The dissociation constant of aniline, `C_6H_5NH_2` is `4.3 xx 10^(-10) mol L^(-1)`. Calculate the pH of a 0.01 M solution of aniline in water.

Answer

To calculate the pH of a 0.01 M solution of aniline, we will follow these steps:

### Step 1: Write the dissociation equation for aniline.
Aniline (C₆H₅NH₂) can dissociate in water as follows:
\[ C_6H_5NH_2 + H_2O \rightleftharpoons C_6H_5NH_3^+ + OH^- \]

### Step 2: Set up the expression for the dissociation constant (Kₐ).
The dissociation constant (Kₐ) for the reaction is given by:
\[ K_a = \frac{[C_6H_5NH_3^+][OH^-]}{[C_6H_5NH_2]} \]
Where:
- \([C_6H_5NH_3^+]\) is the concentration of the conjugate acid,
- \([OH^-]\) is the concentration of hydroxide ions,
- \([C_6H_5NH_2]\) is the concentration of aniline.

### Step 3: Define the initial concentrations and changes.
Let the initial concentration of aniline be \(C = 0.01 \, \text{M}\).
When \( \alpha \) is the degree of dissociation, at equilibrium:
- \([C_6H_5NH_2] = C - C\alpha \approx C\) (since \(\alpha\) is small),
- \([C_6H_5NH_3^+] = C\alpha\),
- \([OH^-] = C\alpha\).

### Step 4: Substitute into the Kₐ expression.
Substituting the equilibrium concentrations into the Kₐ expression:
\[ K_a = \frac{(C\alpha)(C\alpha)}{C - C\alpha} \approx \frac{C\alpha^2}{C} = C\alpha^2 \]
Given that \(K_a = 4.3 \times 10^{-10} \, \text{mol L}^{-1}\) and \(C = 0.01 \, \text{M}\):
\[ 4.3 \times 10^{-10} = 0.01 \cdot \alpha^2 \]

### Step 5: Solve for \(\alpha\).
Rearranging gives:
\[ \alpha^2 = \frac{4.3 \times 10^{-10}}{0.01} = 4.3 \times 10^{-8} \]
Taking the square root:
\[ \alpha = \sqrt{4.3 \times 10^{-8}} \approx 2.07 \times 10^{-4} \]

### Step 6: Calculate the concentration of hydroxide ions \([OH^-]\).
Using \([OH^-] = C\alpha\):
\[ [OH^-] = 0.01 \times 2.07 \times 10^{-4} = 2.07 \times 10^{-6} \, \text{M} \]

### Step 7: Calculate pOH.
Using the formula for pOH:
\[ pOH = -\log[OH^-] = -\log(2.07 \times 10^{-6}) \approx 5.68 \]

### Step 8: Calculate pH.
Using the relationship \(pH + pOH = 14\):
\[ pH = 14 - pOH = 14 - 5.68 = 8.32 \]

### Final Answer:
The pH of a 0.01 M solution of aniline in water is approximately **8.32**.

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