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The solubility of iron(II) hydroxide, \(\mathrm{Fe}(\mathrm{OH})_{2}\), is \(1.43 \times 10^{-3} \mathrm{~g} / \mathrm{L}\) at \(25^{\circ} \mathrm{C}\) (a) Write a balanced equation for the solubility equilibrium. (b) Write the expression for \(K_{\mathrm{sp}}\) and calculate its value. (c) \(\mathrm{A} 60.0 \mathrm{~mL}\) sample of \(3.00 \times 10^{-3} \mathrm{M} \mathrm{FeSO}_{4}\) solution is added to \(60.0 \mathrm{~mL}\) of \(4.00 \times 10^{-6} \mathrm{M} \mathrm{NaOH}\) solution. Does a precipitate of \(\mathrm{Fe}(\mathrm{OH})_{2}\) form? (Hint: Use the concentration values for \(\mathrm{Fe}^{2+}\) and \(\mathrm{OH}^{-}\) and plug them into the \(K_{\mathrm{sp}}\) expression. If the value you get is larger than \(K_{\mathrm{sp}}\) ' precipitation will occur.)

Step by step solution

01

Write the solubility equilibrium equation

For iron(II) hydroxide, the solubility equilibrium reaction can be written as follows: \[ \mathrm{Fe}(\mathrm{OH})_{2} \rightleftharpoons \mathrm{Fe^{2+}} + 2\mathrm{OH^-}\]

02

Write the Ksp expression and calculate its value

The solubility product constant (Ksp) expression for the reaction can be written as: \[ K_{sp} = [\mathrm{Fe^{2+}}][\mathrm{OH^-}]^2 \] Given that the solubility of Fe(OH)鈧� is 1.43 x 10鈦宦� g/L, we must first convert this to moles per liter (mol/L): \[ \dfrac{1.43 \times 10^{-3} \mathrm{~g/L}}{89.86 \mathrm{~g/mol}} = 1.59 \times 10^{-5} \mathrm{~mol/L} \] Now, we can use this value to find the concentration of specifically Fe虏鈦� and OH鈦� ions: \[ [\mathrm{Fe^{2+}}] = 1.59 \times 10^{-5} \mathrm{~M} \] \[ [\mathrm{OH^-}] = 2 \times 1.59 \times 10^{-5} \mathrm{~M} = 3.18 \times 10^{-5} \mathrm{~M} \] Now, plug these values into the Ksp expression: \[ K_{sp} = (1.59 \times 10^{-5})(3.18 \times 10^{-5})^2 = 1.61 \times 10^{-14} \]

03

Determine if a precipitate forms

Now we need to find the concentrations of Fe虏鈦� and OH鈦� ions after mixing the two solutions. Total volume = 60 mL + 60 mL = 120 mL The new concentration of Fe虏鈦� ions is: \[ [\mathrm{Fe^{2+}}]_{new} = \dfrac{3.00 \times 10^{-3} \mathrm{M} \times 60.0 \mathrm{~mL}}{120.0 \mathrm{~mL}} = 1.50 \times 10^{-3} \mathrm{M} \] The new concentration of OH鈦� ions is: \[ [\mathrm{OH^-}]_{new} = \dfrac{4.00 \times 10^{-6} \mathrm{M} \times 60.0 \mathrm{~mL}}{120.0 \mathrm{~mL}} = 2.00 \times 10^{-6} \mathrm{M} \] Let's use these new concentration values to find the ion product (Q), and compare it to the Ksp calculated earlier: \[ Q = (1.50 \times 10^{-3})(2.00 \times 10^{-6})^2 = 6.0 \times 10^{-15} \] Since Q < Ksp (6.0 脳 10鈦宦光伒 < 1.61 脳 10鈦宦光伌), the solution is unsaturated, and no precipitate will form.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Balanced Chemical Equation

Understanding a balanced chemical equation is essential to predict how compounds will react in different scenarios. In solubility equilibria, we deal with substances dissolving in water to form ions. For iron(II) hydroxide,

• The balanced equation for its solubility is: \[\mathrm{Fe}(\mathrm{OH})_{2} \rightleftharpoons \mathrm{Fe^{2+}} + 2\mathrm{OH^-}\]
• This indicates one iron cation (\(\mathrm{Fe^{2+}}\)) and two hydroxide anions (\(\mathrm{OH^-}\)).

In the equation, the double arrow (\(\rightleftharpoons\)) shows that the reaction can move in both directions - dissolving iron(II) hydroxide and forming a precipitate.This symmetry is crucial to maintaining equilibrium between the ions and the solid form, thus defining the system's behavior.

Solubility Product Constant (Ksp)

The solubility product constant, \(K_{sp}\), quantifies the solubility of a compound. It represents the maximum product of the ion concentrations that will dissolve in water before starting to form a precipitate. For iron(II) hydroxide:

• The expression for \(K_{sp}\) is given by: \[K_{sp} = [\mathrm{Fe^{2+}}][\mathrm{OH^-}]^2\]
• The value is derived from the equilibrium concentrations of ions. In this case, using the provided concentrations: \[K_{sp} = (1.59 \times 10^{-5})(3.18 \times 10^{-5})^2 = 1.61 \times 10^{-14}\]

A lower \(K_{sp}\) value hints at lower solubility, meaning iron(II) hydroxide is not very soluble in water at \(25^{\circ} \mathrm{C}\). This concept is key in predicting whether a compound will precipitate.

Precipitation Reaction

A precipitation reaction occurs when the product of the ion concentrations, called the ion product (\(Q\)), exceeds \(K_{sp}\). In such cases, the excess solute will form a solid precipitate. However, if \(Q\) is less than \(K_{sp}\), no precipitate forms.In the given scenario:

• The concentration of \(\mathrm{Fe^{2+}}\) and \(\mathrm{OH^-}\) ions results in \(Q = 6.0 \times 10^{-15}\).
• Comparing \(Q\) and \(K_{sp}\): Since \(Q < K_{sp}\) (6.0 脳 10鈦宦光伒 < 1.61 脳 10鈦宦光伌), the solution is unsaturated.

This means no iron(II) hydroxide precipitate will form, as the ion concentration remains below the threshold needed for precipitation.

Ionic Equilibrium

Ionic equilibrium describes the balance of ion concentrations in a solution, crucial in understanding reactions in aqueous environments.
It involves both the dynamic process of dissolving and precipitating ions.For iron(II) hydroxide:

• The ions in solution, \([\mathrm{Fe^{2+}}]\) and \([\mathrm{OH^-}]\), drift towards equilibrium.
• At equilibrium, the rates of dissolving and precipitating equalize.

Maintaining ionic equilibrium ensures that calculations like \(Q\) and \(K_{sp}\) align. This balance is pivotal to predict system behavior and whether a precipitate will occur. Understanding these interactions aids in solving complex chemical problems accurately.