Dilute sodium cyanide solution is slowly dripped into a slowly stirred silver nitrate solution. A white precipitate forms temporarily but dissolves as the addition of sodium cyanide continues. Use chemical equations to explain this observation. Silver cyanide is similar to silver chloride in its solubility.

Short Answer

Expert verified
When AgNO3 and NaCN react, AgCN precipitate forms and then dissolves in excess NaCN due to the formation of a soluble complex ion, [Ag(CN)_2]^-.

Step by step solution

01

Identify the Reaction Participants

The first step is to identify the chemical substances involved in the reaction. Here, we have sodium cyanide (NaCN) and silver nitrate (AgNO3).
02

Write the Initial Precipitation Reaction

Write down the chemical equation for the initial reaction between silver nitrate and sodium cyanide, where silver cyanide precipitate (AgCN) forms:\[ AgNO_3(aq) + NaCN(aq) \rightarrow AgCN(s) + NaNO_3(aq) \]
03

Recognize the Solubility of Silver Cyanide

Understand that silver cyanide initially forms a precipitate because it is not very soluble. The equation shows it as a solid (s).
04

Describe the Complex Ion Formation

As more sodium cyanide is added, a soluble complex ion is formed, causing the precipitate to dissolve:\[ AgCN(s) + 2 NaCN(aq) \rightarrow [Ag(CN)_2]^- + Na^+ \]
05

Explain Solubility of Complex Ion

This complex ion, [Ag(CN)_2]^-, is highly soluble in water which is why the precipitate of AgCN dissolves in excess NaCN.

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Key Concepts

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

Chemical Equations
Understanding chemical reactions hinges on the ability to write and balance chemical equations. These equations concisely represent a chemical process, where the reactants transform into products. In the example described, the reaction involves sodium cyanide (NaCN) and silver nitrate (AgNO3), producing silver cyanide (AgCN) and sodium nitrate (NaNO3). The balanced chemical equation for the reaction is:
\[ AgNO_3(aq) + NaCN(aq) \rightarrow AgCN(s) + NaNO_3(aq) \]
It illustrates the principle of conservation of mass, indicating that the number of atoms for each element is the same on both sides. Chemical equations also depict the physical states of the substances (solid, liquid, aqueous, or gas) which is crucial in predicting the outcomes of chemical reactions.
Precipitation Reactions
Precipitation reactions involve the formation of an insoluble solid, called a precipitate, when two aqueous solutions combine. In the provided exercise, when a solution of sodium cyanide is added to silver nitrate, a white precipitate of silver cyanide forms, as illustrated by the reaction equation above. However, it's key to note that precipitation is influenced by the solubility of the compounds formed during the reaction. In some cases, like with the addition of excess sodium cyanide, the precipitate may dissolve, indicating the involvement of additional chemical processes, such as complex ion formation.
Complex Ion Formation
Complex ions are charged species consisting of a central metal ion surrounded by one or more ligands, often resulting in increased solubility of typically insoluble metals. The observed phenomenon in our exercise is an excellent example. After the initial formation of solid silver cyanide, additional sodium cyanide acts as a ligand to form a soluble complex ion:
\[ AgCN(s) + 2 NaCN(aq) \rightarrow [Ag(CN)_2]^- + Na^+ \]
This process highlights the dynamic nature of chemical reactions where the product from one reaction can further react under certain conditions, altering the outcome. Complex ion formation can prevent or reverse precipitation, which is why the initial precipitate dissolves upon continuous addition of sodium cyanide.
Chemical Solubility
Chemical solubility is a measure of how much of a solute can dissolve in a solvent to form a homogeneous mixture. In many cases, ionic compounds like silver cyanide might have low solubility, leading to precipitation when two soluble salts react. However, solubility can be altered significantly by complex ion formation. As the solution becomes enriched with the ligand, sodium cyanide in this case, the formation of a soluble complex ion, [Ag(CN)2]-, increases the solubility of silver cyanide. This change in solubility upon the formation of complex ions is a crucial concept when predicting the behavior of ionic species in various chemical environments.

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Most popular questions from this chapter

Predict the products of the following reactions and balance the equations. (a) \(\mathrm{Zn}\) is added to a solution of \(\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) in acid. (b) \(\mathrm{FeCl}_{2}\) is added to a solution containing an excess of \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) in hydrochloric acid. (c) \(\mathrm{Cr}^{2+}\) is added to \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) in acid solution. (d) Mn is heated with CrO \(_{3}\). (e) CrO is added to \(2 \mathrm{HNO}_{3}\) in water. (f) \(\mathrm{FeCl}_{3}\) is added to an aqueous solution of \(\mathrm{NaOH}\).

Is it possible for a complex of a metal in the transition series to have six unpaired electrons? Explain.

Give the coordination number for each metal ion in the following compounds: (a) \(\left[\mathrm{Co}\left(\mathrm{CO}_{3}\right)_{3}\right]^{3-}\) (note that \(\mathrm{CO}_{3}^{2-}\) is bidentate in this complex) (b) \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) (c) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Br}_{2}\right]_{2}\left(\mathrm{SO}_{4}\right)_{3}\) (d) \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4}\right]\left[\mathrm{PtCl}_{4}\right]\) (e) \(\left[\mathrm{Cr}(\mathrm{en})_{3}\right]\left(\mathrm{NO}_{3}\right)_{3}\) (f) \(\left[\mathrm{Pd}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Br}_{2}\right]\) (square planar) (g) \(\mathrm{K}_{3}\left[\mathrm{Cu}(\mathrm{Cl})_{5}\right]\) (h) \(\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]\)

Give the oxidation state of the metal, number of \(d\) electrons, and the number of unpaired electrons predicted for \(\left.\mathrm{I} \mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}\).

Sketch the structures of the following complexes. Indicate any cis, trans, and optical isomers. (a) \(\left[\mathrm{Pt}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2} \mathrm{Br}_{2}\right]\) (square planar) (b) \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)(\mathrm{py})(\mathrm{Cl})(\mathrm{Br})\right]\) (square planar, py = pyridine, \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\) ) (c) \(\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}\right]^{+}\) (tetrahedral) (d) \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}\right]^{+}\) (square planar) (e) \(\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\right]\) \((\mathrm{f})\left[\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2} \mathrm{Cl}_{2}\right]^{3-}\) (note that \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) is the bidentate oxalate ion, \(-\mathrm{O}_{2} \mathrm{CCO}_{2}^{-}\) )

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