Question

When a diprotic acid, \(\mathrm{H}_{2} \mathrm{A}\), is titrated with \(\mathrm{NaOH}\), the protons on the diprotic acid are generally removed one at a time, resulting in a pH curve that has the following generic shape: a. Notice that the plot has essentially two titration curves. If the first equivalence point occurs at \(100.0 \mathrm{mL}\) NaOH added, what volume of NaOH added corresponds to the second equivalence point? b. For the following volumes of NaOH added, list the major species present after the OH \(^{-}\) reacts completely. i. \(0 \mathrm{mL} \mathrm{NaOH}\) added ii. between 0 and \(100.0 \mathrm{mL}\) NaOH added iii. \(100.0 \mathrm{mL}\) NaOH added iv. between 100.0 and \(200.0 \mathrm{mL}\) NaOH added v. \(200.0 \mathrm{mL} \mathrm{NaOH}\) added vi. after \(200.0 \mathrm{mL}\) NaOH added c. If the \(\mathrm{pH}\) at \(50.0 \mathrm{mL}\) NaOH added is \(4.0,\) and the \(\mathrm{pH}\) at \(150.0 \mathrm{mL} \mathrm{NaOH}\) added is \(8.0,\) determine the values \(K_{\mathrm{a}_{1}}\) and \(K_{\mathrm{a}_{2}}\) for the diprotic acid.

Short answer

The second equivalence point occurs at \(200.0 \mathrm{mL}\) NaOH added. The major species at different stages are: i. H₂A, ii. HA⁻ and H₂A, iii. HA⁻, iv. A²⁻ and HA⁻, v. A²⁻, and vi. A²⁻ and excess OH⁻. The values of \(K_{a_1}\) and \(K_{a_2}\) for the diprotic acid are \(1\times10^{-4}\) and \(1\times 10^{-8}\), respectively.

Step-by-step solution

Volume of NaOH for the second equivalence point

The second equivalence point occurs when double the volume of NaOH is added as compared to the first equivalence point: Second Equivalence Point Volume = \(2 \times 100.0 \mathrm{mL}\) = \(200.0 \mathrm{mL}\) #b. Major species after OH⁻ reacts completely at different stages#

Major species at different stages

i. \(0 \mathrm{mL} \mathrm{NaOH}\) added: H₂A (diprotic acid itself) ii. Between 0 and \(100.0 \mathrm{mL}\) NaOH added: HA⁻ and H₂A iii. \(100.0 \mathrm{mL}\) NaOH added: HA⁻ iv. Between 100.0 and \(200.0 \mathrm{mL}\) NaOH added: A²⁻ and HA⁻ v. \(200.0 \mathrm{mL}\) NaOH added: A²⁻ vi. After \(200.0 \mathrm{mL}\) NaOH added: A²⁻ and excess OH⁻ #c. Calculating the values of \(K_{a_1}\) and \(K_{a_2}\)# When \(50.0 \mathrm{mL}\) NaOH is added, the pH is 4. At this point, half of the first deprotonation occurs, so we can apply the following equation: \(K_{a_{1}}\) = \([HA^{-}][H^+]\)/\([H_{2}A]\) = \(\dfrac{[H^+]}{([HA^{-}][H^+])}\) Similarly, when \(150.0 \mathrm{mL}\) NaOH is added, the pH is 8. At this point, half of the second deprotonation occurs, so the following equation can be applied: \(K_{a_{2}}\) = \([A^{2-}][H^+]\)/\([HA^{-}]\) = \(\dfrac{[H^+]}{([A^{2-}][H^+])}\) Now we can calculate the values of \(K_{a_1}\) and \(K_{a_2}\) using the given pH values.

Calculating \(K_{a_1}\) and \(K_{a_2}\)

At \(50.0 \mathrm{mL}\) NaOH added (pH = 4): \(K_{a_{1}}\) = \(\dfrac{10^{-4}}{(10^{-4})}=1\times10^{-4}\) At \(150.0 \mathrm{mL}\) NaOH added (pH = 8): \(K_{a_{2}}\) = \(\dfrac{10^{-8}}{(10^{-8})}=1\times10^{-8}\) So, the values of \(K_{a_1}\) and \(K_{a_2}\) for the diprotic acid are \(1\times10^{-4}\) and \(1\times 10^{-8}\), respectively.

Key concepts

Titration Curve

In chemistry, a titration curve is a graphical representation of how the pH of a solution changes as a titrant is added. Specifically, when titrating a diprotic acid, the curve will have two distinct stages, each representing the removal of one proton from the acid molecule.

During the titration of a diprotic acid with a strong base such as NaOH, the curve shows these phases distinctly: the first where the pH rises moderately as the first acidic proton is neutralized, and then a more transitional phase leading up to the second equivalence point where the second proton is neutralized. The titration curve, thus, allows us to visualize the titration process and identify key points such as the equivalence points and buffering regions.

Equivalence Point

The equivalence point refers to the juncture where the amount of titrant added exactly neutralizes the amount of the substance being titrated. In the case of a diprotic acid, two equivalence points are expected and can be identified on the titration curve. From the exercise, we can infer that the second equivalence point occurs at a volume of base twice that of the first equivalence point's volume, because each mole of diprotic acid requires two moles of OH⁻ for complete neutralization. Therefore, if the first equivalence point is at 100.0 mL, the second one occurs at 200.0 mL of NaOH added. Identifying the equivalence points is crucial in determining the concentration of the acid and its dissociation constants.

Acid Dissociation Constant

The acid dissociation constant (Ka) is a quantitative measure of the strength of an acid in solution. It is typically expressed in terms of the concentration of products over the concentrations of reactants at equilibrium. For a diprotic acid, there are two dissociation constants: \( K_{a_1} \) for the first dissociation and \( K_{a_2} \) for the second.

\( K_{a} \) values are often used to predict the extent of dissociation at any given pH, and they play a key role in calculating the pH values at different stages of a titration. In the given exercise, knowing that the pH is 4 at the halfway point to the first equivalence point allows us to determine \( K_{a_1} \) as \(1 \times 10^{-4}\) using the given formula. Similarly, \( K_{a_2} \) can be calculated when the pH is 8 at the halfway point to the second equivalence point.

pH Calculation

The calculation of the pH is an essential aspect of acid-base chemistry. In titrations involving diprotic acids, the pH at various stages of titrant addition can indicate the progress of the reaction. Generally, when the concentration of hydronium ions \( [H^+] \) is known, the pH can be found using the formula pH = -log\footnote{[H^+]\footnote}\br>. This value helps in defining the acidity or basicity of the solution at a particular point in the titration. In our exercise, we have pH values of 4 and 8 corresponding to certain volumes of NaOH added. These values can be used to derive the dissociation constants of the diprotic acid.

Diprotic Acid Species

A diprotic acid is a type of acid that has two potentially dissociable protons. When this kind of acid reacts with a strong base, such as NaOH, the titration involves two displacement reactions, each corresponding to one of the protons. As the base is added, the acid goes through a series of transformations: it starts as \( H_{2}A \) and, after the first equivalence point, most of the acid exists as \( HA^- \). As more base is added and the second equivalence point is approached, the predominant species becomes \( A^{2-} \). Understanding the major species present at different titration stages is critical in analyzing the titration curve and helps to explain the buffering action and pH changes observed during the process.