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  4. The equilibrium constant for the disproportionation of HgCl2 into HgCl+ and HgCl3- is Given textHgCl+ + textCl- leftharpoons textHgCl2 ; textK1 = 3 × 106 HgCl2+Cl-leftharpoons HgCl3-;K2=9.0

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

Given $\text{HgCl}^{+} + \text{Cl}^{-} \rightleftharpoons \text{HgCl}_{2} ; \text{K}_{1} = 3 \times 10^{6}$

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

NTA AbhyasNTA Abhyas 2020Equilibrium

Solution:

$K_{1}=\frac{\left[\right. H g C l_{2} \left]\right.}{\left[\right. H g C l^{+} \left]\right. \left[\right. C l^{-} \left]\right.};K_{2}=\frac{\left[\right. H g C l_{3}^{-} \left]\right.}{\left[\right. H g C l_{2} \left]\right. \left[\right. C l^{-} \left]\right.}$

So $K_{3}=\frac{\left[\right. H g C l^{+} \left]\right. \left[\right. H g C l_{3}^{-} \left]\right.}{\left[\right. H g C l_{2} \left]\right.^{2}}=\frac{K_{2}}{K_{1}}=\frac{9}{3 \times 1 0^{6}}=3\times 10^{- 6}$