Question

A solution is prepared by dissolving \(0.350 \mathrm{~g}\) of benzoic acid, \(\mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\), in water to make \(100.0 \mathrm{~mL}\) of solution. A \(30.00-\mathrm{mL}\) sample of the solution is titrated with \(0.272 \mathrm{M}\) KOH. Calculate the \(\mathrm{pH}\) of the solution (a) before titration. (b) halfway to the equivalence point. (c) at the equivalence point.

Short answer

Answer: (a) 3.25, (b) 4.19, and (c) 8.49.

Step-by-step solution

Calculate initial moles, molarity, and Ka of benzoic acid

First, we need to determine the initial moles and molarity of benzoic acid in the solution. The molecular weight of benzoic acid (HC7H5O2) is 122.12 g/mol. Moles of benzoic acid = Mass / Molecular weight Moles of benzoic acid = 0.350 g / 122.12 g/mol = 0.002865 mol Since the total volume of the solution is 100.0 mL, the molarity of benzoic acid can be calculated: Molarity of benzoic acid = Moles / Volume Molarity of benzoic acid = 0.002865 mol / 0.100 L = 0.02865 M Now we need to find the dissociation constant (Ka) of benzoic acid, which is given as 6.5 x 10^(-5).

Calculate the pH before titration (a)

To calculate the pH before the titration, we can use the initial molarity of benzoic acid and the Ka value: HC7H5O2 <=> H+ + C7H5O2- Ka = [H+][C7H5O2-] / [HC7H5O2] Since initially, all concentrations are equal, we can assume x to be the moles of benzoic acid that ionize: Ka = x^2 / (0.02865 - x) 6.5 x 10^(-5) = x^2 / (0.02865 - x) Solve for x using approximation (as x is small compared to 0.02865): x ≈ sqrt(6.5 * 10^(-5) * 0.02865) ≈ 0.0005608 The concentration of H+ ions is equal to the value of x, so we can calculate the pH using the formula: pH = -log[H+] pH ≈ -log(0.0005608) ≈ 3.25 (A) The pH before titration is approximately 3.25.

Calculate the pH halfway to the equivalence point (b)

Halfway to the equivalence point, half of the moles of benzoic acid have reacted with the KOH. At this point, the pH is equal to pKa: pKa = -log(Ka) pKa = -log(6.5 * 10^(-5)) ≈ 4.19 (B) The pH halfway to the equivalence point is approximately 4.19.

Calculate the pH at the equivalence point (c)

At the equivalence point, all benzoic acid has reacted with KOH. We need to find the moles of KOH in 30.00 mL of the solution: Moles of KOH = Molarity * Volume Moles of KOH = 0.272 M * 0.03000 L = 0.00816 mol At the equivalence point, the moles of benzoic acid are equal to the moles of KOH: Moles of C7H5O2- = 0.00816 mol The total volume of the solution at the equivalence point is 130.00 mL (100.00 mL benzoic acid + 30.00 mL KOH). The concentration of C7H5O2- at the equivalence point can be calculated: Concentration of C7H5O2- = Moles / Volume Concentration of C7H5O2- = 0.00816 mol / 0.1300 L ≈ 0.06277 M To find the pH at the equivalence point, we need to consider the hydrolysis of C7H5O2-: C7H5O2- + H2O <=> HC7H5O2 + OH- The equilibrium constant for this reaction is Kb, which is related to Ka by the relation: Kb = Kw / Ka Here, Kw is the ion product constant of water, which is 1.0 * 10^(-14). Kb ≈ (1.0 * 10^(-14)) / (6.5 * 10^(-5)) ≈ 1.54 * 10^(-10). We can now write the hydrolysis expression: Kb = [HC7H5O2][OH-] / [C7H5O2-] Let y be the moles of C7H5O2- that hydrolyze: 1.54 * 10^(-10) = y^2 / (0.06277 - y) Using approximation (as y is very small compared to 0.06277), we have: y ≈ sqrt(1.54 * 10^(-10) * 0.06277) ≈ 3.10 * 10^(-6) The concentration of OH- ions is equal to the value of y, and we can calculate the pOH using the formula: pOH = -log[OH-] pOH ≈ -log(3.10 * 10^(-6)) ≈ 5.51 Now, we can find the pH using the relation: pH = 14 - pOH pH ≈ 14 - 5.51 ≈ 8.49 (C) The pH at the equivalence point is approximately 8.49.

Key concepts

Acid-Base Titration

Acid-base titration is a process used to determine the concentration of an unknown acid or base by neutralizing it with a standard solution of known concentration. The procedure involves adding the titrant (either an acid or base) to a known volume of the analyte (the substance being analyzed), typically with the aid of a pH indicator or a pH meter to detect the endpoint of the reaction. In the given exercise, benzoic acid is being titrated with KOH, a strong base. The pH changes observed during the titration offer valuable information about the acid dissociation and the equivalence point.

In practice, as the titrant is added slowly, the pH of the solution being titrated will change gradually. We can monitor these changes to provide insights into the nature of the acid-base equilibrium and the strength of the acid or base in question. In educational exercises such as the one provided, students learn to relate the amount of titrant added to the pH of the solution and thereby understand the underlying chemistry.

Dissociation Constant Ka

The dissociation constant Ka is a quantitative measure of the strength of an acid in solution. It is a reflection of the acid's tendency to donate protons to the water, forming hydronium ions. The larger the Ka value, the stronger the acid. Strictly speaking, Ka is the equilibrium constant for the dissociation reaction of the acid.

For benzoic acid \( HC7H5O2 \), represented as \( HA \), the dissociation can be written as:
\[ HA \leftrightarrow H^+ + A^- \]
The dissociation constant expression is given by:
\[ Ka = \frac{[H^+][A^-]}{[HA]} \]
In calculations, it's crucial to make an initial assumption that the concentration of \( H^+ \) produced (x) is much smaller than the initial concentration of the acid, allowing for simplification of the Ka expression. This assumption holds true particularly for weak acids, like benzoic acid. Understanding and calculating Ka values are essential to predict the extent of the acid's dissociation and, subsequently, the pH of the solution.

Hydrolysis of Anions

Hydrolysis of anions occurs when the anion of a weak acid reacts with water to generate the weak acid and hydroxide ions. This reaction is of particular interest after reaching the equivalence point in an acid-base titration involving a weak acid and a strong base. The hydrolysis of anions is the reverse of the acid dissociation process and is governed by its own equilibrium constant (Kb) for weak bases.

For the anion of benzoic acid (C7H5O2-), the hydrolysis reaction is expressed as:
\[ C7H5O2^- + H2O \leftrightarrow HC7H5O2 + OH^- \]
Here, Kb can be calculated if Ka is known, through the relationship:
\[ Kb = \frac{Kw}{Ka} \]
where \( Kw \) is the ion-product constant of water (\( 1.0 \times 10^{-14} \) at 25°C). Having established Kb, we can assess the extent of hydrolysis and calculate the pH of the solution at the equivalence point, which is often necessary for understanding salt solutions resulting from the neutralization of weak acids or weak bases.

Equivalence Point

The equivalence point in a titration is reached when the moles of titrant added are stoichiometrically equivalent to the moles of substance present in the solution. At this stage, the acid has been completely neutralized by the base or vice versa. The pH at the equivalence point varies based on the strength of the acid and base involved in the reaction. For the titration of a weak acid with a strong base, the equivalence point pH is generally above 7 because of the resulting salt's anion hydrolysis, which creates a basic solution.

Determining the equivalence point is critical in titration as it signifies the completion of the reaction. The point can be estimated using a pH indicator or more accurately using a pH meter. In the given exercise, we calculate the pH at the equivalence point by first finding the concentration of the anion produced and then evaluating the hydrolysis that occurs, which ultimately provides the pH of the solution. This concept helps students understand the reaction's endpoint and the nature of the resultant solution after titration.