Question

The equilibrium constant for the disproportionation of $HgC{l}_2$ into $HgC{l}^{+}$ and $HgC{l}_3^{-}$ is ...............
Given, $HgC{l}^{+}+C{l}^{-}\rightleftharpoons HgC{l}_2;K_1=3\times {10}^6$
$HgC{l}_2+C{l}^{-}\rightleftharpoons HgC{l}_3^{-};K_2=9.0$

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

Answer 1

Answer: (D)

$K_1=\frac{\left[HgC{l}_2\right]}{\left[HgC{l}^{+}\right]\left[C{l}^{-}\right]};K_2=\frac{\left[HgC{l}_3^{-}\right]}{\left[HgC{l}_2\right]\left[C{l}^{-}\right]}$
$\therefore K_3=\frac{\left[HgC{l}^{+}\right]\left[HgC{l}_3^{-}\right]}{{\left[HgC{l}_2\right]}^2}=\frac{K_2}{K_1}=\frac{9}{3\times {10}^6}=3\times {10}^{-6}$