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Chapter 16: Problem 60

Write equations for the stepwise formation of each of the following complex ions. a. \(\mathrm{CoF}_{6}^{3-}\) b. \(\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\)

Short Answer

Expert verified
The stepwise formation equations for the complex ion \(\mathrm{CoF}_{6}^{3-}\) are as follows: 1. Co³⁺ + F⁻ → CoF²⁺ 2. CoF²⁺ + F⁻ → CoF₂⁺ 3. CoF₂⁺ + F⁻ → CoF₃ 4. CoF₃ + F⁻ → CoF₄⁻ 5. CoF₄⁻ + F⁻ → CoF₅²⁻ 6. CoF₅²⁻ + F⁻ → CoF₆³⁻ The stepwise formation equations for the complex ion \(\mathrm{Zn}(\mathrm{NH}_{3})_{4}^{2+}\) are as follows: 1. Zn²⁺ + NH₃ → Zn(NH₃)²⁺ 2. Zn(NH₃)²⁺ + NH₃ → Zn(NH₃)₂²⁺ 3. Zn(NH₃)₂²⁺ + NH₃ → Zn(NH₃)₃²⁺ 4. Zn(NH₃)₃²⁺ + NH₃ → Zn(NH₃)₄²⁺

Step by step solution

01

Identify components of the complex ions

In order to write stepwise formation equations for the complex ions, we first need to identify the components involved in the formation of each ion. For complex ion a: \(\mathrm{CoF}_{6}^{3-}\) - Central metal ion: Co (Cobalt) - Ligand: F⁻ (Fluoride ion) - Charge: 3- For complex ion b: \(\mathrm{Zn}(\mathrm{NH}_{3})_{4}^{2+}\) - Central metal ion: Zn (Zinc) - Ligand: NH₃ (Ammonia) - Charge: 2+
02

Write the stepwise formation equations for \(\mathrm{CoF}_{6}^{3-}\)

Now we will write the stepwise formation equations for the cobalt fluoride complex ion \(\mathrm{CoF}_{6}^{3-}\): 1. Co³⁺ + F⁻ → CoF²⁺ 2. CoF²⁺ + F⁻ → CoF₂⁺ 3. CoF₂⁺ + F⁻ → CoF₃ 4. CoF₃ + F⁻ → CoF₄⁻ 5. CoF₄⁻ + F⁻ → CoF₅²⁻ 6. CoF₅²⁻ + F⁻ → CoF₆³⁻ The stepwise formation of \(\mathrm{CoF}_{6}^{3-}\) is completed.
03

Write the stepwise formation equations for \(\mathrm{Zn}(\mathrm{NH}_{3})_{4}^{2+}\)

Now we will write the stepwise formation equations for the zinc ammonia complex ion \(\mathrm{Zn}(\mathrm{NH}_{3})_{4}^{2+}\): 1. Zn²⁺ + NH₃ → Zn(NH₃)²⁺ 2. Zn(NH₃)²⁺ + NH₃ → Zn(NH₃)₂²⁺ 3. Zn(NH₃)₂²⁺ + NH₃ → Zn(NH₃)₃²⁺ 4. Zn(NH₃)₃²⁺ + NH₃ → Zn(NH₃)₄²⁺ The stepwise formation of \(\mathrm{Zn}(\mathrm{NH}_{3})_{4}^{2+}\) is completed.

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Most popular questions from this chapter

The copper(I) ion forms a complex ion with \(\mathrm{CN}^{-}\) according to the following equation: \(\mathrm{Cu}^{+}(a q)+3 \mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{Cu}(\mathrm{CN})_{3}^{2-}(a q) \quad K_{\mathrm{f}}=1.0 \times 10^{11}\) a. Calculate the solubility of \(\operatorname{CuBr}(s)\left(K_{\text {sp }}=1.0 \times 10^{-5}\right)\) in \(1.0 \mathrm{~L}\) of \(1.0 \mathrm{M} \mathrm{NaCN}\).b. Calculate the concentration of \(\mathrm{Br}^{-}\) at equilibrium. c. Calculate the concentration of \(\mathrm{CN}^{-}\) at equilibrium.

A friend tells you: "The constant \(K_{\text {sp }}\) of a salt is called the solubility product constant and is calculated from the concentrations of ions in the solution. Thus, if salt A dissolves to a greater extent than salt \(\mathrm{B}\), salt \(\mathrm{A}\) must have a higher \(K_{\mathrm{sp}}\) than salt \(\mathrm{B} .\) " Do you agree with your friend? Explain.

When \(\mathrm{Na}_{3} \mathrm{PO}_{4}(a q)\) is added to a solution containing a metal ion and a precipitate forms, the precipitate generally could be one of two possibilities. What are the two possibilities?

a. Calculate the molar solubility of \(\mathrm{AgI}\) in pure water. \(K_{\mathrm{sp}}\) for \(\mathrm{AgI}\) is \(1.5 \times 10^{-16}\) b. Calculate the molar solubility of AgI in \(3.0 M \mathrm{NH}_{3}\). The overall formation constant for \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\) is \(1.7 \times 10^{7}\). c. Compare the calculated solubilities from parts a and b. Explain any differences.

Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid-base properties. a. \(\mathrm{PbI}_{2}, K_{\mathrm{sp}}=1.4 \times 10^{-8}\) b. \(\mathrm{CdCO}_{3}, K_{\mathrm{sp}}=5.2 \times 10^{-12}\) c. \(\mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}, K_{\text {sp }}=1 \times 10^{-31}\)
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