Question

For a concetrated solution of a weak electrolyte axby of concentration c , the degree of dissociation is given as

Answer 1

Answer: The dissociation constant for the given monoprotic acid is

Explanation: Dissociation of a weak electrolyte $A_xB_y$ is given by the equation:

                        $A_xB_y\rightleftharpoons xA^{y+}+yB^{x-}$

At $t=0$              c             0        0

At $t=t_{eq}$          $c-c\alpha$      $xc\alpha$       $yc\alpha$

Dissociation constant, $k_a$ is given by:

$k_a=\frac{[A^{y+}]^x[B^{x-}]^y}{[A_xB_y]}$

$k_a=\frac{(xc\alpha)^x(yc\alpha)^y}{c-c\alpha}\\\\k_a=\frac{\alpha^{(x+y)}x^xy^yc^{(x+y-1)}}{(1-\alpha)}\\\\k_a(1-\alpha)=\alpha^{(x+y)}x^xy^yc^{(x+y-1)}\\\\k_a-k_a\alpha=\alpha^{(x+y)}x^xy^yc^{(x+y-1)}\\\\-k_a\alpha=\alpha^{(x+y)}x^xy^yc^{(x+y-1)}-k_a$

$\frac{\alpha}{\alpha^{(x+y)}}=\frac{-(x^xy^yc^{(x+y-1)})+k_a}{k_a}$

$\alpha^{(1-x-y)}=\frac{-(x^xy^yc^{(x+y-1)})+k_a}{k_a}$