What is the pH of a 0.10 M `C_(6)H_(5)O^(-)` solution? The `K_(a)` of `C_(6)H_(5)OH` is` 1.0xx10^(-10)`

Options

A. `10.51`
B. `11.04`
C. `11.50`
D. `12`

Answer

To find the pH of a 0.10 M solution of phenoxide ion (C₆H₅O⁻), we can follow these steps:

### Step 1: Determine the dissociation constant (K_b) of the phenoxide ion.
The phenoxide ion is the conjugate base of phenol (C₆H₅OH). We can calculate K_b using the relationship between K_a (the dissociation constant of the acid) and K_w (the ion product of water):

\[ K_b = \frac{K_w}{K_a} \]

Given:
- \( K_a = 1.0 \times 10^{-10} \)
- \( K_w = 1.0 \times 10^{-14} \)

Substituting the values:

\[ K_b = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-10}} = 1.0 \times 10^{-4} \]

### Step 2: Calculate the concentration of hydroxide ions (OH⁻).
Using the formula for the concentration of hydroxide ions in a weak base solution:

\[ [OH^-] = \sqrt{K_b \times C} \]

Where:
- \( C = 0.10 \, \text{M} \) (concentration of phenoxide ion)

Substituting the values:

\[ [OH^-] = \sqrt{(1.0 \times 10^{-4}) \times (0.10)} = \sqrt{1.0 \times 10^{-5}} = 3.16 \times 10^{-3} \, \text{M} \]

### Step 3: Calculate the concentration of hydrogen ions (H⁺).
Using the relationship between H⁺ and OH⁻ concentrations:

\[ [H^+] = \frac{K_w}{[OH^-]} \]

Substituting the values:

\[ [H^+] = \frac{1.0 \times 10^{-14}}{3.16 \times 10^{-3}} = 3.16 \times 10^{-12} \, \text{M} \]

### Step 4: Calculate the pH.
The pH is calculated using the formula:

\[ pH = -\log[H^+] \]

Substituting the value of [H⁺]:

\[ pH = -\log(3.16 \times 10^{-12}) \]

Using the logarithmic property:

\[ pH = -\log(3.16) - \log(10^{-12}) = -0.50 + 12 = 11.50 \]

### Final Answer:
The pH of the 0.10 M phenoxide ion solution is approximately **11.5**.

---