Question

The equilibrium constant for the disproportionation of $HgCl_{2}$ into $HgCl^{+}$ and $HgCl_{3}^{-}$ of

Given $HgCl^{+}+Cl^{-} \rightleftharpoons HgCl_{2};K_{1}=3\times 10^{6}$

$HgCl_{2}+Cl^{-} \rightleftharpoons HgCl_{3}^{-};K_{2}=9.0$

• A. $27\times 10^{6}$
• B. $3.3\times 10^{- 6}$
• C. $3.2\times 10^{- 7}$
• D. $3\times 10^{- 6}$

Answer 1

Answer: (D)

$K_{1}=\frac{\left[\right. H g C l_{2} \left]\right.}{\left[\right. H g C l^{+} \left]\right. \left[\right. C l^{-} \left]\right.}$

$K_{2}=\frac{\left[\right. H g C l_{3}^{-} \left]\right.}{\left[\right. H g C l_{2} \left]\right. \left[\right. C l^{-} \left]\right.}$

So $K_{3}=\frac{\left[\right. H g C l^{+} \left]\right. \left[\right. H g C l_{3}^{-} \left]\right.}{\left[\right. H g C l_{2} \left]\right.^{2}}$

$=\frac{K_{2}}{K_{1}}=\frac{9}{3 \times 1 0^{6}}=3\times 10^{- 6}$ .