Question

Consider a \(0.60-M\) solution of \(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}\), lactic acid \(\left(K_{\mathrm{a}}=\right.\) \(\left.1.4 \times 10^{-4}\right)\) a. Which of the following are major species in the solution? i. \(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}\) ii. \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{3}^{-}\) iii. \(\mathrm{H}^{+}\) iv. \(\mathrm{H}_{2} \mathrm{O}\) v. \(\mathrm{OH}^{-}\) b. Complete the following ICE table in terms of \(x\), the amount \((\mathrm{mol} / \mathrm{L})\) of lactic acid that dissociates to reach equilibrium. c. What is the equilibrium concentration for \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{3}^{-}\) ? d. Calculate the \(\mathrm{pH}\) of the solution.

Short answer

The major species in the 0.60 M lactic acid solution are \(\mathrm{HC}_{3}\mathrm{H}_{5}\mathrm{O}_{3}\), \(\mathrm{C}_{3}\mathrm{H}_{5}\mathrm{O}_{3}^{-}\), \(\mathrm{H}^{+}\), and \(\mathrm{H}_{2}\mathrm{O}\). The equilibrium concentration of \(\mathrm{C}_{3}\mathrm{H}_{5} \mathrm{O}_{3}^{-}\) is approximately \(9.11 \times 10^{-3} \ \mathrm{mol/L}\). The pH of the solution is approximately 2.04.

Step-by-step solution

Identify major species

To begin with, we need to identify the major species present in the solution. Lactic acid (\(\mathrm{HC}_3\mathrm{H}_5\mathrm{O}_3\)) is a weak acid, so it can donate an \(\mathrm{H}^+\) ion in the solution. The major species in the 0.60 M solution will be: i. \(\mathrm{HC}_{3}\mathrm{H}_{5}\mathrm{O}_{3}\), because it is present in its greatest concentration. ii. \(\mathrm{C}_{3}\mathrm{H}_{5}\mathrm{O}_{3}^{-}\), because it is the conjugate base after \(\mathrm{HC}_3\mathrm{H}_5\mathrm{O}_3\) donates an \(\mathrm{H}^+\) ion. iii. \(\mathrm{H}^{+}\), which comes from the dissociation of the weak acid. iv. \(\mathrm{H}_{2}\mathrm{O}\), because it is always present in an aqueous solution. v. \(\mathrm{OH}^{-}\) is present in very low concentration since we are dealing with an acidic solution.

Set up the ICE table

Now we need to complete the ICE table, where I represents the initial concentration, C represents the change in concentration, and E represents the equilibrium concentration. Species: \(\qquad \quad | \qquad \mathrm{HC}_3\mathrm{H}_5\mathrm{O}_3 \qquad | \quad \mathrm{C}_3\mathrm{H}_5\mathrm{O}_3^- \qquad \quad | \qquad \mathrm{H}^+\) Initial: \(\qquad \quad \ | \qquad \ \ 0.60 \quad | \quad \ \ \ 0 \quad \qquad \qquad | \qquad \ 0 \qquad\) Change: \(\qquad \ \ \, | \qquad -x \quad \ | \quad \ \ x \qquad \qquad | \qquad \ x \qquad\) Equilib: \(\qquad \ \ | \quad \ 0.60 -x \ \ | \quad \ \ x \qquad \qquad | \qquad \ x \qquad\) Note that \(\mathrm{OH}^-\) does not react with lactic acid and so it is not included in the ICE table.

Ka equation

Using the \(K_a = 1.4 \times 10^{-4}\), we can set up the equation for the dissociation of lactic acid at equilibrium. \[K_a = \frac{[\mathrm{C}_3\mathrm{H}_5\mathrm{O}_3^-][\mathrm{H}^+]}{[\mathrm{HC}_3\mathrm{H}_5\mathrm{O}_3]}\] Substitute the variables from the equilibrium row of the ICE table: \(1.4 \times 10^{-4} = \frac{x \cdot x}{0.60 - x}\)

Solve for x

As Ka is very small (1.4×10^(-4)), we can approximate x to be considerably small as the equilibrium will shift more towards the reactants. Therefore, we can assume (0.60 - x) ≈ 0.60. So now we simplify and solve for x. \(1.4 \times 10^{-4} = \frac{x^2}{0.60}\) \(x^2 = 1.4 \times 10^{-4} \times 0.60\) \(x = \sqrt{1.4 \times 10^{-4} \times 0.60}\) Solving for x gives: \(x \approx 9.11 \times 10^{-3} \ \mathrm{mol/L}\) So, the equilibrium concentration of \(\mathrm{C}_3\mathrm{H}_5\mathrm{O}_3^-\) is \(9.11 \times 10^{-3} \ \mathrm{mol/L}\).

Calculate the pH

Now that we have the concentration of \(\mathrm{H}^+\), we can calculate the pH of the solution. \(\mathrm{pH} = -\log_{10} {[\mathrm{H}^+]}\) \(\mathrm{pH} = -\log_{10} {(9.11 \times 10^{-3}\ \mathrm{mol/L})}\) After calculating the value, we get: \(\mathrm{pH} \approx 2.04\) So, the pH of the 0.60 M lactic acid solution is approximately 2.04.

Key concepts

Lactic Acid

Lactic acid is a naturally occurring weak organic acid. Its chemical formula is \( \mathrm{HC}_3\mathrm{H}_5\mathrm{O}_3 \). Lactic acid is found in sour milk, and it is also produced by muscle cells during intense exercise when oxygen levels are low. In the context of acid-base chemistry, lactic acid is classified as a weak acid. This means that it does not completely dissociate in water, unlike strong acids like hydrochloric acid (HCl).
In chemical reactions, weak acids like lactic acid will partially release protons (\( \mathrm{H}^+ \)) into the solution, establishing an equilibrium between the undissociated acid and its dissociated form, which is its conjugate base, \( \mathrm{C}_3\mathrm{H}_5\mathrm{O}_3^- \). Understanding this behavior is crucial for calculating pH and analyzing equilibrium in acid-base reactions.

ICE Table

ICE stands for Initial, Change, and Equilibrium. An ICE table is a handy tool used in chemistry to track the concentrations of species in a reaction as they move toward equilibrium. It is particularly useful for reactions where equilibria are established, such as in weak acid/base dissociations.
When setting up an ICE table, you:

• Identify the species involved in the reaction.
• Determine the initial concentrations.
• Define the change that occurs as equilibrium is approached, usually represented by \( x \).
• Calculate the equilibrium concentrations using the initial values and changes defined.

This method provides a clear framework to apply the dissociation constant (\( K_a \)) and solve for the equilibrium concentrations. In the case of lactic acid, the change