Question

Consider the titration of 50.0 \(\mathrm{mL}\) of 0.10\(M \mathrm{H}_{3} \mathrm{A}\left(K_{\mathrm{a}},=\right.\) \(5.0 \times 10^{-4}, K_{\mathrm{a}_{2}}=1.0 \times 10^{-8}, K_{\mathrm{a}_{2}}=1.0 \times 10^{-11}\) ) titrated by 0.10\(M \mathrm{KOH}\) a. Calculate the pH of the resulting solution at 125 \(\mathrm{mL}\) of KOH added. b. At what volume of KOH added does pH \(=3.30 ?\) c. At 75.0 \(\mathrm{mL}\) of KOH added, is the solution acidic or basic?

Short answer

The pH of the resulting solution at 125 mL of KOH added is approximately 12.63. To reach a pH of 3.30, approximately 0.909 mL of KOH must be added. At 75.0 mL of KOH added, the solution is acidic.

Step-by-step solution

Calculate moles of H₃A and OH-

To find the pH, we first need to calculate the moles of H₃A and OH- in the solution at 125 mL of KOH added. We have 50.0 mL of 0.10 M H₃A and 125 mL of 0.10 M KOH. Moles of H₃A = 0.10 M * 50.0 mL * (1 L/1000 mL) = 0.005 mol Moles of OH- = 0.10 M * 125 mL * (1 L/1000 mL) = 0.0125 mol

Determine the neutralization reaction

The strong base KOH will react with the weak acid H₃A to form water and the conjugate base. Since we have added more moles of OH- than H₃A, the solution will have excess OH-, meaning it will be basic. H₃A + OH- → H₂A- + H₂O

Calculate moles of remaining species

Since all the moles of H₃A (0.005 mol) will react with 0.005 mol of OH-, we are left with 0.0075 mol of OH- (0.0125 - 0.005). The solution is basic, so we can find the concentration of OH- in the solution. Total volume = initial volume of H₃A + volume of KOH = 50 mL + 125 mL = 175 mL = 0.175 L Concentration of OH- = (0.0075 mol)/0.175 L = 0.04286 M

Calculate the pH

Since the solution is basic, we can use the concentration of OH- to determine the pOH and finally the pH. pOH = -log(0.04286) = 1.367 pH = 14 - pOH = 14 - 1.367 = 12.633 The pH of the resulting solution at 125 mL of KOH added is approximately 12.63. b. At what volume of KOH added does pH = 3.30?

Calculate the hydrogen ion concentration

To find the volume of KOH added when the pH is 3.30, we first need to determine the hydrogen ion concentration: pH = 3.30 H⁺ concentration = 10^(-pH) = 10^(-3.3) = 5.01 × 10⁻⁴ M

Determine reactions at the equivalence point

At the equivalence point, all moles of H₃A are converted to H₂A-, so the solution will be at its second acidic dissociation point. We will use the second dissociation constant, Ka₂. H₂A- + H₂O → HA²⁻ + H₃O⁺ Ka₂ = [HA²⁻][H₃O⁺]/[H₂A-] Using Ka₂ = 1 × 10⁻⁸ and substituting [H₃O⁺] = 5.01 × 10⁻⁴ M, we can find the ratio of moles of H₂A¯ to moles of HA²⁻ at pH = 3.30.

Calculate the volume of KOH needed

Since we have found the ratio of H₂A¯ to HA²⁻ at pH = 3.30, we can calculate the volume of KOH needed to reach this pH. 1 × 10⁻⁸ = (5.01 × 10⁻⁴)^2 / x x = [HA²⁻]/[H₂A¯] = 1 × 10^4 [H₃A] = 0.10 M, and at pH = 3.30, [H₃A] = 0.10 - x 0.10 - x = x(1 × 10^4) x = [H₃A] = 9.09 × 10⁻⁵ moles (at pH = 3.30) Since we are given the concentration of KOH, we can find the volume of KOH needed to reach pH = 3.30. V_KOH (in L) = moles [H₃A]/[KOH] V_KOH = (9.09 × 10⁻⁵)/(0.10) = 9.09 × 10⁻⁴ L, which is equivalent to 0.909 mL Therefore, the volume of KOH added to reach pH = 3.30 is approximately 0.909 mL. c. At 75.0 mL of KOH added, is the solution acidic or basic?

Determine the moles of H₃A and OH-

First, find the moles of H₃A and OH- in the solution: Moles of H₃A = 0.005 mol (as previously calculated for 50 mL of 0.10 M) Moles of OH- = 0.10 M * 75.0 mL * (1 L/1000 mL) = 0.0075 mol Since moles of H₃A > moles of OH-, the solution is acidic.

Key concepts

pH calculation

The concept of pH is crucial in chemistry, as it measures the acidity or basicity of a solution. The pH scale ranges from 0 to 14:

• A pH less than 7 means the solution is acidic.
• A pH of 7 indicates a neutral solution.
• A pH greater than 7 means the solution is basic.

To calculate the pH from the concentration of hydrogen ions (H^+), you can use the formula:\[pH = -\log[H^+]\]For example, if you know the concentration of H^+ is 5 \times 10^{-4} \text{ M}, the pH is calculated as follows:\[pH = -\log(5 \times 10^{-4}) \approx 3.30\]In a basic solution, like the one formed in the KOH and H_3A titration, instead of calculating pH directly, you can find \text{pOH} first, and then derive the pH using:\[pH = 14 - \text{pOH}\]Understanding these calculations is essential for analyzing the acidity or basicity of solutions during titrations.

neutralization reaction

A neutralization reaction occurs when an acid and a base react to form water and a salt. This type of chemical reaction generally involves the following equation:\[\mathrm{HA} + \mathrm{BOH} \rightarrow \mathrm{BA} + \mathrm{H_2O}\]In the exercise provided, H_3A, a weak acid, reacts with KOH, a strong base.

• As KOH is added to H_3A, it reacts to neutralize the acid, forming water and the conjugate base (H_2A^-).
• This process continues until all the H_3A is converted, or excess KOH is added, creating a basic solution due to leftover OH^- ions.

In titration processes, the point at which the amount of acid equals the amount of base is called the equivalence point. In acid-base titrations, finding this point is crucial, as it helps to determine concentrations and the PH of the solution.

• At equivalence, assuming complete reaction, the moles of H^+ from the acid equals the moles of OH^- from the base.
• In the problem, determining excess OH^- means any additional base will increase the pH, indicating a basic solution.

weak acids

Weak acids are characterized by their partial dissociation in water, which means they do not release all their hydrogen ions into the solution.

• This property contrasts sharply with strong acids, which completely dissociate in water.
• The strength of a weak acid is represented by its acid dissociation constant (K_a), which quantifies its tendency to donate H^+ ions in the solution.

In the case of H_3A, a triprotic acid, it dissociates in steps, having three distinct K_a values corresponding to the release of each hydrogen ion:

• \(K_{\text{a}_1} = 5.0 \times 10^{-4}\) for the first hydrogen.
• \(K_{\text{a}_2} = 1.0 \times 10^{-8}\) for the second hydrogen.
• \(K_{\text{a}_3} = 1.0 \times 10^{-11}\) for the third hydrogen.

The gradual nature of dissociation in weak acids impacts the titration curve and necessitates careful analysis to determine pH changes. Understanding the behavior of weak acids during titration helps in assessing their buffering capacity and predicting the pH of solutions during different stages of reactions.