Question

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

• A. directly proportional to $\left[B^{-}\right]$
• B. inversely proportional to $\left[B^{-}\right]$
• C. directly proportional to the square of $\left[B^{-}\right]$
• D. inversely proportional to the square of $\left[B^{-}\right]$
• E. independent of $\left[B^{-}\right]$

Answer 1

Answer: (D)

inversely proportional to the square of $\left[B^{-}\right]$

$AB\stackrel{K_1}{\rightleftharpoons }{A}^{+}+B^{-}$
$\therefore K_1=\frac{\left[A^{+}\right]\left[B^{-}\right]}{\left[AB\right]}....\left(i\right)$

$AB+B^{-}\stackrel{K_2}{\rightleftharpoons }A{B}_2^{-}$
and $K_2=\frac{\left[A{B}_2^{-}\right]}{\left[AB\right]\left[B^{-}\right]}....\left(ii\right)$

On dividing Eq. (i) from Eq. (ii),

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

or $\frac{\left[A^{+}\right][B^{-}{]}^2}{\left[A{B}_2^{-}\right]}=K^{\prime }$

where $K^{\prime }=\frac{K_1}{K_2}=constant$

or $\frac{\left[A^{+}\right]}{\left[A{B}_2^{-}\right]}=K^{\prime }\times \frac{1}{[B^{-1}{]}^2}$

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

Hence, option D is correct.