Problem 26 - General Chemistry: Principles and Modern Applications

Question

Caproic acid, \(\mathrm{HC}_{6} \mathrm{H}_{11} \mathrm{O}_{2},\) found in small amounts in coconut and palm oils, is used in making artificial flavors. A saturated aqueous solution of the acid contains \(11 \mathrm{g} / \mathrm{L}\) and has \(\mathrm{pH}=2.94 .\) Calculate \(K_{\mathrm{a}}\) for the acid. $$\mathrm{HC}_{6} \mathrm{H}_{11} \mathrm{O}_{2}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{C}_{6} \mathrm{H}_{11} \mathrm{O}_{2}^{-} \quad K_{\mathrm{a}}=?$$

Step-by-Step Solution

Answer

The Ka for the Caproic acid is approximately \(1.5 \times 10^{-5}\).

Step 1: Conversion of pH To Hydronium Ion Concentration

To find the hydronium ion concentration, use the definition of pH, which is -log[H3O+]. It implies that [H3O+] = 10^(-pH). In this case, the pH is given as 2.94, so [H3O+] = 10^(-2.94).

Step 2: Calculate Concentration of Caproic Acid

The concentration of Caproic acid is given in grams per liter, which should be converted to moles per liter (Molarity) for the calculation of Ka. The molar mass of HC6H11O2 is 116.16 g/mol. Hence, the molarity is 11 g/L divided by 116.16 g/mol which yields approximately 0.095 M.

Step 3: Calculate Ka

Equilibrium expression for the reaction is given by Ka = [H3O+][C6H11O2-]/[HC6H11O2]. From step 1, we know [H3O+] and from step 2, we know [HC6H11O2]. Since it's a monoprotic acid, [H3O+] = [C6H11O2-]. Therefore, Ka would be equal to ([H3O+]^2) / [HC6H11O2]. Substituting the known values and calculating gives the value of Ka.

Key Concepts

pH Calculation

Understanding pH is essential for the study of acid-base chemistry. The pH scale is used to measure the acidity or basicity of an aqueous solution. The formula to calculate pH is \( pH = -\log[H_3O^+] \), where \( [H_3O^+] \) is the concentration of hydronium ions in moles per liter. A pH lower than 7 indicates an acidic solution, while a pH higher than 7 indicates a basic solution. A pH of 7 is neutral, like pure water.

In the context of the caproic acid problem, converting the given pH of 2.94 to the hydronium ion concentration involves using the inverse logarithmic relationship: \( [H_3O^+] = 10^{-pH} \). When we calculate \( [H_3O^+] = 10^{-2.94} \), we obtain the hydronium ion concentration necessary for further calculations. The hydronium ion concentration gives us valuable insight into the acid's dissociation in the solution.

Molarity

Molarity, denoted as 'M', is the number of moles of a solute per liter of solution. It's a measure of concentration that's crucial for quantitatively describing chemical reactions occurring in solutions. The formula for molarity is \( M = \frac{\text{moles of solute}}{\text{liters of solution}} \).

To find the molarity of caproic acid, we first need the molar mass of the compound (\( \text{HC}_6\text{H}_{11}\text{O}_2 \)), which is 116.16 g/mol. Given the solution's concentration of 11 g/L, we divide this by the molar mass to find the molarity: \( M = \frac{11 \text{ g/L}}{116.16 \text{ g/mol}} \), resulting in approximately 0.095 M. This step converts the mass concentration to a more usable form for chemical equilibrium calculations.

Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the rates of the forward and reverse reactions are equal, leading to no net change in the amounts of reactants and products. It is reached when a dynamic balance is achieved, meaning that while particles continue to react, the overall concentrations remain constant. Understanding equilibrium is fundamental when calculating the constant values that describe reaction extents, such as the acid dissociation constant.

In our example, caproic acid dissociation reaches equilibrium in water, and the equilibrium expression relates the concentrations of all species involved: \( \text{HC}_6\text{H}_{11}\text{O}_2 + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{C}_6\text{H}_{11}\text{O}_2^- \). Knowledge of these concentrations allows us to solve for the unknown equilibrium constant, which quantifies the extent of acid dissociation.

Ka (Acid-Base Equilibrium)

The acid dissociation constant, denoted as Ka, is a numerical value that measures the strength of an acid in a solution. It is the equilibrium constant for the dissociation reaction of the acid into its conjugate base and hydrogen ions. The general expression for Ka of a monoprotic acid HA is given by \( Ka = \frac{[H^+][A^-]}{[HA]} \), where \( [H^+] \) and \( [A^-] \) are the concentrations of the hydrogen ions and conjugate base, respectively, and \( [HA] \) is the concentration of the undissociated acid.

For the calculation involving caproic acid, because it's a monoprotic acid, the concentrations of hydronium ions \( [H_3O^+] \) and the conjugate base \( [C_6H_{11}O_2^-] \) are equal. The formula simplifies to \( Ka = \frac{[H_3O^+]^2}{[HC_6H_{11}O_2]} \). By inserting the values we've calculated from the hydronium ion concentration and the molarity of caproic acid, we can determine the Ka, which tells us how much the acid dissociates at equilibrium.