Chapter 14: Problem 46

The reaction below has an equilibrium constant of $4.9 \times 10^{11}$ . $$\mathrm{Fe}(\mathrm{OH})_{2}(s)+2 \mathrm{H}_{3} \mathrm{O}^{+}(a q) \rightleftarrows \mathrm{Fe}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l)$$ Write the equilibrium constant expression, and determine the concentration of $\mathrm{Fe}^{2+}$ ions in equilibrium when the hydronium ion concentration is $1.0 \times 10^{-7} \mathrm{mol} / \mathrm{L}$

Short Answer

Expert verified

The concentration of $Fe^{2+}$ ions at equilibrium is 4.9 mol/L

Step by step solution

Write the Equilibrium Constant Expression

The equilibrium constant expression for the given reaction is written as the ratio of the product's concentrations raised to their stoichiometric coefficients to the reactants'. Since solids and liquids are omitted in equilibrium expressions, the expression becomes: \[K = \frac{[Fe^{2+}]}{[H_{3}O^{+}]^2}\]

Calculate the Concentration of $Fe^{2+}$

Rearrange the equilibrium expression in order to calculate $[Fe^{2+}]$, it becomes: \[[Fe^{2+}] = K \times [H_{3}O^{+}]^2\] Then substitute the given values of $K = 4.9 \times 10^{11}$ and $ [H_{3}O^{+}] = 1.0 \times 10^{-7} mol/L$ into the equation: \[[Fe^{2+}] = 4.9 \times 10^{11} \times \left(1.0 \times 10^{-7}\right)^2\]

Solve for the Concentration of $Fe^{2+}$

Solving for $[Fe^{2+}]$ yields: \[[Fe^{2+}] = 4.9\]

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Short Answer

Step by step solution

Write the Equilibrium Constant Expression

Calculate the Concentration of \(Fe^{2+}\)

Solve for the Concentration of \(Fe^{2+}\)

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