JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
For a concentrated solution of a weak electrolyte ( \(K _{ eq }=\) equilibrium constant) \(A _2 B _3\) of concentration ' \(c\) ', the degree of dissociation " \(\alpha\) ' is
- A \(\left(\frac{ K _{ eq }}{108 c ^4}\right)^{\frac{1}{5}}\)
- B \(\left(\frac{ K _{ eq }}{6 c ^5}\right)^{\frac{1}{5}}\)
- C \(\left(\frac{K_{e q}}{5 c^4}\right)^{\frac{1}{5}}\)
- D \(\left(\frac{ K _{ eq }}{25 c ^2}\right)^{\frac{1}{5}}\)
Answer & Solution
Correct Answer
(A) \(\left(\frac{ K _{ eq }}{108 c ^4}\right)^{\frac{1}{5}}\)
Step-by-step Solution
Detailed explanation
\(A _2 B _3 \text { (aq.) } \rightleftharpoons 2 A _{\text {(aq.) }}^{3+}+3 B _{\text {(aq) }}^{2-}\) \(K _{ eq }=\frac{\left[ A ^{3+}\right]^2\left[ B ^{2-}\right]^3}{\left[ A _2 B _3\right]}=\frac{4 c ^2 \alpha^2 \times 27 c ^3 \alpha^3}{ c (1-\alpha)}\)…
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