Question

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. \( \left(\frac{K}{c^{x+y-1}x^xy^y}\right)^{\frac{1}{x+y}} \)
• B. \( \left(\frac{Kx^xy^y}{c^{x+y-1}}\right)^{\frac{1}{x+y}} \)
• C. \( \left(\frac{c^{x+y-1}x^xy^y}{K}\right)^{x+y} \)
• D. \( \left(\frac{c^{x+y-1}}{Kx^xy^y}\right)^{\frac{1}{x+y}} \)

Answer 1

The Correct Option is A

Concept: For weak electrolyte: \[ A_xB_y \rightleftharpoons xA^{y+} + yB^{x-} \] If initial concentration is \(c\) and degree of dissociation is \(\alpha\): \[ [A_xB_y] = c(1-\alpha) \] \[ [A^{y+}] = cx\alpha \] \[ [B^{x-}] = cy\alpha \] Step 1: {Write equilibrium expression.} \[ K=\frac{(cx\alpha)^x(cy\alpha)^y}{c(1-\alpha)} \] \[ K=\frac{c^{x+y}x^xy^y\alpha^{x+y}}{c} \] \[ K=c^{x+y-1}x^xy^y\alpha^{x+y} \] Step 2: {Solve for \(\alpha\).} \[ \alpha^{x+y}=\frac{K}{c^{x+y-1}x^xy^y} \] \[ \alpha=\left(\frac{K}{c^{x+y-1}x^xy^y}\right)^{\frac{1}{x+y}} \]