For a weak electrolyte \( A B \) as dilution increases : \( (\alpha= \) degree of ionisation) (1) \( \alpha \) increases, \( \left[A^{+}\right] \)increases. \( A \) decreases, \( \left[A^{+}\right] \)decreasest \( \rightarrow A^{+}+B_{\alpha}^{-} \) (3) \( \alpha \) increases, \( \left[A^{+}\right] \)increases but \( [B] \) decreases (4) \( \alpha \) increases, \( \left[A^{+}\right] \)decreases

Question

For a weak electrolyte \( A B \) as dilution increases : \( (\alpha= \) degree of ionisation)

 (1) \( \alpha \) increases, \( \left[A^{+}\right] \)increases. 

(2) \( \alpha \) decreases, \( \left[A^{+}\right] \)decreases

 (3) \( \alpha \) increases, \( \left[A^{+}\right] \)increases but [B^-] decreases

 (4) \( \alpha\) increases, \( \left[A^{+}\right] \)decreases

Answer 1

\( A B \rightleftharpoons A^{+}(a q)+B^{-}(a q) \)

         C(1–\( \alpha \) )           C\( \alpha \)            C\( \alpha \)

as dilution increases for weak electrolyte,

\( \alpha \uparrow \quad c \alpha \downarrow \)

Hence, option (4) is correct

Answer 2

\(k=\frac{c^2\alpha^2}{c(1-\alpha)}\)

∵ AB is weak electrolyte then α << 1

∴ 1 - α ≈ 1

k = cα²

\(\alpha=\sqrt{\frac{k}{c}}\)

Here we can conclude that as concentration (c) of electrolyte decreases, α increases.

∵ [A⁺] cα , it means as degree of ionisation increases. concentration of A⁺ also increases. it 'α' decrease, then A⁺ also decrease.

Hence, option (A) and (B) both are right answer.