Problem 109 - Chemistry The Central Science

Question

Butyric acid is responsible for the foul smell of rancid butter. The \(\mathrm{pK}_{a}\) of butyric acid is 4.84 (a) Calculate the pK \(_{b}\) for the butyrate ion. (b) Calculate the pH of a 0.050 \(M\) solution of butyric acid. (c) Calculate the pH of a 0.050\(M\) solution of sodium butyrate.

Step-by-Step Solution

Answer

(a) The pKb for the butyrate ion is 9.16. (b) The pH of a 0.050 M solution of butyric acid is 2.87. (c) The pH of a 0.050 M solution of sodium butyrate is 11.13.

Step 1: (a) Calculate the pKb for the butyrate ion

To find the pKb for the butyrate ion, we can use the relationship between the dissociation constants of an acid and its conjugate base: \(K_a \times K_b = K_w\) \(K_w\) is the ion-product constant of water, and its value is \(1.0 \times 10^{-14}\). First, we need to find the value of Ka for butyric acid using the given pKa: \(K_a = 10^{-pKa}\) \(K_a = 10^{-4.84}\) Now, we can find the value of Kb for the butyrate ion: \(K_b = \frac{K_w}{K_a}\) \(K_b = \frac{1.0 \times 10^{-14}}{10^{-4.84}}\) Finally, we can find the pKb for the butyrate ion: \(pK_b = -\log{K_b}\)

Step 2: (b) Calculate the pH of a 0.050 M solution of butyric acid

To calculate the pH of the butyric acid solution, we can use the equilibrium expression for the dissociation of butyric acid: \(K_a = \frac{[H^+][A^-]}{[HA]}\) Since the initial concentration of H+ and A- is 0 and the concentration of HA is 0.050 M, we can use an ICE table to find the equilibrium concentrations. Note that x is the dissociation value: \(HA \longleftrightarrow H^+ + A^-\) Initial: \(0.050 \qquad 0 \qquad 0\) Change: \(-x \qquad +x \qquad +x\) Equilibrium: \(0.050 - x \qquad x \qquad x\) Now, we can substitute the values in the Ka expression: \(K_a = \frac{x^2}{0.050 - x}\) We can use the quadratic formula, or we can assume that x is very small compared to 0.050 to simplify our calculation: \(K_a = \frac{x^2}{0.050}\) Solve for x, which represents the [H+]: Now, we can use the pH formula: \(pH = -\log{[H^+]}\)

Step 3: (c) Calculate the pH of a 0.050 M solution of sodium butyrate

Sodium butyrate is the salt of a weak acid (butyric acid) and a strong base (NaOH). When sodium butyrate dissolves in water, it produces the conjugate base of butyric acid, the butyrate ion, which can accept a proton (H+) from water: \(A^- (aq) + H_2O (l) \longleftrightarrow HA (aq) + OH^- (aq)\) The equilibrium constant for the reaction is Kb. Since the concentration of sodium butyrate is 0.050 M, we can assume that the initial concentration of A- is also 0.050 M (considering 100% dissociation). We can use the ICE table method and the Kb expression to calculate the [OH-] concentration: \(A^- + H_2O \longleftrightarrow HA + OH^-\) Initial: \(0.050 \qquad \qquad\qquad 0\) Change: \(-x \qquad\qquad\qquad\qquad+x\) Equilibrium: \(0.050 - x \qquad\qquad\qquad x\) \(K_b = \frac{[HA][OH^-]}{[A^-]}\) Now, we can substitute the values in the Kb expression: \(K_b = \frac{x^2}{0.050 - x}\) We can use the quadratic formula, or we can assume that x is very small compared to 0.050 to simplify our calculation: \(K_b = \frac{x^2}{0.050}\) Solve for x, which represents the [OH-]: Now, we can use the pOH formula and then find the pH: \(pOH = -\log{[OH^-]}\) \(pH = 14 - pOH\)

Key Concepts

pKa and pKb calculations

One important aspect of understanding acid-base equilibria is the relationship between a weak acid and its conjugate base. This is where the concepts of \( pK_a \) and \( pK_b \) come into play. For a given acid, \( K_a \), the acid dissociation constant, quantifies the strength of the acid in solution. Likewise, \( K_b \), the base dissociation constant, does the same for a base.

The relationship between \( K_a \) and \( K_b \) can be understood by knowing that their product is equal to the ion product constant of water, \( K_w \), which is \( 1.0 \times 10^{-14} \) at 25 °C. Therefore, to find \( pK_b \) from \( pK_a \), we can use the formula:

• \( K_a \times K_b = K_w \)
• \( \text{p}K_a + \text{p}K_b = 14 \)

If you know \( pK_a \), you can calculate \( K_a \) using \( K_a = 10^{-\text{p}K_a} \). Subsequently, \( K_b \) can be calculated as \( K_b = \frac{K_w}{K_a} \), and \( pK_b \) can be determined as \( pK_b = -\log{K_b} \). This relationship highlights how the strength of an acid is inversely related to the strength of its conjugate base.

pH of acidic solutions

Calculating the pH of an acidic solution is a fundamental skill in chemistry that helps understand how acidic a solution is. The pH is a measure of the hydrogen ion concentration \([H^+]\) in a solution. For a weak acid like butyric acid, the pH depends on the dissociation in water.

Use the following steps for calculation:

• Determine the initial concentration of the acid. In our case, for butyric acid, it's given as 0.050 M.
• Use an ICE table (Initial, Change, Equilibrium) to set up the equilibrium concentrations.
• Apply the equilibrium expression, \( K_a = \frac{[H^+][A^-]}{[HA]} \), using the initial concentration and the dissociation (x value) found from the ICE table.
• If \( x \), the change in concentration, is small relative to the initial concentration, this simplification \( K_a = \frac{x^2}{0.050} \) can be used to solve for x.
• The pH is calculated as \( pH = -\log{[H^+]} \), where \([H^+]\) is the value of x found from the equilibrium calculation.

This step-by-step process shows how to handle the dissociation of weak acids and find their pH in solution.

ICE table method

The ICE (Initial, Change, Equilibrium) table is a powerful tool for visualizing and solving equilibrium problems in chemistry. It helps keep track of concentration changes as a reaction approaches equilibrium. Here's how to use one:

• **Initial Concentrations**: Start with the initial concentrations of reactants and products. Typically, the products are zero at the start.
• **Change**: Account for the change in concentrations that will occur as the system moves towards equilibrium. Represent this change with variables, such as \( -x \) or \(+x \), corresponding to how much each species is expected to change.
• **Equilibrium Concentrations**: Calculate the concentrations at equilibrium by adding together the initial concentration and the change for each species.

Once the equilibrium concentrations are identified, they can be plugged into the equilibrium constant expression \( K \) to solve for the change variable \( x \). Since many reactions reach completion to a small extent, assuming \( x \ll \text{initial concentrations} \) often simplifies calculations, making the math more manageable. This technique is particularly helpful for weak acid or base equilibria where the percent dissociation is small.