Question

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

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

Answer 1

Answer: (D)

inversely proportional to the square of the concentration of $B^{-}$

$AB\rightleftharpoons A^{+}+B^{-}$

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

i.e. $\left[AB\right]=\left[A^{+}\right]\left[B^{-}\right]$

$AB+B^{-}\rightleftharpoons {A{B}_2}^{-}$

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

So, $K_2=\frac{\left[A{B}_2^{-}\right]}{\left[A^{+}\right]\left[B^{-}\right]\left[B^{-}\right]}$

Now, $\frac{\left[A^{+}\right]}{\left[A{B}_2^{-}\right]}=\frac{K_1}{[B^{-}{]}^2\times K_2}$

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