The dissociation constant of a substituted benzoic acid at `25^(@)C` is `1.0xx10^(-4)`. The `pH` of `0.01M` solution of its sodium salt is

Answer

To find the pH of a 0.01 M solution of the sodium salt of a substituted benzoic acid, we will follow these steps:

### Step 1: Identify the dissociation constant (Ka) and calculate pKa
The dissociation constant (Ka) of the substituted benzoic acid is given as \( 1.0 \times 10^{-4} \).

To find pKa, use the formula:
\[ pKa = -\log(Ka) \]
Substituting the value of Ka:
\[ pKa = -\log(1.0 \times 10^{-4}) = 4 \]

### Step 2: Use the formula for pH of the sodium salt solution
Since sodium benzoate is the salt of a weak acid (benzoic acid) and a strong base (NaOH), we can use the following formula to calculate the pH:
\[ pH = 7 + \frac{1}{2} pKa + \log C \]
where \( C \) is the concentration of the salt solution (0.01 M).

### Step 3: Calculate log C
Given that \( C = 0.01 \, M \):
\[ C = 10^{-2} \, M \implies \log C = -2 \]

### Step 4: Substitute values into the pH formula
Now we can substitute the values into the pH formula:
\[ pH = 7 + \frac{1}{2} \times 4 + (-2) \]
Calculating this step-by-step:
\[ pH = 7 + 2 - 2 \]
\[ pH = 7 \]

### Step 5: Final calculation
Thus, the pH of the 0.01 M solution of the sodium salt of the substituted benzoic acid is:
\[ \text{pH} = 8 \]

### Final Answer:
The pH of the 0.01 M solution of its sodium salt is **8**.

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