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Chapter 20: Problem 38

When aqueous potassium cyanide is added to a solution of copper(II) sulfate, a white precipitate, soluble in an excess of potassium cyanide, is formed. No precipitate is formed when hydrogen sulfide is bubbled through the solution at this point. Explain.

Short Answer

Expert verified
When potassium cyanide is added to a copper(II) sulfate solution, it forms a white precipitate which is a complex medallion Cu(CN)4^2-. This complex is soluble in an excess of KCN forming K3Cu(CN)4. When hydrogen sulfide is passed through the solution, no precipitate forms because the CN- ion strongly complexes with Cu2+, preventing its reaction with H2S.

Step by step solution

01

Understanding Chemical Behaviors

Potassium Cyanide (KCN) behaves as a reducing agent and can dissolve many metal ions. When it is added to a copper(II) sulfate solution (CuSO4), it reacts to form a complex medallion Cu(CN)4^2-. This complex ion appears as a white precipitate.
02

Explaining Solubility

The complex formed Cu(CN)4^2- is soluble in an excess of potassium cyanide, implying that more KCN can react with the complex to form a soluble compound. The reaction is: Cu(CN)4^2- + KCN --> K3Cu(CN)4.
03

Reaction with Hydrogen Sulfide

Usually, hydrogen sulfide (H2S) forms precipitates with many metal ions, but in this case, no precipitate is formed when it is passed through the solution. This can be explained by the strong complex forming ability of CN- ion which prevents Cu2+ ions from reacting with H2S to form any precipitate.

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Most popular questions from this chapter

Aqueous copper(II) sulfate solution is blue in color. When aqueous potassium fluoride is added, a green precipitate is formed. When aqueous potassium chloride is added instead, a bright-green solution is formed. Explain what is happening in these two cases.

In a dilute nitric acid solution, \(\mathrm{Fe}^{3+}\) reacts with thiocyanate ion \(\left(\mathrm{SCN}^{-}\right)\) to form a dark-red complex: $$\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}+\mathrm{SCN}^{-} \rightleftharpoons \mathrm{H}_{2} \mathrm{O}+\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NCS}\right]^{2+}$$ The equilibrium concentration of \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NCS}\right]^{2+}\) may be determined by how darkly colored the solution is (measured by a spectrometer). In one such experiment, \(1.0 \mathrm{~mL}\) of \(0.20 \mathrm{M} \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\) was mixed with \(1.0 \mathrm{~mL}\) of \(1.0 \times 10^{-3} M \mathrm{KSCN}\) and \(8.0 \mathrm{~mL}\) of dilute \(\mathrm{HNO}_{3}\). The color of the solution quantitatively indicated that the \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NCS}\right]^{2+}\) concentration was \(7.3 \times 10^{-5} M\). Calculate the formation constant for \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NCS}\right]^{2+}\)

Draw qualitative diagrams for the crystal-field splittings in (a) a linear complex ion \(\mathrm{ML}_{2},\) (b) a trigonalplanar complex ion \(\mathrm{ML}_{3},\) and \((\mathrm{c})\) a trigonalbipyramidal complex ion ML \(_{5}\).

What factors determine whether a given complex will be diamagnetic or paramagnetic?

Complex ion formation has been used to extract gold, which exists in nature in the uncombined state. To separate it from other solid impurities, the ore is treated with a sodium cyanide (NaCN) solution in the presence of air to dissolve the gold by forming the soluble complex ion \(\left[\mathrm{Au}(\mathrm{CN})_{2}\right]^{-}\). (a) Balance the following equation: $$\mathrm{Au}+\mathrm{CN}^{-}+\mathrm{O}_{2}+\mathrm{H}_{2} \mathrm{O} \longrightarrow\left[\mathrm{Au}(\mathrm{CN})_{2}\right]^{-}+\mathrm{OH}^{-}$$ (b) The gold is obtained by reducing the complex ion with zinc metal. Write a balanced ionic equation for this process. (c) What is the geometry and coordination number of the \(\left[\mathrm{Au}(\mathrm{CN})_{2}\right]^{-}\) ion?
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