Chemistry - Chemical Equilibrium Question with Solution | TestHub

The equilibrium is:

$$HgO(s)+4I^{-}(aq)+H_{2}O(l)\rightleftharpoons Hg{I}_4^{2-}(aq)+2O{H}^{-}(aq)$$

According to Le Chatelier's Principle:

A. Addition of HI(aq): HI is a strong acid, increasing [I⁻] and decreasing [OH⁻]. Both shifts favor the forward reaction, increasing [HgI₄²⁻].

B. Addition of HgO(s): HgO is a solid, so its concentration does not affect the equilibrium position.

C. Addition of H₂O(l): Dilution by adding water decreases concentrations of all aqueous species. The reaction quotient $Q$ becomes $Q=\frac{[Hg{I}_4^{2-}{]}_0/f\cdot ([O{H}^{-}{]}_0/f{)}^2}{([I^{-}{]}_0/f{)}^4}=fK$. Since $Q>K$, the equilibrium shifts backward, decreasing [HgI₄²⁻].

D. Addition of KOH(aq): Increases [OH⁻], shifting the equilibrium backward, decreasing [HgI₄²⁻].

Therefore, adding H₂O(l) or KOH(aq) will decrease the equilibrium concentration of HgI₄²⁻(aq).