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Chapter 16: Problem 85

Nanotechnology has become an important field, with applications ranging from high-density data storage to the design of "nano machines." One common building block of nanostructured architectures is manganese oxide nanoparticles. The particles can be formed from manganese oxalate nanorods, the formation of which can be described as follows: \(\mathrm{Mn}^{2+}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) \rightleftharpoons \mathrm{MnC}_{2} \mathrm{O}_{4}(a q) \quad K_{1}=7.9 \times 10^{3}\) \(\mathrm{MnC}_{2} \mathrm{O}_{4}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) \rightleftharpoons \mathrm{Mn}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}^{2-}(a q)\) $$ K_{2}=7.9 \times 10^{1} $$ Calculate the value for the overall formation constant for \(\mathrm{Mn}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}^{2-}\) $$ K=\frac{\left[\mathrm{Mn}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}^{2-}\right]}{\left[\mathrm{Mn}^{2+}\right]\left[\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\right]^{2}} $$

Short Answer

Expert verified
The overall formation constant (K) for \(\mathrm{Mn}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}^{2-}\) is calculated by multiplying the individual equilibrium constants: \(K = K_{1} \times K_{2} = (7.9 \times 10^{3}) \times (7.9 \times 10^{1}) = 6.24 \times 10^{5}\).

Step by step solution

01

Write down the reactions and equilibrium constants

Reaction 1: \(\mathrm{Mn}^{2+}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) \rightleftharpoons \mathrm{MnC}_{2} \mathrm{O}_{4}(a q)\) \(K_{1}=7.9 \times 10^{3}\) Reaction 2: \(\mathrm{MnC}_{2} \mathrm{O}_{4}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) \rightleftharpoons \mathrm{Mn}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}^{2-}(a q)\) \(K_{2}=7.9 \times 10^{1}\)
02

Add the reactions

Add the two reactions so that you are left only with the desired ions on each side of the equilibrium: \(\mathrm{Mn}^{2+}(a q) + 2\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) \rightleftharpoons \mathrm{Mn}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}^{2-}(a q)\)
03

Calculate the overall formation constant (K)

When combining two reactions, the overall equilibrium constant K is the product of the individual equilibrium constants: \(K = K_{1} \times K_{2} = (7.9 \times 10^{3}) \times (7.9 \times 10^{1})\) Calculate the product: \(K = 6.24 \times 10^{5}\) This overall formation constant represents the affinity of \(\mathrm{Mn}^{2+}\) for the oxalate ion in forming the complex ion \(\mathrm{Mn}\left(\mathrm{C}_{2}\mathrm{O}_{4}\right)_{2}^{2-}\).

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Most popular questions from this chapter

A solution is formed by mixing \(50.0 \mathrm{~mL}\) of \(10.0 \mathrm{M} \mathrm{NaX}\) with \(50.0 \mathrm{~mL}\) of \(2.0 \times 10^{-3} \mathrm{M} \mathrm{CuNO}_{3}\). Assume that \(\mathrm{Cu}^{+}\) forms complex ions with \(\mathrm{X}^{-}\) as follows: $$ \begin{aligned} \mathrm{Cu}^{+}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \operatorname{CuX}(a q) & K_{1}=1.0 \times 10^{2} \\ \mathrm{CuX}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{2}-(a q) & K_{2}=1.0 \times 10^{4} \\ \mathrm{CuX}_{2}^{-}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{3}{ }^{2-}(a q) & K_{3}=1.0 \times 10^{3} \end{aligned} $$ with an overall reaction \(\mathrm{Cu}^{+}(a q)+3 \mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{3}^{2-}(a q) \quad K=1.0 \times 10^{9}\) Calculate the following concentrations at equilibrium. a. \(\mathrm{CuX}_{3}^{2-}\) b. \(\mathrm{CuX}_{2}^{-}\) c. \(\mathrm{Cu}^{+}\)

A solution contains \(1.0 \times 10^{-5} \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4} .\) What is the minimum concentration of \(\mathrm{AgNO}_{3}\) that would cause precipitation of solid \(\mathrm{Ag}_{3} \mathrm{PO}_{4}\left(K_{\mathrm{sp}}=1.8 \times 10^{-18}\right) ?\)

Which of the following will affect the total amount of solute that can dissolve in a given amount of solvent? a. The solution is stirred. b. The solute is ground to fine particles before dissolving. c. The temperature changes.

a. Using the \(K_{\mathrm{sp}}\) value for \(\mathrm{Cu}(\mathrm{OH})_{2}\left(1.6 \times 10^{-19}\right)\) and the overall formation constant for \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\left(1.0 \times 10^{13}\right)\), calculate the value for the equilibrium constant for the following reaction: \(\mathrm{Cu}(\mathrm{OH})_{2}(s)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)\) b. Use the value of the equilibrium constant you calculated in part a to calculate the solubility (in \(\mathrm{mol} / \mathrm{L}\) ) of \(\mathrm{Cu}(\mathrm{OH})_{2}\) in \(5.0 \mathrm{M} \mathrm{NH}_{3} .\) In \(5.0 \mathrm{M} \mathrm{NH}_{3}\) the concentration of \(\mathrm{OH}^{-} \mathrm{is}\) \(0.0095 M .\)

What mass of \(\operatorname{ZnS}\left(K_{\text {sp }}=2.5 \times 10^{-22}\right.\) ) will dissolve in \(300.0 \mathrm{~mL}\) of \(0.050 \mathrm{M} \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2} ?\) Ignore the basic properties of \(\mathrm{S}^{2-}\).
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