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)₂ is 1.43 x 10⁻³ 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 Fe²⁺ and OH⁻ 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 Fe²⁺ and OH⁻ ions after mixing the two solutions. Total volume = 60 mL + 60 mL = 120 mL The new concentration of Fe²⁺ 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⁻ 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 × 10⁻¹⁵ < 1.61 × 10⁻¹⁴), the solution is unsaturated, and no precipitate will form.

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