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Chapter 15: Problem 65

When aqueous KI is added gradually to mercury(II) nitrate, an orange precipitate forms. Continued addition of KI causes the precipitate to dissolve. Write balanced equations to explain these observations. (Hint: \(\mathrm{Hg}^{2+}\) reacts with \(\mathrm{I}^{-}\) to form \(\mathrm{Hg} \mathrm{I}_{4}^{2-} .\) )

Short Answer

Expert verified
The two balanced equations to explain the observations are as follows: 1. Formation of the orange precipitate: \(Hg(NO_3)_2(aq) + 2KI(aq) \rightarrow HgI_2(s) + 2KNO_3(aq)\) 2. Dissolution of the orange precipitate: \(HgI_2(s) + 2I^-(aq) \rightarrow HgI_4^{2-}(aq)\)

Step by step solution

01

Write the balanced equation for the formation of the orange precipitate.

For the formation of the orange precipitate, the reaction involves the combination of aqueous KI (potassium iodide) with mercury(II) nitrate (Hg(NO₃)₂). Adding KI gradually to Hg(NO₃)₂ will lead to the formation of the orange precipitate, which is mercury(II) iodide (HgI₂). Potassium nitrate (KNO₃) will also be formed as a byproduct. We can write the balanced chemical equation as follows: Hg(NO₃)₂(aq) + 2KI(aq) → HgI₂(s) + 2KNO₃(aq) The equation is balanced since we have the same number of each element on both sides.
02

Write the balanced equation for the dissolution of the orange precipitate.

With further addition of aqueous KI, the orange precipitate dissolves. This is due to the formation of a soluble complex ion, HgI₄²⁻, as per the hint given in the exercise. The balanced equation for this reaction can be written as follows: HgI₂(s) + 2I⁻(aq) → HgI₄²⁻(aq) The equation is balanced because we have equal numbers of each element on both sides of the equation.

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