Question

The two equilibria $AB\rightleftharpoons A^{+}+B^{-}$ and $AB+B^{-}\rightleftharpoons A{B}_2^{-}$ are simultaneously maintained in a solution with equilibrium constants, $K_1$ and $K_2$ respectively. The ratio of $A^{+}$ to ${AB}_2^{-}$ in the solution is

• A. inversely proportional to the square of the concentration of $B^{-}$
• B. directly proportional to the square of the concentration of $B^{-}$
• C. inversely proportional to the concentration of $B^{-}$
• D. directly proportional to the concentration of $B^{-}$

Answer 1

The correct option is A inversely proportional to the square of the concentration of $B^{-}$

Given the reactions:
$AB\rightleftharpoons A^{+}+B^{-}$ and $AB+B^{-}\rightleftharpoons A{B}_2^{-}$

$K_1=\frac{\left[A^{+}\right]\left[B^{-}\right]}{\left[AB\right]};K_2=\frac{\left[{AB}_2^{-}\right]}{\left[AB\right]\left[B^{-}\right]}$

$\therefore \frac{K_1}{K_2}=\frac{\left[A^{+}\right]{\left[B^{-}\right]}^2}{{\left[{AB}_2\right]}^{-}}$

$\therefore \frac{\left[A^{+}\right]}{\left[{AB}_2\right]}\propto \frac{1}{{\left[B^{-}\right]}^2}$

Hence, the ratio of $A^{+}$ to ${AB}_2^{-}$ in the solution is inversely proportional to the square of the concentration of $B^{-}$.