Question

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

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

Step-by-step solution

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: \[K_{overall} = \frac{[\mathrm{Fe}(\mathrm{CN})_{6}^{3-}]}{[\mathrm{Fe}^{3+}][\mathrm{CN}^-]^6}\]

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 \(0.11 M\).

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 \(6 \times (1.5 \times 10^{-3})\).

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-: \[[\mathrm{CN}^{-}]_{eq} = [\mathrm{CN}^{-}]_{initial} - 6 \times [\mathrm{Fe}(\mathrm{CN})_{6}^{3-}]_{eq} \] \[[\mathrm{CN}^{-}]_{eq} = 0.11 - 6 \times (1.5 \times 10^{-3}) = 0.101M \]

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 \(K_{overall}\): \[K_{overall} = \frac{[\mathrm{Fe}(\mathrm{CN})_{6}^{3-}]}{[\mathrm{Fe}^{3+}][\mathrm{CN}^-]^6}\] \[K_{overall} = \frac{1.5 \times 10^{-3}}{(8.5 \times10^{-40})(0.101)^6}\]

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 \(2.2 \times 10^{41}\).

Key concepts

Equilibrium Expression

In chemistry, an equilibrium expression is used to express the state of balance within a reversible reaction. For the formation of a complex ion, such as \(\mathrm{Fe}(\mathrm{CN})_6^{3-}\), the equilibrium expression relates the concentrations of products and reactants at equilibrium. This expression is based on the balanced chemical equation for the reaction. The general form for the equilibrium expression, denoted as \(K_{overall}\) in this scenario, is written as the ratio of the product concentration to the reactant concentrations raised to the power of their stoichiometric coefficients.
For the given reaction: \(\mathrm{Fe}^{3+}(aq) + 6\mathrm{CN}^-(aq) \rightleftharpoons \mathrm{Fe}(\mathrm{CN})_6^{3-}\), the equilibrium expression is: \[K_{overall} = \frac{[\mathrm{Fe}(\mathrm{CN})_6^{3-}]}{[\mathrm{Fe}^{3+}][\mathrm{CN}^-]^6}\]
Writing this expression is the first step in calculating the formation constant, which provides valuable insights into the extent of the reaction under equilibrium conditions.

Complex Ion

A complex ion is an ion composed of a central metal atom or ion (in this case, \(\mathrm{Fe}^{3+}\)) bonded to surrounding molecules or ions known as ligands (such as \(\mathrm{CN}^-\)). Complex ions like \(\mathrm{Fe}(\mathrm{CN})_6^{3-}\) are crucial in understanding coordination chemistry. The formation of these ions occurs through coordinate covalent bonding, where the ligands donate pairs of electrons to the metal center.
The significant point about complex ions is that they often result in unique chemical behavior, affecting solubility and stability of compounds. The formation of complex ions like \(\mathrm{Fe}(\mathrm{CN})_6^{3-}\) is highly favored in the presence of suitable ligands, resulting in a high formation constant \(K_{overall}\), indicating a stable and predominantly forward reaction towards the formation of the complex ion under equilibrium conditions.

Concentration Change

Concentration change during a reaction involves how the amounts of reactants and products vary. In our context, the change in the concentration of \(\mathrm{CN}^-\) is significant. Since 6 moles of \(\mathrm{CN}^-\) react to form 1 mole of the complex ion \(\mathrm{Fe}(\mathrm{CN})_6^{3-}\), the change in concentration of \(\mathrm{CN}^-\) is substantial and results in a decrease.
This change is quantified as six times the equilibrium concentration of the complex ion, which helps in calculating the new equilibrium concentration of \(\mathrm{CN}^-\). Such calculations are essential to manage the stoichiometry of reactions and determine the actual concentrations of species involved.
The reaction's progress alters the concentrations of reacting ions, providing a direct influence on the equilibrium state, which directly affects the calculation of the formation constant \(K_{overall}\).

Equilibrium Concentration

Equilibrium concentration refers to the concentration of reactants and products when the rate of the forward reaction equals the rate of the reverse reaction. In these conditions, the system is at a balanced state, and concentrations remain constant.
In this problem, the equilibrium concentrations of \(\mathrm{Fe}^{3+}\) and \(\mathrm{Fe}(\mathrm{CN})_6^{3-}\) are provided as \(8.5 \times 10^{-40} M\) and \(1.5 \times 10^{-3} M\) respectively. The equilibrium concentration of \(\mathrm{CN}^-\) can be calculated using the initial concentration and the change in concentration: \[[\mathrm{CN}^-]_{eq} = [\mathrm{CN}^-]_{initial} - 6 \times [\mathrm{Fe}(\mathrm{CN})_6^{3-}]_{eq}\] \[[\mathrm{CN}^-]_{eq} = 0.11 - 6 \times (1.5 \times 10^{-3}) = 0.101M\]
These calculated concentrations are then plugged into the equilibrium expression derived for the reaction. Understanding equilibrium concentration helps in quantifying the extent of a reaction and is essential in calculating the formation constant of complex ions.