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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 \(\mathrm{HgI}_{4}^{2-}.)\)

Short Answer

Expert verified
In the first step, the formation of the orange precipitate occurs due to a double displacement reaction between mercury (II) nitrate (Hg(NO鈧)鈧) and potassium iodide (KI), resulting in insoluble mercury (II) iodide (HgI鈧) and potassium nitrate (KNO鈧): Hg(NO鈧)鈧 + 2 KI 鈫 HgI鈧 + 2 KNO鈧 In the second step, the dissolution of the precipitate occurs as more KI is added, causing the Hg虏鈦 ions to react with I鈦 ions and form HgI鈧劼测伝 ions: HgI鈧 + 2 I鈦 鈫 HgI鈧劼测伝

Step by step solution

01

Write the equation for the formation of the orange precipitate

When aqueous potassium iodide (KI) is added to mercury (II) nitrate (Hg(NO鈧)鈧), a double displacement reaction occurs. This reaction forms mercury (II) iodide (HgI鈧) and potassium nitrate (KNO鈧). The balanced equation for this reaction is: Hg(NO鈧)鈧 + 2 KI 鈫 HgI鈧 + 2 KNO鈧 Since we observe an orange precipitate, we know that the mercury (II) iodide (HgI鈧) is insoluble in water and precipitates out.
02

Write the equation for the dissolution of the orange precipitate

According to the hint, when more KI is added to the solution, the Hg虏鈦 ions react with the I鈦 ions to form HgI鈧劼测伝 ions. This causes the previously formed orange precipitate of HgI鈧 to dissolve. We'll write a balanced equation for this process: HgI鈧 + 2 I鈦 鈫 HgI鈧劼测伝 This equation shows that for every molecule of HgI鈧, two iodide ions (I鈦) are needed to form HgI鈧劼测伝, allowing the precipitate to dissolve. Using these two balanced equations, we now understand both the formation and dissolution of the orange precipitate observed in this exercise.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Precipitation Reaction
In chemistry, a precipitation reaction occurs when two soluble reactants combine to form an insoluble product, known as a precipitate. This process involves the interaction of ions in aqueous solutions, leading to the formation of a solid that settles out of the solution.
Precipitation reactions are a type of double displacement reaction where the cations and anions of two different compounds switch places.
In the provided exercise, when potassium iodide (KI) is added to mercury (II) nitrate (Hg(NO鈧)鈧), the ions in these solutions rearrange to form mercury (II) iodide (HgI鈧), which is the orange precipitate. This precipitate results because HgI鈧 is insoluble in water.
When ions form an insoluble compound, they leave the aqueous solution by forming a solid, which is what happens in this reaction. Observing the formation of a precipitate can be visually exciting and serves as a key indicator of a chemical change. Understanding precipitation reactions is crucial for experiments in both laboratory and industrial settings.
Balanced Equations
A balanced chemical equation is essential for accurately representing a chemical reaction. It ensures that the number of atoms for each element is the same on both sides of the equation, adhering to the Law of Conservation of Mass.
This means that the mass and matter are conserved during the reaction. For the reactions in the exercise, balancing the equations was vital to show how the reactants lead to the formation and dissolution of the precipitate.
The initial formation of the orange precipitate is represented by the balanced equation:
  • Hg(NO鈧)鈧 + 2 KI 鈫 HgI鈧 + 2 KNO鈧
In this equation, notice how each element appears an equal number of times on both sides. Mercury (Hg), iodine (I), and potassium (K) elements are balanced, making the equation valid.
For the dissolution of the precipitate, the following balanced equation is used:
  • HgI鈧 + 2 I鈦 鈫 HgI鈧劼测伝
This equation similarly balances ions, proving that the transformation of substances is consistent with the conservation laws. Understanding the balancing of equations is fundamental for any chemistry student as it forms the basis for quantitative chemical analysis.
Double Displacement Reaction
Double displacement reactions, also known as metathesis reactions, involve the exchange of ions between two reactants to form new compounds. These reactions generally occur in aqueous solutions. During a double displacement reaction, the anions and cations switch partners, leading to the formation of either a precipitate, a gas, or occasionally a neutral molecule like water.
In the provided exercise, the reaction between aqueous KI and mercury (II) nitrate is a classic example of a double displacement reaction. The potassium (K+) from KI exchanges partners with mercury (Hg虏鈦) from Hg(NO鈧)鈧, leading to the formation of HgI鈧 and KNO鈧.
The general form for a double displacement reaction can be described as:
  • AB + CD 鈫 AD + CB
Here, A and C are cations, while B and D are anions, with AD and CB being the new compounds formed.
Double displacement reactions are easily identifiable by the swapping of components and the frequently visible changes like the formation of a precipitate, which can be seen in this exercise with the orange precipitate of HgI鈧.

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Most popular questions from this chapter

Calculate the equilibrium concentrations of \(\mathrm{NH}_{3}, \mathrm{Cu}^{2+}\) \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)^{2+}, \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{2}^{2+}, \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{3}^{2+},\) and \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) in a solution prepared by mixing \(500.0 \mathrm{mL}\) of \(3.00M\) \(\mathrm{NH}_{3}\) with \(500.0 \mathrm{mL}\) of \(2.00 \times 10^{-3} \mathrm{M} \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) . The step wise equilibria are $$\mathrm{Cu}^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{CuNH}_{3}^{2+}(a q) \quad K_{1}=1.86 \times 10^{4}$$ $$\mathrm{CuNH}_{3}^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{2}^{2+}(a q) \quad K_{2}=3.88 \times 10^{3}$$ $$\mathrm{Cu}\left(\mathrm{NH}_{2}\right)_{2}^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{2}\right)_{3}^{2+}(a q) \quad K_{3}=1.00 \times 10^{3}$$ $$\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{3}^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) \quad K_{4}=1.55 \times 10^{2}$$

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