The user wants me to solve for the pH of a sodium salt solution of a weak acid.

Step-by-step Solution:

1. Identify the nature of the salt: The given salt is the sodium salt of a substituted benzoic acid. This is a salt formed from a weak acid (substituted benzoic acid, let's denote it as HA) and a strong base (NaOH). The salt's formula can be represented as NaA.

2. Dissociation and Hydrolysis: When this salt (NaA) is dissolved in water, it dissociates completely into its ions: $\text{NaA}(aq) \rightarrow \text{Na}^+(aq) + \text{A}^-(aq)$ The anion A$^-$, being the conjugate base of the weak acid HA, undergoes hydrolysis in water. The Na$^+$ ion is a spectator ion and does not react with water. The hydrolysis reaction is: $\text{A}^-(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{HA}(aq) + \text{OH}^-(aq)$ This reaction produces hydroxide ions (OH$^-$), which makes the solution basic (pH > 7).

3. Calculate the Hydrolysis Constant ($K_h$ or $K_b$): The equilibrium constant for this hydrolysis reaction is the hydrolysis constant ($K_h$), which is also the base dissociation constant ($K_b$) for the conjugate base A$^-$. The relationship between the acid dissociation constant ($K_a$) of the acid, the base dissociation constant ($K_b$) of its conjugate base, and the ion product of water ($K_w$) at 25 ${}^\circ$C is: $K_a \times K_b = K_w = 1.0 \times 10^{-14}$ We are given $K_a = 1.0 \times 10^{-4}$. We can now calculate $K_b$: $K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-4}} = 1.0 \times 10^{-10}$

4. Calculate the Hydroxide Ion Concentration ($[\text{OH}^-]$): Let's set up an ICE table for the hydrolysis reaction. The initial concentration of the salt, and thus the anion A$^-$, is C = 0.01 M.

The expression for $K_b$ is:$$ K_b = \frac{[\text{HA}][\text{OH}^-]}{[\text{A}^-]} = \frac{(x)(x)}{0.01 - x} = \frac{x^2}{0.01 - x} $$ Since $K_b$ is very small ($1.0 \times 10^{-10}$), the degree of hydrolysis, x, will be very small compared to the initial concentration (0.01 M). Therefore, we can make the approximation $0.01 - x \approx 0.01$.$$ 1.0 \times 10^{-10} \approx \frac{x^2}{0.01} $$ Now, we can solve for x, which represents $[\text{OH}^-]$:$$ x^2 = (1.0 \times 10^{-10}) \times (0.01) = 1.0 \times 10^{-12} $$ $$ x = \sqrt{1.0 \times 10^{-12}} = 1.0 \times 10^{-6} \, \text{M} $$ So, $[\text{OH}^-] = 1.0 \times 10^{-6}$ M.

5. Calculate pOH: The pOH of the solution is the negative logarithm of the hydroxide ion concentration: $\text{pOH} = -\log[\text{OH}^-] = -\log(1.0 \times 10^{-6}) = 6$

6. Calculate pH: At 25 ${}^\circ$C, the relationship between pH and pOH is: $\text{pH} + \text{pOH} = 14$ $\text{pH} = 14 - \text{pOH} = 14 - 6 = 8$

Alternative Method (Direct Formula):

The pH of a solution of a salt of a weak acid and a strong base can also be calculated directly using the formula: $\text{pH} = 7 + \frac{1}{2}(\text{p}K_a + \log C)$ First, calculate p$K_a$: $\text{p}K_a = -\log K_a = -\log(1.0 \times 10^{-4}) = 4$ The concentration C is 0.01 M. $\log C = \log(0.01) = \log(10^{-2}) = -2$ Now, substitute these values into the formula: $\text{pH} = 7 + \frac{1}{2}(4 + (-2)) = 7 + \frac{1}{2}(2) = 7 + 1 = 8$ Both methods yield the same result.

The final answer is 8.