The ionization constant of phenol is `1.0xx10^(-10)`. What is the concentration of phenolate ion in `0.05 M` solution of phenol? What will be its degree of ionization if the solution is also `0.01 M` in sodium phenolate?

Explanation

To solve the problem, we need to find the concentration of phenolate ion in a 0.05 M solution of phenol and determine its degree of ionization when the solution is also 0.01 M in sodium phenolate.

### Step-by-Step Solution:

1. **Understanding the Ionization of Phenol:**
The ionization of phenol (C6H5OH) can be represented as:
\[ \text{C}_6\text{H}_5\text{OH} \rightleftharpoons \text{C}_6\text{H}_5\text{O}^- + \text{H}^+ \]
Here, phenol ionizes to form phenolate ions (C6H5O^-) and hydrogen ions (H^+).

2. **Setting Up the Equilibrium Expression:**
The ionization constant (Ka) for phenol is given as \(1.0 \times 10^{-10}\). The equilibrium expression can be written as:
\[ K_a = \frac{[\text{C}_6\text{H}_5\text{O}^-][\text{H}^+]}{[\text{C}_6\text{H}_5\text{OH}]} \]

3. **Initial Concentrations:**
In a 0.05 M solution of phenol, initially:
- \([\text{C}_6\text{H}_5\text{OH}] = 0.05\) M
- \([\text{C}_6\text{H}_5\text{O}^-] = 0\) M
- \([\text{H}^+] = 0\) M

4. **Change in Concentrations:**
Let \(x\) be the amount of phenol that ionizes. At equilibrium:
- \([\text{C}_6\text{H}_5\text{OH}] = 0.05 - x\)
- \([\text{C}_6\text{H}_5\text{O}^-] = x\)
- \([\text{H}^+] = x\)

5. **Substituting into the Equilibrium Expression:**
Substituting these values into the equilibrium expression gives:
\[ 1.0 \times 10^{-10} = \frac{x \cdot x}{0.05 - x} \]
Assuming \(x\) is very small compared to 0.05, we can approximate:
\[ 1.0 \times 10^{-10} \approx \frac{x^2}{0.05} \]

6. **Solving for \(x\):**
Rearranging gives:
\[ x^2 = 1.0 \times 10^{-10} \times 0.05 \]
\[ x^2 = 5.0 \times 10^{-12} \]
\[ x = \sqrt{5.0 \times 10^{-12}} \approx 2.24 \times 10^{-6} \text{ M} \]

7. **Concentration of Phenolate Ion:**
Therefore, the concentration of the phenolate ion in the 0.05 M solution of phenol is:
\[ [\text{C}_6\text{H}_5\text{O}^-] = 2.24 \times 10^{-6} \text{ M} \]

8. **Degree of Ionization Calculation:**
Now, if the solution is also 0.01 M in sodium phenolate, the initial concentration of phenolate ion is:
\[ [\text{C}_6\text{H}_5\text{O}^-] = 0.01 \text{ M} \]
The total concentration of phenolate ion at equilibrium will be:
\[ [\text{C}_6\text{H}_5\text{O}^-] = 0.01 + x \approx 0.01 \text{ M} \]

9. **New Equilibrium Expression:**
The new equilibrium expression becomes:
\[ 1.0 \times 10^{-10} = \frac{x \cdot x}{0.05 - x} \]
Again, assuming \(x\) is small:
\[ 1.0 \times 10^{-10} \approx \frac{x^2}{0.05} \]
Solving for \(x\):
\[ x^2 = 1.0 \times 10^{-10} \times 0.05 \]
\[ x^2 = 5.0 \times 10^{-12} \]
\[ x = \sqrt{5.0 \times 10^{-12}} \approx 2.24 \times 10^{-6} \text{ M} \]

10. **Calculating Degree of Ionization:**
The degree of ionization (\(\alpha\)) is given by:
\[ \alpha = \frac{x}{[\text{C}_6\text{H}_5\text{OH}]_0} = \frac{2.24 \times 10^{-6}}{0.05} = 4.48 \times 10^{-5} \]

### Final Results:
- The concentration of phenolate ion in a 0.05 M solution of phenol is approximately \(2.24 \times 10^{-6}\) M.
- The degree of ionization when the solution is also 0.01 M in sodium phenolate is approximately \(4.48 \times 10^{-5}\).