Given information:
- Dissociation constant (Ka) of a substituted benzoic acid at 25C = 1.0 × 10^-4
- Concentration of sodium salt = 0.01 M

To find: pH of the solution

Solution:
Step 1: Write the dissociation reaction of substituted benzoic acid in water
Substituted benzoic acid (HA) + H2O ⇌ H3O+ + A-

Step 2: Write the expression for the dissociation constant (Ka)
Ka = [H3O+][A-]/[HA]

Step 3: Use the given value of Ka to find the concentration of H3O+ and A-
Ka = [H3O+][A-]/[HA]
1.0 × 10^-4 = [H3O+][A-]/[HA]

Assuming that the concentration of H3O+ formed from water is negligible compared to that from HA,
[H3O+] = √(Ka[HA])

[H3O+] = √(1.0 × 10^-4 × 0.01) = 1.0 × 10^-3 M
[A-] = [HA] - [H3O+] = 0.01 - 1.0 × 10^-3 = 0.009 M

Step 4: Calculate the pH of the solution
pH = -log[H3O+]
pH = -log(1.0 × 10^-3)
pH = 3

Step 5: Calculate the pH of the original solution (sodium salt)
Since the sodium salt is formed by the reaction of the substituted benzoic acid (HA) with NaOH, it is a basic salt. Therefore, it hydrolyzes in water according to the following reaction:
A- + H2O ⇌ HA + OH-

The equilibrium constant for this reaction (Kb) can be calculated using the expression:
Kb = Kw/Ka = 1.0 × 10^-14/1.0 × 10^-4 = 1.0 × 10^-10

Assuming that the concentration of OH- formed from water is negligible compared to that from A-,
[OH-] = √(Kb[A-])

[OH-] = √(1.0 × 10^-10 × 0.009) = 3.0 × 10^-6 M

pOH = -log[OH-] = -log(3.0 × 10^-6) = 5.52

pH + pOH = 14
pH = 14 - pOH = 8 (Answer)

Therefore, the pH of the 0.01 M solution of the sodium salt of substituted benzoic acid at 25C is 8.

Note:
- The dissociation constant (Ka) of a weak acid gives an idea about the strength of the acid. The smaller the value of Ka, the weaker the acid.
- The hydrolysis of a salt depends on the acidic or basic nature of the salt. If the salt is formed by the reaction of a strong acid and a weak base, it is an acidic salt and hydrolyzes in water to give an acidic solution. If the salt is formed by the reaction of a