Question

For a concentrated solution of a weak electrolyte $A_x{B}_y$ By of concentration $C$ , the degree of dissociation $\alpha$ is given as

• A. $\alpha =\sqrt{K_{eq}/C(x+y)}$
• B. $\alpha =\sqrt{K_{eq}C/(xy)}$
• C. $\alpha ={(K_{eq}/C^{x+y-1}{x}^x{y}^y)}^{1/(x+y)}$
• D. $\alpha =(K_{eq}/Cxy)$
• E. $\alpha =(K_{eq}/C^{xy})$

Answer 1

Answer: (C)

The weak electrolyte $A_x{B}_y$ dissociates as follows
$A_x{B}_y\rightleftharpoons x{A}^{y+}+y{B}^{x-}$ $\begin{array}{cccc}C& 0& 0& Initially\\ C(1-\alpha )& xC\alpha & yC\alpha & Atequilibrium\end{array}$
where, $\alpha$ degree of dissociation
$C=$ concentration
$K_{eq}=\frac{{[A^{y+}]}^x{[B^{x-}]}^y}{[A_x{B}_y]}$
$=\frac{{[xC\alpha ]}^x{[yC\alpha ]}^y}{C(1-\alpha )}$
$=\frac{x^x.C^x.{\alpha }^x.y^y.C^y.{\alpha }^y}{C}$ $[\because 1-\alpha \approx 1]$
$=x^x.y^y.{\alpha }^{x+y}.C^{x+y-1}$
$a^{x+y}=\frac{K_{eq}}{x^x.y^y.C^{x+y-1}}$
$\alpha ={\left(\frac{K_{eq}}{x^x.y^y.C^{x+t-1}}\right)}^{\left(\frac{1}{x+y}\right)}$