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Chapter 16: 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 \(\mathrm{Mg}(\mathrm{OH})_{2}\) is approximately \(1.24 \times 10^{-4}\) mol/L.

Step by step solution

01

1. Write the dissolution equation for Mg(OH)2

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

2. Set up the expression for the solubility product constant (Ksp)

Based on the balanced dissolution equation, we can represent the solubility product constant as follows: \[ K_{\mathrm{sp}} = [\mathrm{Mg^{2+}}][\mathrm{OH^-}]^2 \]
03

3. Define the concentration of ions in terms of x

Let's denote the molar solubility of \(\mathrm{Mg}(\mathrm{OH})_{2}\) as x. Then, the concentration of Mg^2+ ions will be x, and the concentration of OH- ions will be 2x. Therefore, we can rewrite the Ksp expression as: \[ K_{\mathrm{sp}} = (x)(2x)^2 \]
04

4. Solve the expression for x

Now we will solve for x, which represents the molar solubility of Mg(OH)2. Plug in the given Ksp value (8.9 x 10^(-12)) and solve for x: \[ 8.9 \times 10^{-12} = x(2x)^2 \] Simplify the expression: \[ 8.9 \times 10^{-12} = 4x^3 \] Divide both sides of the equation by 4: \[ x^3 = \dfrac{8.9 \times 10^{-12}}{4} \] Now, take the cube root of both sides to solve for x: \[ x = \sqrt[3]{\dfrac{8.9 \times 10^{-12}}{4}} \]
05

5. Find the molar solubility of Mg(OH)2

Finally, calculate x using a calculator: \[ x \approx 1.24 \times 10^{-4} \] Therefore, the molar solubility of \(\mathrm{Mg}(\mathrm{OH})_{2}\) is approximately \(1.24 \times 10^{-4}\) mol/L.

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Most popular questions from this chapter

Consider a solution made by mixing \(500.0 \mathrm{~mL}\) of \(4.0 \mathrm{M} \mathrm{NH}_{3}\) and \(500.0 \mathrm{~mL}\) of \(0.40 \mathrm{M} \mathrm{AgNO}_{3} \cdot \mathrm{Ag}^{+}\) reacts with \(\mathrm{NH}_{3}\) to form \(\begin{aligned} \mathrm{AgNH}_{3}{ }^{+} \text {and } \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}+: \\ \mathrm{Ag}^{+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{AgNH}_{3}{ }^{+}(a q) & K_{1}=2.1 \times 10^{3} \\ \mathrm{AgNH}_{3}^{+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q) & K_{2}=8.2 \times 10^{3} \end{aligned}\) Determine the concentration of all species in solution.

The overall formation constant for \(\mathrm{HgI}_{4}{ }^{2-}\) is \(1.0 \times 10^{30} .\) That is, $$ 1.0 \times 10^{30}=\frac{\left[\mathrm{HgI}_{4}{ }^{2-}\right]}{\left[\mathrm{Hg}^{2+}\right]\left[\mathrm{I}^{-}\right]^{4}} $$ What is the concentration of \(\mathrm{Hg}^{2+}\) in \(500.0 \mathrm{~mL}\) of a solution that was originally \(0.010 \mathrm{M} \mathrm{Hg}^{2+}\) and \(0.78 \mathrm{M} \mathrm{I}^{-}\) ? The reaction is $$ \mathrm{Hg}^{2+}(a q)+4 \mathrm{I}^{-}(a q) \rightleftharpoons \mathrm{HgI}_{4}^{2-}(a q) $$

Calculate the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}, K_{\text {sp }}=5.9 \times 10^{-11}\).

When \(\mathrm{Na}_{3} \mathrm{PO}_{4}(a q)\) is added to a solution containing a metal ion and a precipitate forms, the precipitate generally could be one of two possibilities. What are the two possibilities?

The active ingredient of Pepto-Bismol is the compound bismuth subsalicylate, which undergoes the following dissociation when added to water: \(\begin{aligned} \mathrm{C}_{7} \mathrm{H}_{5} \mathrm{BiO}_{4}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{7} \mathrm{H}_{4} \mathrm{O}_{3}{ }^{2-}(a q) \\\ &+\mathrm{Bi}^{3+}(a q)+\mathrm{OH}^{-}(a q) \quad K=? \end{aligned}\) If the maximum amount of bismuth subsalicylate that reacts by this reaction is \(3.2 \times 10^{-19} \mathrm{~mol} / \mathrm{L}\), calculate the equilibrium constant for the preceding reaction.
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