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When an aqueous solution of KCN is added to a solution containing \(\mathrm{Ni}^{2+}\) ions, a precipitate forms, which redissolves on addition of more \(\mathrm{KCN}\) solution. Write reactions describing what happens in this solution. [Hint: \(\mathrm{CN}^{-}\) is a Bronsted-Lowry base \(\left(K_{\mathrm{b}} \approx 10^{-5}\right)\) and a Lewis base.]

Short Answer

Expert verified
First, the precipitation reaction occurs between Ni²⁺ ions and CN⁻ ions, forming a solid precipitate: \[ \mathrm{Ni^{2+}(aq) + 2CN^-(aq) \rightarrow Ni(CN)_2(s)} \] Upon addition of more KCN, the CN⁻ ions act as Bronsted-Lowry and Lewis bases, generating OH⁻ ions and water-soluble complex species, causing the precipitate to redissolve: \[ \mathrm{Ni(CN)_2(s) + 4CN^-(aq) + H_2O(l) \leftrightharpoons [Ni(CN)_6]^{4-}(aq) + HCN(aq) + OH^-(aq)} \]

Step by step solution

01

Formation of precipitate

First, we have the precipitation reaction between Ni²⁺ ions and CN⁻ ions, which results in the formation of a solid precipitate. The reaction can be represented as follows: \[ \mathrm{Ni^{2+}(aq) + 2CN^-(aq) \rightarrow Ni(CN)_2(s)} \]
02

Dissolution of precipitate

Since CN⁻ is a Bronsted-Lowry base, it can accept a proton from water molecules, forming OH⁻ ions. The generated OH⁻ ions can react with a water-soluble complex species formed by Ni(CN)₂ and additional CN⁻ ions. The reactions in this step are: First, the base reaction: \[ \mathrm{CN^-(aq) + H_2O(l) \leftrightharpoons HCN(aq) + OH^-(aq)} \] Then, the dissolution reaction: \[ \mathrm{Ni(CN)_2(s) + 4CN^-(aq) \leftrightharpoons [Ni(CN)_6]^{4-}(aq)} \]
03

Combined reaction

We can now combine these reactions to represent the overall process when more KCN solution is added: \[ \mathrm{Ni(CN)_2(s) + 4CN^-(aq) + H_2O(l) \leftrightharpoons [Ni(CN)_6]^{4-}(aq) + HCN(aq) + OH^-(aq)} \] These are the reactions describing what happens in the solution during the process. The precipitate forms at first and then redissolves when more KCN is added, showing the behavior of CN⁻ ions as both Bronsted-Lowry and Lewis bases.

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Key Concepts

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

Bronsted-Lowry base
A Bronsted-Lowry base is any substance that can accept a proton (H⁺). In our scenario, the cyanide ion (\(\mathrm{CN}^-\)) acts as a Bronsted-Lowry base. When it encounters water, \(\mathrm{CN}^-\) can accept protons from water molecules (\(\mathrm{H_2O}\)). This reaction produces hydroxide ions (\(\mathrm{OH}^-\)) and hydrogen cyanide (\(\mathrm{HCN}\)).
  • The reaction: \[ \mathrm{CN^-(aq) + H_2O(l) \leftrightharpoons HCN(aq) + OH^-(aq)} \] is balanced and shows the base property of \(\mathrm{CN}^-\).
  • \(\mathrm{K_b}\) for \(\mathrm{CN^-}\) is approximately \(10^{-5}\), indicating it's a weak base. Although it's not as strong as typical bases like hydroxide itself, it effectively interacts with water.
This ability to accept protons and produce hydroxide ions is important. It helps change the conditions of the solution, playing a critical role in dissolving the precipitate formed during the reaction.
Lewis base
A Lewis base, unlike the Bronsted-Lowry base, is defined by its ability to donate a pair of electrons. The \(\mathrm{CN}^-\) ion also acts as a Lewis base. It donates its electron pair to \(\mathrm{Ni^{2+}}\) ions, thereby forming a complex. This electron pair donation leads to the formation of \(\mathrm{Ni(CN)_2}\), a solid precipitate.
  • In the initial reaction, \[ \mathrm{Ni^{2+}(aq) + 2CN^-(aq) \rightarrow Ni(CN)_2(s)} \] \(\mathrm{CN}^-\) provides an electron pair to \(\mathrm{Ni^{2+}}\), illustrating its nature as a Lewis base.
Though \(\mathrm{CN}^-\) starts as a precipitate forming agent, when further \(\mathrm{CN^-}\) ions are added, it can further coordinate with \(\mathrm{Ni^{2+}}\) to form a soluble complex.
This highlights the dual role of \(\mathrm{CN^-}\) as both a Bronsted-Lowry and Lewis base, enabling varying interactions for complexation reactions.
Precipitation reaction
A precipitation reaction occurs when ions in solution form an insoluble solid (the precipitate). In this exercise, \(\mathrm{Ni^{2+}}\) ions meet \(\mathrm{CN^-}\) ions, resulting in the precipitation reaction:
\[ \mathrm{Ni^{2+}(aq) + 2CN^-(aq) \rightarrow Ni(CN)_2(s)} \]
  • The product, \(\mathrm{Ni(CN)_2}\), is an insoluble solid that precipitates out of the solution.
The formation of a precipitate indicates a chemical change where specific conditions favor the formation of specific compounds.
In such reactions, solubility rules and the presence of a complexation agent like \(\mathrm{CN^-}\) can determine whether precipitation occurs or not.
Dissolution reaction
Dissolution reactions involve breaking down a precipitate back into its ionic components. With additional \(\mathrm{CN^-}\), \(\mathrm{Ni(CN)_2}\) redissolves by forming a complex, shown in:
\[ \mathrm{Ni(CN)_2(s) + 4CN^-(aq) \leftrightharpoons [Ni(CN)_6]^{4-}(aq)} \]
  • This equilibrium shows the breakdown of \(\mathrm{Ni(CN)_2}\) into a soluble species \(\mathrm{[Ni(CN)_6]^{4-}}\), suggesting complex formation.
  • The reaction reaches equilibrium as long as there are enough \(\mathrm{CN^-}\) ions.
The interacting \(\mathrm{CN^-}\) ions stabilize \(\mathrm{Ni}^{2+}\) in solution, shifting conditions from solid to liquid form. This dissolving action highlights how adding more reactant shifts equilibrium and underscores the reversible nature of many chemical reactions.

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Most popular questions from this chapter

Name the following complex ions. a. \(\operatorname{Ru}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}^{2+}\) b. \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}\) c. \(\mathrm{Mn}\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\right)_{3}^{2+}\) d. \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{55} \mathrm{NO}_{2}^{2+}\) e. \(\mathrm{Ni}(\mathrm{CN})_{4}^{2-}\) f. \(\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}^{+}\) g. \(\mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}^{3-}\) h. \(\mathrm{Co}(\mathrm{SCN})_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}^{+}\)

Which of the following ligands are capable of linkage isomerism? Explain your answer. $$\mathrm{SCN}^{-}, \mathrm{N}_{3}^{-}, \mathrm{NO}_{2}^{-}, \mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}, \mathrm{OCN}^{-}, \mathrm{I}^{-}$$

One of the classic methods for the determination of the manganese content in steel involves converting all the manganese to the deeply colored permanganate ion and then measuring the absorption of light. The steel is first dissolved in nitric acid, producing the manganese(II) ion and nitrogen dioxide gas. This solution is then reacted with an acidic solution containing periodate ion; the products are the permanganate and iodate ions. Write balanced chemical equations for both of these steps.

Which of the following molecules exhibit(s) optical isomerism? a. \(c i s-P t\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\) b. trans-Ni(en) \(_{2} \mathrm{Br}_{2}\) (en is ethylenediamine) c. \(c i s-\mathrm{Ni}(\mathrm{en})_{2} \mathrm{Br}_{2}(\text { en is ethylenediamine })\)

\(\mathrm{CoCl}_{4}^{2-}\) forms a tetrahedral complex ion and \(\mathrm{Co}(\mathrm{CN})_{6}^{3-}\) forms an octahedral complex ion. What is wrong about the following statements concerning each complex ion and the \(d\) orbital splitting diagrams? a. \(\mathrm{CoCl}_{4}^{2-}\) is an example of a strong-field case having two unpaired electrons. b. Because \(\mathrm{CN}^{-}\) is a weak-field ligand, \(\mathrm{Co}(\mathrm{CN})_{6}^{3-}\) will be a low-spin case having four unpaired electrons.

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