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Chapter 16: Problem 65

Write equations for the step wise formation of each of the following complex ions. a. \(N i(C N)_{4}^{2-}\) b. \(V\left(C_{2} O_{4}\right)_{3}^{3-}\)

Short Answer

Expert verified
The stepwise formation of the complex ions is as follows: a. Formation of \(Ni(CN)_{4}^{2-}\): \(Ni^{2+} + CN^- \rightarrow NiCN\) \(NiCN + CN^- \rightarrow Ni(CN)_{2}\) \(Ni(CN)_{2} + CN^- \rightarrow Ni(CN)_{3}^-\) \(Ni(CN)_{3}^- + CN^- \rightarrow Ni(CN)_{4}^{2-}\) b. Formation of \(V(C_{2}O_{4})_{3}^{3-}\): \(V^{3+} + C_{2}O_{4}^{2-} \rightarrow VC_{2}O_{4}^{+}\) \(VC_{2}O_{4}^{+} + C_{2}O_{4}^{2-} \rightarrow V(C_{2}O_{4})_{2}^{-}\) \(V(C_{2}O_{4})_{2}^{-} + C_{2}O_{4}^{2-} \rightarrow V(C_{2}O_{4})_{3}^{3-}\)

Step by step solution

01

Identify the central metal ion and the ligands

For the complex ion \(Ni(CN)_{4}^{2-}\), the central metal ion is Nickel (Ni) and the ligand is Cyanide (CN^-).
02

Formation of the first coordination bond

In the first step, one Cyanide ligand will form a coordination bond with the Nickel ion: \[Ni^{2+} + CN^- \rightarrow NiCN\]
03

Formation of the second coordination bond

In the next step, another Cyanide ligand will form a coordination bond with the Nickel ion: \[NiCN + CN^- \rightarrow Ni(CN)_{2}\]
04

Formation of the third coordination bond

Now, the third Cyanide ligand will form a coordination bond with the Nickel ion: \[Ni(CN)_{2} + CN^- \rightarrow Ni(CN)_{3}^-\]
05

Formation of the fourth coordination bond

Finally, the last Cyanide ligand will form a coordination bond with the Nickel ion, resulting in the formation of the complex ion: \[Ni(CN)_{3}^- + CN^- \rightarrow Ni(CN)_{4}^{2-}\] #b. Formation of \(V(C_{2}O_{4})_{3}^{3-}\)#
06

Identify the central metal ion and the ligands

For the complex ion \(V(C_{2}O_{4})_{3}^{3-}\), the central metal ion is Vanadium (V) and the ligand is Oxalate (\(C_{2}O_{4}^{2-}\)).
07

Formation of the first coordination bond

In the first step, one Oxalate ligand will form a coordination bond with the Vanadium ion: \[V^{3+} + C_{2}O_{4}^{2-} \rightarrow VC_{2}O_{4}^{+}\]
08

Formation of the second coordination bond

In the next step, the second Oxalate ligand will form a coordination bond with the Vanadium ion: \[VC_{2}O_{4}^{+} + C_{2}O_{4}^{2-} \rightarrow V(C_{2}O_{4})_{2}^{-}\]
09

Formation of the third coordination bond

Now, the third Oxalate ligand will form a coordination bond with the Vanadium ion, resulting in the formation of the complex ion: \[V(C_{2}O_{4})_{2}^{-} + C_{2}O_{4}^{2-} \rightarrow V(C_{2}O_{4})_{3}^{3-}\]

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Most popular questions from this chapter

The equilibrium constant for the following reaction is \(1.0 \times 10^{23} :\) $$\mathrm{Cr}^{3+}(a q)+\mathrm{H}_{2} \mathrm{EDTA}^{2-}(a q) \rightleftharpoons \mathrm{CrEDTA}^{-}(a q)+2 \mathrm{H}^{+}(a q)$$ EDTA is used as a complexing agent in chemical analysis. Solutions of EDTA, usually containing the disodium salt $\mathrm{Na}_{2} \mathrm{H}_{2} \mathrm{EDTA}$ , are used to treat heavy metal poisoning. Calculate \(\left[\mathrm{Cr}^{3+}\right]\) at equilibrium in a solution originally \(0.0010 M\) in \(\mathrm{Cr}^{3+}\) and \(0.050 M\) in $\mathrm{H}_{2} \mathrm{EDTA}^{2-}\( and buffered at \)\mathrm{pH}=6.00.$

Aluminum ions react with the hydroxide ion to form the precipitate \(\mathrm{Al}(\mathrm{OH})_{3}(s),\) but can also react to form the soluble complex ion \(\mathrm{Al}(\mathrm{OH})_{4}^{-}.\) In terms of solubility, All \((\mathrm{OH})_{3}(s)\) will be more soluble in very acidic solutions as well as more soluble in very basic solutions. a. Write equations for the reactions that occur to increase the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) in very acidic solutions and in very basic solutions. b. Let's study the pH dependence of the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) in more detail. Show that the solubility of \(\mathrm{Al}(\mathrm{OH})_{3},\) as a function of \(\left[\mathrm{H}^{+}\right],\) obeys the equation $$S=\left[\mathrm{H}^{+}\right]^{3} K_{\mathrm{sp}} / K_{\mathrm{w}}^{3}+K K_{\mathrm{w}} /\left[\mathrm{H}^{+}\right]$$ where \(S=\) solubility \(=\left[\mathrm{Al}^{3+}\right]+\left[\mathrm{Al}(\mathrm{OH})_{4}^{-}\right]\) and \(K\) is the equilibrium constant for $$\mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{Al}(\mathrm{OH})_{4}^{-}(a q)$$ c. The value of \(K\) is 40.0 and \(K_{\mathrm{sp}}\) for \(\mathrm{Al}(\mathrm{OH})_{3}\) is \(2 \times 10^{-32}\) Plot the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}\) in the ph range \(4-12.\)

a. Calculate the molar solubility of \(\mathrm{AgBr}\) in pure water. \(K_{\mathrm{sp}}\) for \(\mathrm{AgBr}\) is \(5.0 \times 10^{-13}\) . b. Calculate the molar solubility of \(\mathrm{AgBr}\) in \(3.0M\) \(\mathrm{NH}_{3} .\) The overall formation constant for \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}+\) is \(1.7 \times 10^{7}\) that is, $$\mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \longrightarrow \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q) \quad K=1.7 \times 10^{7}$$ c. Compare the calculated solubilities from parts a and b. Explain any differences. d. What mass of \(\mathrm{AgBr}\) will dissolve in \(250.0 \mathrm{mL}\) of $3.0 M\( \)\mathrm{NH}_{3}?$ e. What effect does adding \(\mathrm{HNO}_{3}\) have on the solubilities calculated in parts a and b?

What is the maximum possible concentration of \(\mathrm{Ni}^{2+}\) ion in water at \(25^{\circ} \mathrm{C}\) that is saturated with \(0.10 M\) $\mathrm{H}_{2} \mathrm{S}\( and maintained at \)\mathrm{pH} 3.0\( with \)\mathrm{HCl} ?$

Consider 1.0 L of an aqueous solution that contains 0.10 M sulfuric acid to which 0.30 mole of barium nitrate is added. Assuming no change in volume of the solution, determine the pH, the concentration of barium ions in the final solution, and the mass of solid formed.
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