The dissociation constant of a substituted benzoic acid at 25^oC is
1.0 × 10^-4. The pH of a 0.01 M solution of its sodium salt is

To determine the pH of a 0.01 M solution of the sodium salt of a substituted benzoic acid with a dissociation constant (Ka) of 1.0 × 10^-4, we can follow a systematic approach. The sodium salt of the acid will dissociate in water to produce the conjugate base of the acid, which will then interact with water to establish an equilibrium that affects the pH of the solution.

Understanding the Concept of pH and Dissociation Constants

The dissociation constant (Ka) is a measure of the strength of an acid in solution. For benzoic acid, the dissociation can be represented as:

• HA ⇌ H⁺ + A^-

Here, HA is the benzoic acid, H⁺ is the hydrogen ion, and A^- is the conjugate base (the salt). The dissociation constant is given by:

• Ka = [H⁺][A^-]/[HA]

Calculating the pH of the Sodium Salt Solution

When we dissolve the sodium salt of the benzoic acid in water, it dissociates completely:

• NaA → Na⁺ + A^-

In a 0.01 M solution, the concentration of A^- will also be 0.01 M. The conjugate base A^- can react with water to produce hydroxide ions (OH^-), which will affect the pH:

• A^- + H₂O ⇌ HA + OH^-

To find the pH, we first need to calculate the concentration of hydroxide ions produced in this equilibrium. We can use the relationship between Ka and Kb (the base dissociation constant) for the conjugate base:

• Kw = Ka × Kb

Where Kw is the ion product of water (1.0 × 10^-14 at 25 °C). Rearranging gives us:

• Kb = Kw / Ka = (1.0 × 10^-14) / (1.0 × 10^-4) = 1.0 × 10^-10

Setting Up the Equilibrium Expression

Now we can set up the equilibrium expression for the base dissociation:

• Kb = [HA][OH^-]/[A^-]

Let x be the concentration of OH^- produced at equilibrium. Initially, we have:

• [A^-] = 0.01 M
• [HA] = 0
• [OH^-] = 0

At equilibrium, we have:

• [A^-] = 0.01 - x
• [HA] = x
• [OH^-] = x

Substituting into the Kb expression gives:

• 1.0 × 10^-10 = (x)(x)/(0.01 - x)

Assuming x is small compared to 0.01, we can simplify this to:

• 1.0 × 10^-10 = x²/0.01

Solving for x:

• x² = 1.0 × 10^-10 × 0.01
• x² = 1.0 × 10^-12
• x = 1.0 × 10^-6

Finding the pH

Now that we have the concentration of OH^-, we can find the pOH:

• pOH = -log[OH^-] = -log(1.0 × 10^-6) = 6

Finally, we can convert pOH to pH using the relationship:

• pH + pOH = 14

Thus:

• pH = 14 - pOH = 14 - 6 = 8

In summary, the pH of a 0.01 M solution of the sodium salt of the substituted benzoic acid is approximately 8. This indicates that the solution is basic due to the presence of the conjugate base, which can accept protons from water, leading to an increase in hydroxide ion concentration.