JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
Given is a concentrated solution of a weak electrolyte \(A_xB_y\) of concentration 'c' and dissociation constant 'K'. The degree of dissociation is given by :
- A \(\left[K \times c^{x+y-1} x^x y^y\right]^{x+y}\)
- B \(\left(\dfrac{K}{c^{x+y-1} x^x y^y}\right)^{\frac{1}{x+y}}\)
- C \(\left(\dfrac{c^{x+y-1} x^x y^y}{K}\right)^{x+y}\)
- D \(\left(\dfrac{K}{c^{x+y-1} x^x y^y}\right)^{\frac{1}{x+y}}\)
Answer & Solution
Correct Answer
(B) \(\left(\dfrac{K}{c^{x+y-1} x^x y^y}\right)^{\frac{1}{x+y}}\)
Step-by-step Solution
Detailed explanation
The dissociation reaction of the weak electrolyte \(A_xB_y\) is: \(A_xB_y \rightleftharpoons xA^{y+} + yB^{x-}\) Initial concentrations: \(c \quad 0 \quad 0\) Equilibrium concentrations: \(c(1-\alpha) \quad cx\alpha \quad cy\alpha\) The dissociation constant \(K\) is given by:…
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