Question

Consider the titration of 100.0 \(\mathrm{mL}\) of 0.10 \(\mathrm{M} \mathrm{H}_{3} \mathrm{AsO}_{4}\) by 0.10 \(M \mathrm{NaOH}\) . What are the major species present at 50.0 \(\mathrm{mL}\) of \(\mathrm{NaOH}\) added? How would you calculate the pH at this point? Answer the same questions for 150.0 \(\mathrm{mL}\) of NaOH added. At what volume of NaOH added does pH \(=\mathrm{p} K_{\mathrm{a}_{\mathrm{i}}} ?\)

Short answer

At 50.0 mL of NaOH added, the major species present are H₃AsO₄ and H₂AsO₄⁻. To calculate the pH at this point, we use the first acidity constant (Kₐ₁) given and the equilibrium concentrations of H₃AsO₄ and H₂AsO₄⁻. At 150.0 mL of NaOH added, the major species present are H₃AsO₄, H₂AsO₄⁻, and HAsO₄²⁻. To calculate the pH at this point, we use the second acidity constant (Kₐ₂) and the respective concentrations of the species. To find the volume of NaOH added, where pH = pKₐᵢ, we must determine the half-equivalence points, where [A⁻] = [HA], for each dissociation reaction and compare the pH and pKₐ values at these points.

Step-by-step solution

Analyzing the dissociation of H₃AsO₄ and corresponding reactions with NaOH

Since H₃AsO₄ is a triprotic acid, we need to consider its dissociation and the corresponding reactions with NaOH, which acts as a strong base. The dissociation reactions are as follows: 1) H₃AsO₄ + OH⁻ → H₂AsO₄⁻ + H₂O 2) H₂AsO₄⁻ + OH⁻ → HAsO₄²⁻ + H₂O 3) HAsO₄²⁻ + OH⁻ → AsO₄³⁻ + H₂O

Calculate the moles of the species present at different volumes of NaOH added.

First, let's determine the initial number of moles of H₃AsO₄ and NaOH at each given volume. Since the initial concentration of H₃AsO₄ (as well as NaOH) is given in M (molarity), we can multiply it by the volume in liters to obtain the number of moles. Moles of H₃AsO₄ = 0.10 M × 100.0 mL × (1 L/1000 mL) = 0.01 mol For 50.0 mL of added NaOH: Moles of NaOH = 0.10 M × 50.0 mL × (1 L/1000 mL) = 0.005 mol For 150.0 mL of added NaOH: Moles of NaOH = 0.10 M × 150.0 mL × (1 L/1000 mL) = 0.015 mol

Determine the dominant reaction for each volume of NaOH added.

For 50.0 mL of NaOH: There is an excess of H₃AsO₄, so we can assume that the stoichiometry is 1:1 for the first reaction: H₃AsO₄ + OH⁻ → H₂AsO₄⁻ + H₂O. For 150.0 mL of NaOH: There is an excess of NaOH, so after the first two moles of OH⁻ react with H₃AsO₄ and form H₂AsO₄⁻, the remaining NaOH will react with H₂AsO₄⁻.

Calculate the pH at each volume of NaOH added using the dominant reactions and pKₐ values.

At 50.0 mL of added NaOH: Since the first dissociation is dominant (1:1 stoichiometry), the solution contains the conjugate base H₂AsO₄⁻ with H₃AsO₄. At this point, we have to use the first acidity constant (Kₐ₁): Kₐ₁ = \([H^+][H₂AsO₄⁻]\)/\([H₃AsO₄]\) With given Kₐ values, we can find the pH value. For 150.0 mL of added NaOH: Here, both the first and second reactions are dominant. At this point, we have the conjugate base HAsO₄²⁻ along with H₃AsO₄ and H₂AsO₄⁻. We use the second acidity constant (Kₐ₂) to calculate the pH: Kₐ₂ = \([H^+][HAsO₄²⁻]\)/\([H₂AsO₄⁻]\)

Find the volume of NaOH added for which pH is equal to pKₐᵢ

As we know that pH = pKₐ at the half-equivalence point, where \([A^-]\) = \([HA]\). For each dissociation reaction, we will find the volume of NaOH required for the half-equivalence point and then compare the pKₐ value to the pH. If pH = pKₐ, then we have the required volume at that point. Repeat the corresponding steps, considering the other dissociation steps (the second and third dissociation reactions). Hence, by following these steps, we can find the major species present, calculate the pH at different volumes of added NaOH, and find the volume of NaOH, where the pH equals the pKₐᵢ value.

Key concepts

Triprotic Acid

A triprotic acid is a type of polyprotic acid that can donate three protons (hydrogen ions, \(H^+\)) per molecule. This ability to donate three protons results in three sequential dissociation reactions, each with its own equilibrium constant, referred to as \(K_a\).

In the case of the exercise, arsenic acid (\(H_3AsO_4\)) is the triprotic acid that undergoes the following dissociation steps:

• \(H_3AsO_4 + OH^- \rightarrow H_2AsO_4^- + H_2O\) (first dissociation)
• \(H_2AsO_4^- + OH^- \rightarrow HAsO_4^{2-} + H_2O\) (second dissociation)
• \(HAsO_4^{2-} + OH^- \rightarrow AsO_4^{3-} + H_2O\) (third dissociation)

Each step progresses when a base like NaOH is added, providing hydroxide ions (\(OH^-\)) to reactors. Understanding these dissociation steps is crucial to predicting the behavior of a triprotic acid in a titration and its resulting pH changes.

pH Calculation

Calculating the pH during the titration of a triprotic acid involves understanding the concentration of protons and different ionic species present in solution. During each dissociation step, the dominant species will change, which directly affects the pH.

To calculate pH:

• Identify the major species present after each addition of NaOH, based on stoichiometry and dissociation steps known for triprotic acids.
• Use the relevant equilibrium constant \(K_a\) for the specific dissociation step that is dominant at a particular point in the titration.
• Apply the formula: \(K_a = \frac{[H^+][A^-]}{[HA]}\) to solve for \([H^+]\), then use \(pH = -\log[H^+]\) to find the pH.

For example, at the first half-equivalence point, you primarily consider the first dissociation. As more OH⁻ is added, subsequent dissociation reactions become significant, and thus, different \(K_a\) values must be used for precise pH calculations.

Half-Equivalence Point

The half-equivalence point in a titration of a polyprotic acid is where half of the acid has been neutralized. At this point, the concentration of the acid equals the concentration of its conjugate base, making it significant for pH calculation.

Key points to consider:

• At the half-equivalence point, \(pH = pK_a\) for that particular dissociation. This is an important property as it allows easy calculation and interpretation of the titration curve.
• For each dissociation step of a triprotic acid, there is a corresponding half-equivalence point, each signifying a stage in the titration where the triprotic acid partially neutralizes.
• To find it experimentally, you need to determine at what volume of \(NaOH\) the pH equals the \(pK_a\), assisting in recognizing the buffer region in a titration curve.

Understanding half-equivalence points aids in evaluating buffer capacities in solutions and designing buffer systems.

Dissociation Reactions

Dissociation reactions are the processes by which an acid releases its protons to form its conjugate base in a solution. For a triprotic acid, like \(H_3AsO_4\), these reactions occur in a stepwise manner, each step possessing unique characteristics.

• Each dissociation is characterized by its own equilibrium constant represent by \(K_a\). This determines the strength of each step of dissociation.
• Successful predictions of acid-base reactions rely on understanding each specific dissociation equilibrium, noting how each subsequent step requires progressively less energy compared to the first.

During a titration:

• The reaction proceeds through each dissociation phase sequentially as a base is added.
• This helps in determining not just the resulting species but also helps ascertain the pH during different phases of titration.
• Practically observing these reactions can be seen by plotting a titration curve, which displays the characteristic inflection points associated with each dissociation transition.

Understanding these mechanisms is essential for accurately modeling the behavior of triprotic systems in chemical equilibria.