Problem 51 - Chemistry: Principles and Reactions

Question

Butyric acid, \(\mathrm{HC}_{4} \mathrm{H}_{7} \mathrm{O}_{2}\), is responsible for the odor of rancid butter and cheese. Its \(K_{\mathrm{a}}\) is \(1.51 \times 10^{-5} .\) Calculate \(\left[\mathrm{H}^{+}\right]\) in solutions prepared by adding enough water to the following to make \(1.30 \mathrm{~L}\). (a) \(0.279 \mathrm{~mol}\) (b) \(13.5 \mathrm{~g}\)

Step-by-Step Solution

Answer

To summarize: (a) For a solution containing 0.279 moles of butyric acid in 1.3 L, the concentration of \(\mathrm{H}^+\) ions is \(9.77 \times 10^{-4} \ \mathrm{M}\). (b) For a solution containing 13.5 grams (0.1532 moles) of butyric acid in 1.3 L, the concentration of \(\mathrm{H}^+\) ions is \(7.45 \times 10^{-4} \ \mathrm{M}\).

Step 1: (a) Prepare the given information for calculation

The information given for part (a) of this exercise is: - Moles of butyric acid (\(\mathrm{HC}_{4} \mathrm{H}_{7} \mathrm{O}_{2}\)): 0.279 mol - Volume of solution: 1.3 L - \(K_\mathrm{a}\) of butyric acid: \(1.51 \times 10^{-5}\) First, we need to calculate the initial concentration of butyric acid in the solution: $$[\mathrm{HC}_{4} \mathrm{H}_{7} \mathrm{O}_{2}]_0 = \frac{0.279 \ \text{mol}}{1.3 \ \text{L}}$$

Step 2: (a) Calculate the initial concentration of butyric acid

Divide the moles of butyric acid by the volume of the solution to find the initial concentration: $$[\mathrm{HC}_{4} \mathrm{H}_{7} \mathrm{O}_{2}]_0 = \frac{0.279 \ \text{mol}}{1.3 \ \text{L}} = 0.2146 \ \mathrm{M}$$

Step 3: (a) Set up the reaction equilibrium expression and ICE table

We can write the reaction of butyric acid with water to release \(\mathrm{H}^+\) ions as follows: \(\mathrm{HC}_{4} \mathrm{H}_{7} \mathrm{O}_{2}(aq) \leftrightharpoons \mathrm{H}^+(aq) + \mathrm{C}_{4} \mathrm{H}_{7} \mathrm{O}_{2}^-(aq)\) Now, we can set up an ICE (Initial, Change, Equilibrium) table to help us keep track of the concentrations of each species in the solution: ||||| |---|---|---|---| ||Initial (M)|Change (M)|Equilibrium (M)| |\(\mathrm{HC}_{4} \mathrm{H}_{7} \mathrm{O}_{2}\)|0.2146|-x|0.2146 - x| |\(\mathrm{H}^+\)|0|+x|x| |\(\mathrm{C}_{4} \mathrm{H}_{7} \mathrm{O}_{2}^-\)|0|+x|x| Now, we can use the given \(K_\mathrm{a}\) value and the equilibrium expression for the reaction: $$K_\mathrm{a} = \frac{[\mathrm{H}^+][\mathrm{C}_{4} \mathrm{H}_{7} \mathrm{O}_{2}^-]}{[\mathrm{HC}_{4} \mathrm{H}_{7} \mathrm{O}_{2}]}$$ Substitute the equilibrium concentrations from the ICE table into the expression: $$1.51 \times 10^{-5} = \frac{x\cdot x}{0.2146-x}$$

Step 4: (a) Solve for x (concentration of \(\mathrm{H}^+\) ions)

Since the \(K_\mathrm{a}\) value is small, we can assume that the dissociation of butyric acid is minimal. Hence, we can simplify the expression by assuming that the subtraction of x in the denominator doesn't significantly change the value: $$1.51 \times 10^{-5} \approx \frac{x^2}{0.2146}$$ Now, solve for x (concentration of \(\mathrm{H}^+\) ions): $$x^2 = (1.51 \times 10^{-5})(0.2146)$$ $$x = \sqrt{(1.51 \times 10^{-5})(0.2146)}$$ $$x = 9.77 \times 10^{-4} \ \mathrm{M}$$ So, the concentration of \(\mathrm{H}^+\) ions in the solution for part (a) is \(9.77 \times 10^{-4} \ \mathrm{M}\).

Step 5: (b) Calculate the moles of butyric acid from grams

For part (b), we are given 13.5 grams of butyric acid instead of moles. We will first convert grams to moles: Molar mass of butyric acid: $$\mathrm{HC}_{4} \mathrm{H}_{7} \mathrm{O}_{2} = 12.01 + 4(12.01) + 7(1.01) + 2(16.00) = 88.11 \ \text{g/mol}$$ $$\text{moles of butyric acid} = \frac{13.5 \ \text{g}}{88.11 \ \text{g/mol}}$$

Step 6: (b) Find the moles of butyric acid

Divide the grams of butyric acid by its molar mass to find the moles: $$\text{moles of butyric acid} = \frac{13.5 \ \text{g}}{88.11 \ \text{g/mol}} = 0.1532 \ \mathrm{mol}$$ Now, we can repeat the same calculation process from part (a), but using 0.1532 moles of butyric acid instead of 0.279 moles.

Step 7: (b) Perform the same calculations as in part (a)

Following the same steps as in part (a), we calculate the initial concentration, set up the ICE table, use the equilibrium expression, and solve for the concentration of \(\mathrm{H}^+\) ions: Initial concentration of butyric acid: $$[\mathrm{HC}_{4} \mathrm{H}_{7} \mathrm{O}_{2}]_0 = \frac{0.1532 \ \text{mol}}{1.3 \ \text{L}} = 0.1178 \ \mathrm{M}$$ The equilibrium expression becomes: $$K_\mathrm{a} = \frac{x^2}{0.1178-x}$$ By simplifying the expression and solving for x (concentration of \(\mathrm{H}^+\) ions), we obtain: $$x^2 = (1.51 \times 10^{-5})(0.1178)$$ $$x = \sqrt{(1.51 \times 10^{-5})(0.1178)}$$ $$x = 7.45 \times 10^{-4} \ \mathrm{M}$$ So, the concentration of \(\mathrm{H}^+\) ions in the solution for part (b) is \(7.45 \times 10^{-4} \ \mathrm{M}\).

Key Concepts

Butyric Acid

Butyric acid, also known as \(\mathrm{HC}_{4} \mathrm{H}_{7} \mathrm{O}_{2}\), is an organic acid that is famous for its unpleasant smell, similar to rancid butter or stinky cheese. This compound is part of a larger family known as carboxylic acids, which are characterized by a carboxyl group (\(-COOH\)) attached to a carbon chain.
Butyric acid plays a role in various biological and industrial processes. For students of chemistry, it's particularly significant due to its weak acidic nature, signified by its relatively small acid dissociation constant, \( K_{\mathrm{a}} = 1.51 \times 10^{-5}\), indicating that it doesn't completely dissociate in water.
This acid's dissociation allows it to participate in equilibrium reactions within solutions, making it an intriguing example when learning about acid-base chemistry. Understanding butyric acid provides a window into how scientists manage and utilize organic acids in various applications, from food science to pharmaceuticals.

Equilibrium Concentration

Equilibrium concentration refers to the concentrations of reactants and products in a chemical reaction that has reached a state of balance, where the forward and reverse reactions occur at equal rates. In the context of butyric acid dissociation, achieving equilibrium is crucial to understanding how the acid behaves in solution.
When butyric acid is dissolved in water, it partially dissociates to form hydronium ions \( \mathrm{H}^+ \) and butyrate ions \( \mathrm{C}_{4} \mathrm{H}_{7} \mathrm{O}_{2}^- \). At equilibrium, the concentrations of these ions and un-dissociated butyric acid remains constant.
Calculating these concentrations requires understanding how the dissociation constant \( K_{\mathrm{a}} \) is used. \( K_{\mathrm{a}} \) represents the acid's tendency to donate protons, and it allows chemists to predict how much of the acid will dissociate at a given initial concentration. The equilibrium concentration of the species in the solution is then found by solving the equilibrium expression, helping us determine characteristics like the solution’s pH.

ICE Table

An ICE table, which stands for Initial, Change, and Equilibrium, is a useful tool in chemistry for keeping track of the concentrations of reactants and products throughout a reaction. It's particularly helpful in equilibrium calculations involving weak acids like butyric acid.
This table organizes data by showing the initial concentrations of the chemical species in the reaction, the change they undergo as the system approaches equilibrium, and the final, equilibrium concentrations.
The process begins with determining the initial concentration of butyric acid in the solution. Then, as the acid dissociates, the ICE table helps track the small changes in concentration that occur and calculates the concentration of \( \mathrm{H}^+ \) at equilibrium. This is crucial for applying the equilibrium expression, where \( x \) is often used to represent the change in concentration of hydrogen ions.
By setting up and solving the algebraic equations from the ICE table, students can grasp not only the specific concentration of ions they are interested in but also reinforce their understanding of concepts like dynamic equilibrium and dissociation in weak acids.