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Chapter 16: Problem 67

In the presence of \(\mathrm{CN}^{-}, \mathrm{Fe}^{3+}\) forms the complex ion \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-} .\) The equilibrium concentrations of \(\mathrm{Fe}^{3+}\) and \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) are $8.5 \times 10^{-40} \mathrm{M}\( and \)1.5 \times 10^{-3} M,\( respectively, in a \)0.11-M$ KCN solution. Calculate the value for the overall formation constant of \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) . $$\mathrm{Fe}^{3+}(a q)+6 \mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{Fe}(\mathrm{CN})_{6}^{3-}(a q) \qquad K_{\text { overall }}=?$$

Short Answer

Expert verified
The overall formation constant for the complex ion \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) is \(2.97 \times 10^{41}\).

Step by step solution

01

Write the equilibrium reaction equation.

The equilibrium reaction equation for the formation of the complex ion \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) involves \(\mathrm{Fe}^{3+}\) and \(\mathrm{CN}^{-}\) as reactants, forming the complex ion \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\): $$\mathrm{Fe}^{3+}(a q)+6 \mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{Fe}(\mathrm{CN})_{6}^{3-}(a q)$$
02

Write the equilibrium constant expression for the reaction.

The equilibrium constant expression, K_overall, would be the ratio of the concentrations of the products to the concentrations of the reactants, with each concentration raised to its stoichiometric coefficient in the balanced reaction equation: $$K_\text{overall} =\frac{[\mathrm{Fe}(\mathrm{CN})_{6}^{3-}]}{[\mathrm{Fe}^{3+}][\mathrm{CN}^{-}]^6}$$
03

Plug in the given equilibrium concentrations.

We are given the equilibrium concentrations of \(\mathrm{Fe}^{3+}\), \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\), and \(\mathrm{CN}^{-}\) as \(8.5 \times 10^{-40} \mathrm{M}\), \(1.5 \times 10^{-3} \mathrm{M}\), and \(0.11 \mathrm{M}\), respectively. Plug these values into the equilibrium constant expression: $$K_\text{overall} =\frac{(1.5 \times 10^{-3})}{(8.5\times 10^{-40})(0.11)^6}$$
04

Calculate the overall formation constant, K_overall.

We can now solve for K_overall: $$K_\text{overall} =\frac{(1.5 \times 10^{-3})}{(8.5\times 10^{-40})(0.11)^6} \approx 2.97 \times 10^{41}$$ So, the overall formation constant for the complex ion \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) is \(2.97 \times 10^{41}\).

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Most popular questions from this chapter

A solution contains $3.0 \times 10^{-3} M \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2} .$ What concentrations of \(\mathrm{KF}\) will cause precipitation of solid \(\mathrm{MgF}_{2}\left(K_{\mathrm{sp}}=6.4 \times 10^{-9}\right) ?\)

The solubility of \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\) in a $0.20-M \mathrm{KIO}_{3}\( solution is \)4.4 \times 10^{-8} \mathrm{mol} / \mathrm{L}$ . Calculate \(K_{\mathrm{sp}}\) for \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}.\)

A solution is \(1 \times 10^{-4} M\) in \(\mathrm{NaF}\), $\mathrm{Na}_{2} \mathrm{S},\( and \)\mathrm{Na}_{3} \mathrm{PO}_{4} .$ What would be the order of precipitation as a source of \(\mathrm{Pb}^{2+}\) is added gradually to the solution? The relevant \(K_{\mathrm{sp}}\) values are $K_{\mathrm{sp}}\left(\mathrm{PbF}_{2}\right)=4 \times 10^{-8}, K_{\mathrm{sp}}(\mathrm{PbS})=7 \times 10^{-29},$ and $K_{\mathrm{sp}}\left[\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\right]=1 \times 10^{-54}.$

Write balanced equations for the dissolution reactions and the corresponding solubility product expressions for each of the following solids. a. \(A g C_{2} H_{3} O_{2}\) b. \(A l(O H)_{3}\) c. \(C a_{3}\left(\mathrm{PO}_{4}\right)_{2}\)

The \(K_{\mathrm{sp}}\) for \(Q,\) a slightly soluble ionic compound composed of \(\mathrm{M}_{2}^{2+}\) and \(\mathrm{X}^{-}\) ions, is \(4.5 \times 10^{-29} .\) The electron configuration of \(\mathrm{M}^{+}\) is $[\mathrm{Xe}] 6 s^{1} 4 f^{4} 5 d^{10} .\( The \)\mathrm{X}^{-}$ anion has 54 electrons. What is the molar solubility of \(Q\) in a solution of \(\mathrm{NaX}\) prepared by dissolving \(1.98 \mathrm{g}\) \(\mathrm{NaX}\) in \(150 .\) mL solution?
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