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Chapter 16: Problem 61

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

Short Answer

Expert verified
The overall formation constant of the complex ion Fe(CN)_{6}^{3-} is approximately \(2.2 \times 10^{41}\).

Step by step solution

01

Write the equilibrium expression for the formation of the complex ion

Using the balanced equation, we can write the equilibrium expression for this reaction as: \[K_{overall} = \frac{[\mathrm{Fe}(\mathrm{CN})_{6}^{3-}]}{[\mathrm{Fe}^{3+}][\mathrm{CN}^-]^6}\]
02

Calculate the initial concentration of CN-

The problem states that the solution contains \(0.11 M\) of KCN. As 1 mole of KCN dissociates to 1 mole of CN- (\(\mathrm{K^+}\) and \(\mathrm{CN^-}\)), therefore, the initial concentration of CN- is also \(0.11 M\).
03

Find the change in concentration of CN-

Since 6 moles of CN- form 1 mole of Fe(CN)_{6}^{3-}, the change in concentration of CN- will be 6 times the equilibrium concentration of Fe(CN)_{6}^{3-}. That is, the concentration of CN- decreases by \(6 \times (1.5 \times 10^{-3})\).
04

Calculate the equilibrium concentration of CN-

Subtract the change in concentration of CN- (as calculated in step 3) from the initial concentration of CN- (as calculated in step 2) to determine the equilibrium concentration of CN-: \[[\mathrm{CN}^{-}]_{eq} = [\mathrm{CN}^{-}]_{initial} - 6 \times [\mathrm{Fe}(\mathrm{CN})_{6}^{3-}]_{eq} \] \[[\mathrm{CN}^{-}]_{eq} = 0.11 - 6 \times (1.5 \times 10^{-3}) = 0.101M \]
05

Calculate K_{overall} using the equilibrium concentrations

Plug in the equilibrium concentrations of the species into the equilibrium expression we derived in step 1 and solve for \(K_{overall}\): \[K_{overall} = \frac{[\mathrm{Fe}(\mathrm{CN})_{6}^{3-}]}{[\mathrm{Fe}^{3+}][\mathrm{CN}^-]^6}\] \[K_{overall} = \frac{1.5 \times 10^{-3}}{(8.5 \times10^{-40})(0.101)^6}\]
06

Compute the value of K_{overall}

Evaluate the expression to get the value of K_{overall}: \[K_{overall} \approx 2.2 \times 10^{41}\] Hence, the overall formation constant of the complex ion Fe(CN)_{6}^{3-} is approximately \(2.2 \times 10^{41}\).

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Most popular questions from this chapter

Sodium tripolyphosphate \(\left(\mathrm{Na}_{5} \mathrm{P}_{3} \mathrm{O}_{10}\right)\) is used in many synthetic detergents. Its major effect is to soften the water by complexing \(\mathrm{Mg}^{2+}\) and \(\mathrm{Ca}^{2+}\) ions. It also increases the efficiency of surfactants, or wetting agents that lower a liquid's surface tension. The \(K\) value for the formation of \(\mathrm{MgP}_{3} \mathrm{O}_{10}^{3-}\) is \(4.0 \times 10^{8} .\) The reaction is \(\mathrm{Mg}^{2+}+\mathrm{P}_{3} \mathrm{O}_{10}^{5-} \rightleftharpoons \mathrm{MgP}_{3} \mathrm{O}_{10}{ }^{3-} .\) Calculate the concentration of \(\mathrm{Mg}^{2+}\) in a solution that was originally \(50 . \mathrm{ppm} \mathrm{Mg}^{2+}(50 . \mathrm{mg} / \mathrm{L}\) of solution) after 40. g \(\mathrm{Na}_{5} \mathrm{P}_{3} \mathrm{O}_{10}\) is added to \(1.0 \mathrm{~L}\) of the solution.

Calculate the solubility of \(\mathrm{AgCN}(s)\left(K_{\mathrm{sp}}=2.2 \times 10^{-12}\right)\) in a solution containing \(1.0 M \mathrm{H}^{+} .\left(K_{\mathrm{a}}\right.\) for \(\mathrm{HCN}\) is \(6.2 \times 10^{-10}\).)

Calculate the molar solubility of \(\mathrm{Mg}(\mathrm{OH})_{-}, K_{\mathrm{sp}}=8.9 \times 10^{-12}\).

The concentration of \(\mathrm{Pb}^{2+}\) in a solution saturated with \(\mathrm{PbBr}_{2}(s)\) is \(2.14 \times 10^{-2} M .\) Calculate \(K_{\text {sp }}\) for \(\mathrm{PbBr}_{2}\).

Write equations for the stepwise formation of each of the following complex ions. a. \(\mathrm{Ni}(\mathrm{CN})_{4}^{2-}\) b. \(\mathrm{V}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}{ }^{3-}\)
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