Question

The pH of \(0.01\) M aqueous aniline solution at \(298\) K is \(8.3\). Its degree of dissociation \((\alpha)\) is \[ (K_b\text{ of aniline}=4\times10^{-10}; \ \text{antilog}(0.7)=5.0; \ \text{antilog}(0.3)=2.0; \ \text{antilog}(0.4)=2.5) \]

• A. \(10^{-4}\)
• B. \(4\times10^{-4}\)
• C. \(1.5\times10^{-4}\)
• D. \(2\times10^{-4}\)

Hint:

For a weak base, \[ \boxed{ [\mathrm{OH^-}]=C\alpha. } \] First calculate \[ \boxed{\mathrm{pOH}=14-\mathrm{pH}} \] and then obtain \([\mathrm{OH^-}]\).

Answer 1

The Correct Option is D

Step 1: Calculate the hydroxide ion concentration. Given, \[ \mathrm{pH}=8.3. \] Hence, \[ \mathrm{pOH}=14-8.3=5.7. \] Therefore, \[ [\mathrm{OH^-}] = 10^{-5.7} = 2\times10^{-6}\ \text{M}. \]

Step 2: Find the degree of dissociation. For a weak base, \[ [\mathrm{OH^-}] = C\alpha. \] Given, \[ C=0.01=10^{-2}\ \text{M}. \] Thus, \[ \alpha = \frac{2\times10^{-6}}{10^{-2}} = 2\times10^{-4}. \]

Step 3: Write the answer. Hence, \[ \boxed{2\times10^{-4}}. \] Thus, \[ \boxed{(D)} \] is the correct answer.