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Chapter 16: Problem 35

Calculate the solubility (in moles per liter) of \(\mathrm{Fe}(\mathrm{OH})_{3}\left(K_{\mathrm{sp}}=4 \times\right.\) \(10^{-33}\) ) in each of the following. a. water b. a solution buffered at \(\mathrm{pH}=5.0\) c. a solution buffered at \(\mathrm{pH}=11.0\)

Short Answer

Expert verified
The solubility of \(\mathrm{Fe}(\mathrm{OH})_{3}\) is approximately: a. \(3.33 \times 10^{-9}\) moles per liter in water. b. \(4.0 \times 10^{-6}\) moles per liter in a solution buffered at \(\mathrm{pH}=5.0\). c. \(4.0 \times 10^{-24}\) moles per liter in a solution buffered at \(\mathrm{pH}=11.0\).

Step by step solution

01

Write the balanced chemical equation for the dissolution of \(\mathrm{Fe}(\mathrm{OH})_{3}\) in water

The balanced chemical equation for the dissolution of \(\mathrm{Fe}(\mathrm{OH})_{3}\) in water is: \(\mathrm{Fe}(\mathrm{OH})_{3(s)} \rightleftharpoons \mathrm{Fe}^{3+}_{(aq)} + 3\mathrm{OH}^-_{(aq)}\)
02

Write the expression for the solubility product

The \(K_{sp}\) expression for the dissolution of \(\mathrm{Fe}(\mathrm{OH})_{3}\) is: \(K_{sp} = [\mathrm{Fe}^{3+}] [(\mathrm{OH}^-)^3]\)
03

Determine the concentrations of ions in terms of solubility

Let the solubility of \(\mathrm{Fe}(\mathrm{OH})_{3}\) in water be \(s\) mol/L. Then \([\mathrm{Fe}^{3+}] = s\) mol/L and \([\mathrm{OH}^-] = 3s\) mol/L
04

Plug the concentrations into the \(K_{sp}\) expression and solve for the solubility

Now we need to substitute the given \(K_{sp}\) value and the expressions for concentrations of ions in the \(K_{sp}\) expression, which is: \(4 \times 10^{-33} = s(3s)^3\) Now we need to solve this equation for the solubility \(s\): \(4 \times 10^{-33} = 27s^4\) \(s^4 = \frac{4 \times 10^{-33}}{27}\) \(s = \sqrt[4]{\frac{4 \times 10^{-33}}{27}} \approx 3.33 \times 10^{-9}\) mol/L The solubility of \(\mathrm{Fe}(\mathrm{OH})_{3}\) in water is approximately \(3.33 \times 10^{-9}\) moles per liter. b. Solution buffered at \(\mathrm{pH}=5.0\)
05

Calculate the hydroxide ion concentration

To find the hydroxide ion concentration, we first need to calculate the hydronium ion concentration using the given pH: \(\mathrm{pH} = -\log[H^+]\) \(5.0 = -\log[H^+]\) \(H^+ = 10^{-5}\) M Now, we can use the ion product of water, \(K_w = [\mathrm{H}^+][\mathrm{OH}^-]\), to find the concentration of hydroxide ions given that \(K_w = 1.0 \times 10^{-14}\) for water at 25°C, and the concentration of hydronium ions has been calculated: \([\mathrm{OH}^-] = \frac{K_w}{[\mathrm{H}^+]} = \frac{1.0 \times 10^{-14}}{10^{-5}} = 1.0 \times 10^{-9}\) M
06

Calculate the solubility

Now, we can go back to our \(K_{sp}\) expression for \(\mathrm{Fe}(\mathrm{OH})_{3}\) in the presence of a solution buffered at \(\mathrm{pH}=5.0\): \(K_{sp} = [\mathrm{Fe}^{3+}] [\mathrm{OH}^-]^3\) Given that \([\mathrm{OH}^-]\) is now \(1.0 \times 10^{-9}\) M, and letting the solubility in this case be \(s_b\): \(4 \times 10^{-33} = [\mathrm{Fe}^{3+}] (1.0 \times 10^{-9})^3\) \(4 \times 10^{-33} = s_b\ (1.0 \times 10^{-27} )\) \(s_b = \frac{4 \times 10^{-33}}{1.0 \times 10^{-27}} = 4.0 \times 10^{-6}\) mol/L The solubility of \(\mathrm{Fe}(\mathrm{OH})_{3}\) in a solution buffered at \(\mathrm{pH}=5.0\) is \(4.0 \times 10^{-6}\) moles per liter. c. Solution buffered at \(\mathrm{pH}=11.0\)
07

Calculate the hydroxide ion concentration

As in part (b), we start by calculating the hydronium ion concentration from the given pH and then using the ion product of water to find the hydroxide ion concentration: \(\mathrm{pH} = -\log[H^+]\) \(11.0 = -\log[H^+]\) \(H^+ = 10^{-11}\) M \([\mathrm{OH}^-] = \frac{K_w}{[\mathrm{H}^+]} = \frac{1.0 \times 10^{-14}}{10^{-11}} = 1.0\times 10^{-3}\) M
08

Calculate the solubility

Now, we use the \(K_{sp}\) expression for \(\mathrm{Fe}(\mathrm{OH})_{3}\) in the presence of a solution buffered at \(\mathrm{pH}=11.0\): \(K_{sp} = [\mathrm{Fe}^{3+}] [\mathrm{OH}^-]^3\) Given that \([\mathrm{OH}^-]\) is now \(1.0 \times 10^{-3}\) M, and letting the solubility in this case be \(s_c\): \(4 \times 10^{-33} = [\mathrm{Fe}^{3+}] (1.0 \times 10^{-3})^3\) \(4 \times 10^{-33} = s_c\ (1.0 \times 10^{-9})\) \(s_c = \frac{4 \times 10^{-33}}{1.0 \times 10^{-9}} = 4.0 \times 10^{-24}\) mol/L The solubility of \(\mathrm{Fe}(\mathrm{OH})_{3}\) in a solution buffered at \(\mathrm{pH}=11.0\) is \(4.0 \times 10^{-24}\) moles per liter.

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Most popular questions from this chapter

\(K_{\mathrm{f}}\) for the complex ion \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}{ }^{+}\) is \(1.7 \times 10^{7} . K_{\text {sp }}\) for \(\mathrm{AgCl}\) is \(1.6 \times 10^{-10}\). Calculate the molar solubility of AgCl in \(1.0 \mathrm{M} \mathrm{NH}_{3}\).

Describe how you could separate the ions in each of the following groups by selective precipitation. a. \(\mathrm{Ag}^{+}, \mathrm{Mg}^{2+}, \mathrm{Cu}^{2+}\) c. \(\mathrm{Pb}^{2+}, \mathrm{Bi}^{3+}\) b. \(\mathrm{Pb}^{2+}, \mathrm{Ca}^{2+}, \mathrm{Fe}^{2+}\)

The \(K_{\mathrm{s} p}\) for \(Q\), a slightly soluble ionic compound composed of \(\mathrm{M}_{2}^{2+}\) and \(\mathrm{X}^{-}\) ions, is \(4.5 \times 10^{-29} .\) The electron configuration of \(\mathrm{M}^{+}\) is \([\mathrm{Xe}] 6 s^{1} 4 f^{14} 5 d^{10}\). The \(\mathrm{X}^{-}\) anion has 54 electrons. What is the molar solubility of \(Q\) in a solution of \(\mathrm{NaX}\) prepared by dissolving \(1.98 \mathrm{~g}\) NaX in \(150 . \mathrm{mL}\) solution?

Approximately \(0.14 \mathrm{~g}\) nickel(II) hydroxide, \(\mathrm{Ni}(\mathrm{OH})_{2}(s)\), dissolves per liter of water at \(20^{\circ} \mathrm{C}\). Calculate \(K_{\mathrm{sp}}\) for \(\mathrm{Ni}(\mathrm{OH})_{2}(s)\) at this temperature.

Calculate the final concentrations of \(\mathrm{K}^{+}(a q), \mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q)\), \(\mathrm{Ba}^{2+}(a q)\), and \(\mathrm{Br}^{-}(a q)\) in a solution prepared by adding \(0.100 \mathrm{~L}\) of \(0.200 \mathrm{M} \mathrm{K}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) to \(0.150 \mathrm{~L}\) of \(0.250 \mathrm{M} \mathrm{BaBr}_{2}\). (For \(\mathrm{BaC}_{2} \mathrm{O}_{4}\) \(\left.K_{\mathrm{sp}}=2.3 \times 10^{-\mathrm{s}} .\right)\)
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