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Chapter 15: Problem 31

Calculate the molar solubility of \(\mathrm{Mg}(\mathrm{OH})_{2}, K_{\mathrm{sp}}=8.9 \times 10^{-12}\).

Short Answer

Expert verified
The molar solubility of \(\mathrm{Mg(OH)_2}\) is \(1.3 \times 10^{-4}\,\text{mol/L}\).

Step by step solution

01

Write the dissolution equation for \(\mathrm{Mg(OH)_2}\)

The dissolution equation for \(\mathrm{Mg(OH)_2}\) is: \[ \mathrm{Mg(OH)_{2}(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^{-}(aq)} \]
02

Define the solubility product expression

The solubility product expression uses the product of the concentrations of dissolved ions raised to their stoichiometric coefficients. For this dissolution reaction, the expression is: \[ K_{sp} = [\mathrm{Mg^{2+}}][\mathrm{OH^-}]^2 \]
03

Express the concentrations of ions in terms of molar solubility

Let's call the molar solubility of \(\mathrm{Mg(OH)_2}\) "x". This means that x moles of \(\mathrm{Mg(OH)_2}\) dissolve per liter of solution. Dissolving x moles of \(\mathrm{Mg(OH)_2}\) will also produce x moles of \(\mathrm{Mg^{2+}}\) ions and 2x moles of \(\mathrm{OH^-}\) ions. We can then express the concentrations as follows: \[ [\mathrm{Mg^{2+}}] = x \quad \text{and} \quad [\mathrm{OH^-}]=2x \]
04

Substitute the concentrations into the solubility product expression and solve for x

Substitute the concentrations of ions in terms of x into the solubility product expression: \[ K_{sp} = x(2x)^2 \] Now, plug in the given \(K_{sp}=8.9 \times 10^{-12}\) and solve for x: \[ 8.9 \times 10^{-12} = x(4x^2) \] \[ 8.9 \times 10^{-12} = 4x^3 \] \[ x^3=\frac{8.9 \times 10^{-12}}{4} \] \[ x^3=2.225 \times 10^{-12} \] \[ x=\sqrt[3]{2.225 \times 10^{-12}} \] \[ x=1.3 \times 10^{-4} \] So, the molar solubility of \(\mathrm{Mg(OH)_2}\) is \(1.3 \times 10^{-4}\,\text{mol/L}\).

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Most popular questions from this chapter

A friend tells you: "The constant \(K_{\mathrm{sp}}\) of a salt is called the solubility product constant and is calculated from the concentrations of ions in the solution. Thus, if salt A dissolves to a greater extent than salt \(\mathbf{B}\), salt \(\mathbf{A}\) must have a higher \(K_{\mathrm{sp}}\) than salt \(\mathbf{B}\)." Do you agree with your friend? Explain.

A solution is formed by mixing \(50.0 \mathrm{mL}\) of \(10.0 \mathrm{M}\) NaX with \(50.0 \mathrm{mL}\) of \(2.0 \times 10^{-3} \mathrm{M} \mathrm{CuNO}_{3} .\) Assume that \(\mathrm{Cu}^{+}\) forms complex ions with \(X^{-}\) as follows: $$\mathrm{Cu}^{+}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}(a q) \quad K_{1}=1.0 \times 10^{2}$$ $$\mathrm{CuX}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{2}^{-}(a q) \quad K_{2}=1.0 \times 10^{4}$$ $$\mathrm{CuX}_{2}^{-}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{3}^{2-}(a q) \quad K_{3}=1.0 \times 10^{3}$$ with an overall reaction $$\mathrm{Cu}^{+}(a q)+3 \mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{3}^{2-}(a q) \quad K=1.0 \times 10^{9}$$ Calculate the following concentrations at equilibrium. a. \(\mathrm{CuX}_{3}^{2-}\) b. \(\mathrm{CuX}_{2}^{-}\) c. \(\mathrm{Cu}^{+}\)

A 50.0 -mL sample of \(0.0413\) \(M\) \(\mathrm{AgNO}_{3}(a q)\) is added to \(50.0 \mathrm{mL}\) of 0.100 \(M \mathrm{NaIO}_{3}(a q) .\) Calculate the \(\left[\mathrm{Ag}^{+}\right]\) at equilibrium in the resulting solution. \(\left[K_{\mathrm{sp}} \text { for } \mathrm{AgIO}_{3}(s)=3.17 \times 10^{-8} .\right]\)

Calculate the solubility of solid \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\left(K_{\mathrm{sp}}=1.3 \times 10^{-32}\right)\) in a \(0.20-M \mathrm{Na}_{3} \mathrm{PO}_{4}\) solution.

Order the following solids (a-d) from least soluble to most soluble. Ignore any potential reactions of the ions with water. a. \(\mathrm{AgCl} \quad K_{\mathrm{sp}}=1.6 \times 10^{-10}\) b. \(\mathrm{Ag}_{2} \mathrm{S} \quad K_{\mathrm{sp}}=1.6 \times 10^{-49}\) c. \(\mathrm{CaF}_{2} \quad K_{\mathrm{sp}}=4.0 \times 10^{-11}\) d. CuS \(\quad K_{\mathrm{sp}}=8.5 \times 10^{-45}\)
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