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Chapter 16: Problem 86

Calculate the mass of manganese hydroxide that dissolves to form 1300 mL of a saturated manganese hydroxide solution. For $\mathrm{Mn}(\mathrm{OH})_{2}, K_{\mathrm{sp}}=2.0 \times 10^{-13}.$

Short Answer

Expert verified
The mass of manganese hydroxide that dissolves to form 1300 mL of a saturated manganese hydroxide solution is approximately \(4.25 \times 10^{-6}\) grams.

Step by step solution

01

Write the balanced chemical equation and the expression for \(K_{sp}\)

Write the balanced chemical equation for the dissolution of manganese hydroxide in water: \[\mathrm{Mn}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Mn}^{2+}(aq) + 2\mathrm{OH}^-(aq)\] Write the expression for the solubility product constant, \(K_{sp}\), for manganese hydroxide: \[K_{sp} = [\mathrm{Mn}^{2+}][\mathrm{OH}^-]^2\]
02

Define the solubility of manganese hydroxide

Let the solubility of \(\mathrm{Mn}(\mathrm{OH})_{2}\) in water be \(x\) moles per liter, then the equilibrium concentrations of the ions will be: - \([\mathrm{Mn}^{2+}] = x\) moles/L - \([\mathrm{OH}^-] = 2x\) moles/L
03

Substitute the solubilities into the \(K_{sp}\) expression and solve for x

Substitute the solubility expressions into the \(K_{sp}\) expression: \[K_{sp} = (x)(2x)^2\] Solve for x: \[2.0 \times 10^{-13} = x(4x^2)\] \[2.0 \times 10^{-13} = 4x^3\] \[x^3 = \frac{2.0 \times 10^{-13}}{4}\] \[x^3 = 5.0 \times 10^{-14}\] \[x = \sqrt[3]{5.0 \times 10^{-14}}\] Calculate x: \[x \approx 3.68 \times 10^{-5}\ \text{moles/L}\] So, the solubility of \(\mathrm{Mn}(\mathrm{OH})_{2}\) in water is approximately \(3.68 \times 10^{-5}\) moles per liter.
04

Calculate the mass of \(\mathrm{Mn}(\mathrm{OH})_{2}\) dissolved in 1300 mL of the solution

The moles of \(\mathrm{Mn}(\mathrm{OH})_{2}\) dissolved in 1300 mL of the solution is given by: \[n_{\mathrm{Mn}(\mathrm{OH})_{2}} = (\text{solubility}) \times (\text{volume})\] \[n_{\mathrm{Mn}(\mathrm{OH})_{2}} = (3.68 \times 10^{-5}\ \text{moles/L}) \times (0.0013\ \text{L})\] \[n_{\mathrm{Mn}(\mathrm{OH})_{2}} \approx 4.78 \times 10^{-8}\ \text{moles}\] Next, convert the moles to mass: \[m_{\mathrm{Mn}(\mathrm{OH})_{2}} = n_{\mathrm{Mn}(\mathrm{OH})_{2}} \times M_{\mathrm{Mn}(\mathrm{OH})_{2}}\] \[m_{\mathrm{Mn}(\mathrm{OH})_{2}} \approx (4.78 \times 10^{-8}\ \text{moles}) \times (88.94\ \text{g/mol})\] \[m_{\mathrm{Mn}(\mathrm{OH})_{2}} \approx 4.25 \times 10^{-6}\ \text{g}\] The mass of manganese hydroxide that dissolves to form 1300 mL of a saturated manganese hydroxide solution is approximately \(4.25 \times 10^{-6}\) grams.

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Most popular questions from this chapter

A solution is prepared by adding 0.10 mole of \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6} \mathrm{Cl}_{2}\) to 0.50 \(\mathrm{L}\) of \(3.0 M\) \(\mathrm{NH}_{3}\) . Calculate \(\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\right]\) and \(\left[\mathrm{Ni}^{2+}\right]\) in this solution. \(K_{\text { overall }}\) for \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) is \(5.5 \times 10^{8} .\) That is, $$5.5 \times 10^{8}=\frac{\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\right]}{\left[\mathrm{Ni}^{2+}\right]\left[\mathrm{NH}_{3}\right]^{6}}$$ for the overall reaction $$\mathrm{Ni}^{2+}(a q)+6 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}(a q)$$

The solubility of \(\mathrm{PbCl}_{2}\) increases with an increase in temperature. Is the dissolution of \(\mathrm{PbCl}_{2}(s)\) in water exothermic or endothermic? Explain.

As sodium chloride solution is added to a solution of silver nitrate, a white precipitate forms. Ammonia is added to the mixture and the precipitate dissolves. When potassium bromide solution is then added, a pale yellow precipitate appears. When a solution of sodium thiosulfate is added, the yellow precipitate dissolves. Finally, potassium iodide is added to the solution and a yellow precipitate forms. Write equations for all the changes mentioned above. What conclusions can you draw concerning the sizes of the \(K_{\mathrm{sp}}\) values for \(\mathrm{AgCl}, \mathrm{AgBr},\) and \(\mathrm{AgI?}\)

For which salt in each of the following groups will the solubility depend on \(\mathrm{pH}\) ? $\begin{array}{ll}{\text { a. AgF, AgCl, AgBr }} & {\text { c. } \operatorname{Sr}\left(\mathrm{NO}_{3}\right)_{2}, \operatorname{Sr}\left(\mathrm{NO}_{2}\right)_{2}} \\ {\text { b. } \mathrm{Pb}(\mathrm{OH})_{2}, \mathrm{PbCl}_{2}} & {\text { d. } \mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}, \mathrm{Ni}(\mathrm{CN})_{2}}\end{array}$

Nitrate salts are generally considered to be soluble salts. One of the least soluble nitrate salts is barium nitrate. Approximately 15 g of \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) will dissolve per liter of solution. Calculate the \(K_{\mathrm{sp}}\) value for barium nitrate.
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