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Q6.52:
What is the pH of 0.001M aniline solution? The ionization constant of aniline can be taken from Table 6.7. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.

Solution :

Step 1: Identifying Constants and Initial Setup

Aniline ($C_6H_5NH_2$) is a weak base. Per standard chemical data (Table 6.7), the ionization constant ($K_b$) of aniline is $4.27 \times 10^{-10}$.

  • Initial concentration ($C$) = $0.001 \, M = 10^{-3} \, M$
  • Ionization constant ($K_b$) = $4.27 \times 10^{-10}$

Step 2: Calculating Degree of Ionization ($\alpha$)

For a weak base, the degree of ionization is calculated using the Ostwald Dilution Law approximation since $\alpha \ll 1$:

$K_b = C\alpha^2 \implies \alpha = \sqrt{\frac{K_b}{C}}$

Substituting the values: $\alpha = \sqrt{\frac{4.27 \times 10^{-10}}{10^{-3}}} = \sqrt{4.27 \times 10^{-7}} = \sqrt{42.7 \times 10^{-8}}$

$\alpha \approx 6.53 \times 10^{-4}$ [Dimensionless quantity representing the fraction of molecules ionized]

Step 3: Calculating Hydroxyl Ion Concentration and pH

The concentration of $[OH^-]$ is given by $[OH^-] = C \cdot \alpha$:

$[OH^-] = (10^{-3} \, M) \cdot (6.53 \times 10^{-4}) = 6.53 \times 10^{-7} \, M$

Calculate $pOH$ [Per the definition $pOH = -\log[OH^-]$]:

$pOH = -\log(6.53 \times 10^{-7}) = 7 - \log(6.53) \approx 7 - 0.815 = 6.185$

Calculate $pH$ [Per the relation $pH + pOH = 14$]:

$pH = 14 - 6.185 = 7.815$

Step 4: Calculating Ionization Constant of the Conjugate Acid ($K_a$)

The relationship between the $K_b$ of a base and the $K_a$ of its conjugate acid is governed by the autoprotolysis constant of water ($K_w$):

$K_a \times K_b = K_w$ where $K_w = 1.0 \times 10^{-14}$ at $298 \, K$.

$K_a = \frac{1.0 \times 10^{-14}}{4.27 \times 10^{-10}}$

$K_a \approx 2.34 \times 10^{-5}$

Visual Representation of Ionization Equilibrium

$C_6H_5NH_2 + H_2O \rightleftharpoons C_6H_5NH_3^+ + OH^-$ Initial: 0.001M | 0 | 0 Equil: 0.001(1-$\alpha$) | 0.001$\alpha$ | 0.001$\alpha$
Parameter Calculated Value
Degree of Ionization ($\alpha$) $6.53 \times 10^{-4}$
pH of Solution $7.815$
$K_a$ (Conjugate Acid) $2.34 \times 10^{-5}$

Final Solution: The degree of ionization is $6.53 \times 10^{-4}$, the pH of the solution is $7.815$, and the ionization constant of the conjugate acid is $2.34 \times 10^{-5}$.


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