(a) \(K_a = 1.45 \times 10^{-5}\) (b) pH = 2.98 (c) pH = 8.86.

Step 1: Calculate the Ka from Kb

We know that the relationship between the ionization constants is given by \( K_a \times K_b = K_w \) where \( K_w = 1.0 \times 10^{-14} \) at 25°C.Given \( pK_b = 9.16 \), we can find \( K_b \) using \( K_b = 10^{-pK_b} = 10^{-9.16} \).Now, calculate \( K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{10^{-9.16}} \).

Step 2: Calculate the K_a value

First calculate \( K_b \):\[ K_b = 10^{-9.16} \approx 6.92 \times 10^{-10} \]Next, use the equation for \( K_a \):\[ K_a = \frac{1.0 \times 10^{-14}}{6.92 \times 10^{-10}} \approx 1.45 \times 10^{-5} \]

Step 3: Set up expression for pH calculation of butyric acid

The equation for the ionization of butyric acid (a weak acid) is given by:\[ HA \rightleftharpoons H^+ + A^- \]Use the expression for \( K_a \):\[ K_a = \frac{[H^+][A^-]}{[HA]} \]Assume \([HA] = 0.075 \text{ M}\) initially and \([H^+] = [A^-] = x\).Therefore, \[ K_a = \frac{x^2}{0.075 - x} \approx \frac{x^2}{0.075} \] since \(x\) is small.

Step 4: Solve for [H+] and pH of butyric acid solution

Now plug in the value of \( K_a \) found:\[ 1.45 \times 10^{-5} = \frac{x^2}{0.075} \]\( x^2 = 1.45 \times 10^{-5} \times 0.075 \)\( x^2 = 1.0875 \times 10^{-6} \)\( x = \sqrt{1.0875 \times 10^{-6}} \approx 1.043 \times 10^{-3} \)pH is calculated as:\[ \text{pH} = -\log_{10}(1.043 \times 10^{-3}) \approx 2.98 \]

Step 5: Calculate pH of sodium butyrate solution

Sodium butyrate is a salt of a weak acid and a strong base, so it will slightly hydrolyze in water:\[ A^- + H_2O \rightleftharpoons OH^- + HA \]Expression for \( K_b \):\[ K_b = \frac{[OH^-][HA]}{[A^-]} = \frac{x^2}{0.075 - x} \approx \frac{x^2}{0.075} \]\( K_b = 6.92 \times 10^{-10} \) as calculated earlier.

Step 6: Solve for [OH-] and pOH for sodium butyrate

Plug the values into the expression:\[ 6.92 \times 10^{-10} = \frac{x^2}{0.075} \]Solving for \( x^2 \):\( x^2 = 6.92 \times 10^{-10} \times 0.075 \)\( x^2 = 5.19 \times 10^{-11} \)\( x = \sqrt{5.19 \times 10^{-11}} \approx 7.2 \times 10^{-6} \)OH concentration gives us pOH:\[ \text{pOH} = -\log_{10}(7.2 \times 10^{-6}) \approx 5.14 \]

Step 7: Convert pOH to pH for sodium butyrate

To find pH:\[ \text{pH} = 14 - \text{pOH} = 14 - 5.14 = 8.86 \]

Key Concepts

Ionization Constants

In acid-base chemistry, ionization constants are crucial for understanding how easily an acid or base can donate or accept a proton. These constants measure the strength of acids and bases. Ionization constants include the acid dissociation constant, denoted as \(K_a\), and the base dissociation constant, denoted as \(K_b\). The values of these constants elucidate the extent of ionization of acids and bases in solution. The smaller the \(K_a\) value, the weaker the acid; conversely, a larger \(K_b\) value signifies a stronger base. Knowing these constants allows chemists to predict the behavior of substances in aqueous solutions.

• The relationship between \(K_a\) and \(K_b\) for a conjugate acid-base pair is given by: \(K_a \times K_b = K_w\), where \(K_w\) is the ion product of water, equal to \(1.0 \times 10^{-14}\) at 25°C.

Understanding these relationships is key when calculating one constant from another, particularly in reactions involving weak acids and bases.

Weak Acids

Unlike strong acids, which fully dissociate in water, weak acids only partially ionize. This partial ionization means that only some molecules release protons into the solution. As a result, weak acids have small \(K_a\) values. For instance, butyric acid (the cause of rancid butter's odor) is a typical weak acid, making it a perfect example to understand weak acids' behavior.

• A common calculation involves determining the extent of ionization using the formula for \(K_a\): \(K_a = \frac{[H^+][A^-]}{[HA]}\).
• For butyric acid, we assume \([HA]\) as the initial concentration, while \([H^+]\) and \([A^-]\) are each represented by \(x\), derived from the changes in concentration as ionization occurs.

This simplification is vital to make calculations manageable and to determine the pH of a weak acid solution accurately.

pH Calculation

pH is a measure of the hydrogen ion concentration in a solution, a critical aspect of chemistry crucial for categorizing solutions as either acidic or basic. Calculating the pH of a solution involves logarithms, specifically using the formula: \(\text{pH} = -\log_{10}[H^+]\). For weak acid solutions such as butyric acid, the pH calculation requires determining \([H^+]\). This is done by solving the expression for \(K_a\) where \([H^+] = x\).
In strongly acidic solutions, the value of \([H^+]\) is higher, leading to lower pH values. For the butyric acid example in a \(0.075 \text{ M}\) solution, after solving for \(x\), the pH is calculated to be approximately 2.98—a reflection of its weak acidic nature. Conversely, for salts like sodium butyrate, mistreating the solution as completely basic requires deriving \([OH^-]\) using \(K_b\) and converting to pH using \( \text{pOH} + \text{pH} = 14\).

• pH calculations often involve approximation, assuming \(x\) is small relative to the initial concentration, simplifying mathematical expressions.

Butyrate Ion

The butyrate ion, \(\text{C}_4\text{H}_7\text{O}_2^-\), arises from the deprotonation of butyric acid, a weak acid known for its pungent smell. The ion is a conjugate base, which means it can act to neutralize acids by accepting protons.

• In solutions, the butyrate ion's strength as a base is described by its \(K_b\) value, obtainable from the \(pK_b\).
• Its behavior in water affects the solution's overall pH since it participates in hydrolysis, forming \(OH^-\) ions.
• Calculations involving butyrate ions typically require determining \(K_b\) using the formula \(K_b = 10^{-\text{pK}_b}\), and solving the equilibrium for \(OH^-\) concentration to indirectly find \(pH\).

Thus, the butyrate ion showcases how weak bases work in aqueous solutions and their impact on the pH level.