Bipyridyl (Bipy) is a molecule commonly used as a bidentate ligand. When \(0.17 \mathrm{~mol}\) of bipyridyl is dissolved in \(2.4 \mathrm{~L}\) of a solution that contains \(0.052 \mathrm{M} \mathrm{Fe}^{2+},\left[\mathrm{Fe}(\mathrm{bipy})_{3}\right]^{2+}\left(K_{\mathrm{f}}=1.6 \times 10^{17}\right)\) is formed. What are the con- centrations of all species when equilibrium is established?

Short Answer

Expert verified
Based on the given initial concentration of \(Fe^{2+}\) and the initial moles of bipyridyl (Bipy), the equilibrium concentrations for the formation of the \([Fe(bipy)_3]^{2+}\) complex are approximately: \([Fe^{2+}]_{equilibrium} ≈ 0.052 \mathrm{M}\) \([bipy]_{equilibrium} ≈ 0.07083 \mathrm{M}\) \([[Fe(\mathrm{bipy})_3]^{2+}]_{equilibrium} ≈ 4.38 \times 10^{-8} \mathrm{M}\)

Step by step solution

01

Analyze the reaction

Bipyridyl (bipy) forms a complex \([Fe(bipy)_3]^{2+}\) with \(Fe^{2+}\): \(Fe^{2+} + 3 \mathrm{bipy} \rightleftharpoons [Fe(\mathrm{bipy})_3]^{2+}\) The equilibrium constant for this reaction is given by: \( K_{f} = \frac{[[Fe(\mathrm{bipy})_3]^{2+}]}{[Fe^{2+}][\mathrm{bipy}]^{3}}\)
02

Write the ICE table

To solve for the equilibrium concentrations, we can set up an ICE table (Initial, Change, Equilibrium): |\(Fe^{2+}\) | bipy | \([Fe(\mathrm{bipy})_{3}]^{2+}\)| |----|----|-----| |I | 0.17 mol | 0| |C | -x | -3x | x | |E | 0.052-x "\((1)\)" | 0.17 - 3x "\((2)\)"| x "\((3)\)"| * In order to convert from given moles of bipy to molarity, we need to divide the moles of bipy by the volume of the solution. So initial bipy concentration = \(\frac{0.17 \mathrm{~mol}}{2.4 \mathrm{~L}} = 0.07083 \mathrm{M}\) * Note that we are simplifying the table by assuming the volume of the solution remains constant.
03

Substitute the ICE table values into the equilibrium constant expression

By substituting the equilibrium concentrations from the ICE table into the equilibrium constant expression, we get: \(K_{f} = \frac{x}{(0.052-x)(0.07083-3x)^3}\) Now, we can put the given value \(K_{f} = 1.6 \times 10^{17}\): \(1.6 \times 10^{17} = \frac{x}{(0.052-x)(0.07083-3x)^3}\)
04

Solve for \(x\)

Solving for \(x\) directly in this equation can be quite challenging. So, we can make a simplifying assumption that \(x\) is much smaller than \(0.052\) and \(0.07083/3\) (i.e., x is quite small compared to the initial concentrations that it will not make a significant difference when subtracted). So, we can approximate it as: \(1.6 \times 10^{17} ≈ \frac{x}{(0.052)(0.07083)^3}\) Solving for \(x\), we get: \(x ≈ 4.38 \times 10^{-8} \mathrm{M}\) (Equilibrium concentration of \([Fe(\mathrm{bipy})_3]^{2+}\))
05

Calculate the equilibrium concentrations for Bipy and \(Fe^{2+}\)

Now, we can use equations \((1)\), \((2)\), and \((3)\) to find the equilibrium concentrations: \([Fe^{2+}]_{equilibrium} ≈ 0.052 - x ≈ 0.052 - 4.38 \times 10^{-8} ≈ 0.052 \mathrm{M}\) \([bipy]_{equilibrium} ≈ 0.07083 - 3x ≈ 0.07083 - 3(4.38 \times 10^{-8}) ≈ 0.07083 \mathrm{M}\) Finally, the equilibrium concentrations are: \([Fe^{2+}]_{equilibrium} ≈ 0.052 \mathrm{M}\) \([bipy]_{equilibrium} ≈ 0.07083 \mathrm{M}\) \([[Fe(\mathrm{bipy})_3]^{2+}]_{equilibrium} ≈ 4.38 \times 10^{-8} \mathrm{M}\)

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