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.

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.

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.

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: \(1.4 \times 10^{-4} = \frac{x \cdot x}{0.60 - x}\)

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 \(9.11 \times 10^{-3} \ \mathrm{mol/L}\).

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.