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Chapter 16: Problem 33

Calculate the molar solubility of \(\mathrm{Al}(\mathrm{OH})_{3}, K_{\mathrm{sp}}=2 \times 10^{-32}\).

Short Answer

Expert verified
The molar solubility of \(\mathrm{Al}(\mathrm{OH})_{3}\) is approximately \(1.67 \times 10^{-9}\) moles per liter.

Step by step solution

01

Write the balanced chemical equation

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

Define the solubility variables

Let's denote the molar solubility of \(\mathrm{Al}(\mathrm{OH})_{3}\) as x. Then, the concentration of \(\mathrm{Al}^{3+}\) ions in the solution at equilibrium will also be x, while the concentration of \(\mathrm{OH}^-\) ions in the solution at equilibrium will be 3x, since there are three \(\mathrm{OH}^-\) ions for every one \(\mathrm{Al}^{3+}\) ion.
03

Write the expression for the solubility product constant, \(K_{\mathrm{sp}}\)

The expression for \(K_{\mathrm{sp}}\) of aluminum hydroxide is given by the product of the concentrations of the ions in the solution at equilibrium, raised to the power of their respective stoichiometric coefficients: \[K_{\mathrm{sp}} = [\mathrm{Al}^{3+}] [\mathrm{OH}^-]^3\]
04

Substitute the solubility variables into the \(K_{\mathrm{sp}}\) expression

We are given the value of \(K_{\mathrm{sp}}=2\times10^{-32}\). We can now substitute the solubility variables into the expression for \(K_{\mathrm{sp}}\). The concentration of \(\mathrm{Al}^{3+}\) ions in the solution at equilibrium is x, and the concentration of \(\mathrm{OH}^-\) ions is 3x, so the \(K_{\mathrm{sp}}\) expression becomes: \(2\times10^{-32} = x(3x)^3=27x^4\)
05

Calculate the molar solubility

To find the molar solubility of the aluminum hydroxide, we must solve the equation for x: \[\frac{2\times10^{-32}}{27}=x^4\] First, divide 2 by 27: \[\frac{2}{27}\times10^{-32}\approx7.41\times 10^{-34}=x^4\] Now, take the fourth root of both sides of the equation to solve for x: \[x \approx \sqrt[4]{7.41 \times 10^{-34}}\] \[x \approx 1.67 \times 10^{-9}\] The molar solubility of \(\mathrm{Al}(\mathrm{OH})_{3}\) is approximately \(1.67 \times 10^{-9}\) moles per liter.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solubility Product Constant
The solubility product constant, designated as <em>K<sub>sp</sub></em>, is a key concept in studying the solubility of sparingly soluble salts. It represents the extent to which a compound can dissolve in water. Specifically, <em>K<sub>sp</sub></em> is the product of the molar concentrations of the ions in a saturated solution, each raised to the power of its stoichiometric coefficient as indicated in the balanced chemical equation. This equilibrium constant provides a quantitative measure of how much of a solid salt dissolves to form a saturated solution.<br>For the example of aluminum hydroxide, \(\mathrm{Al}(\mathrm{OH})_3\), a very low <em>K<sub>sp</sub></em> value (\(2 \times 10^{-32}\)) indicates that it is very sparingly soluble in water. The exercise requires understanding that the concentrations of <em>Al<sup>3+</sup></em> and <em>OH<sup>-</sup></em> ions at equilibrium are directly related to this constant. When solving for the molar solubility, which is the solute concentration in a saturated solution, correct application of this constant is crucial.<br><strong>To achieve better comprehension:</strong> envision <em>K<sub>sp</sub></em> as a 'tipping point' of concentration for the ions in solution. If we have a solution where the ion product exceeds <em>K<sub>sp</sub></em>, we know a precipitate will form because the solution is supersaturated. Conversely, if the ion product is less than <em>K<sub>sp</sub></em>, it indicates that more solid can dissolve until equilibrium is reached.
Chemical Equilibrium
Chemical equilibrium is the state reached when the concentrations of reactants and products remain constant over time, because the rate of the forward reaction equals the rate of the reverse reaction. For soluble substances, this dynamic state allows us to determine the molar solubility in a saturated solution.<br>For instance, in the dissolution of \(\mathrm{Al}(\mathrm{OH})_3\), the equilibrium is represented by the equation \(\mathrm{Al}(\mathrm{OH})_{3} (s) \rightleftharpoons \mathrm{Al}^{3+} (aq) + 3 \mathrm{OH}^- (aq)\). At equilibrium, the rate of dissolution equals the rate of precipitation, leading us to calculate the molar solubility using the provided <em>K<sub>sp</sub></em>.<br>To fully grasp the concept, it's crucial to recognize that the equilibrium does not signify that reactants and products are in equal concentrations but that their rates of formation are balanced. When disturbances occur, such as adding more solute or solvent, the system will shift to restore the balance of equilibrium according to Le Chatelier's principle. It is this principle that underpins the changes in solubility in response to variables such as temperature and common ion effect.
Stoichiometry
Stoichiometry is the part of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. Utilizing stoichiometry allows us to predict how much reactants are needed or how much products will be formed under given conditions.<br>In terms of solubility, stoichiometry tells us the ratio in which ions are produced upon dissolution. For aluminum hydroxide, the stoichiometric coefficients imply that one mole of \(\mathrm{Al}(\mathrm{OH})_3\) results in one mole of \(\mathrm{Al}^{3+}\) ions and three moles of \(\mathrm{OH}^-\) ions. Thus, if the molar solubility is <em>x</em>, \(\mathrm{Al}^{3+}\) has a concentration of <em>x</em> and \(\mathrm{OH}^-\) would have a concentration of <em>3x</em> as seen in step 2 of the solution.<br><strong>Practical tip:</strong> When attempting to solve a problem like this, always balance the chemical equation first. Then, establish the stoichiometric relationships between the species involved. Lastly, apply these relationships proportionally to link molar solubility with the solubility product constant. Understanding the stoichiometry enables students to correctly set up the relation between the molar solubility and the concentrations of the ions at equilibrium, a necessity for solving equilibrium problems.

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Most popular questions from this chapter

The \(K_{\text {? }}\) for \(\mathrm{PbI}_{2}(s)\) is \(1.4 \times 10^{-8} .\) Calculate the solubility of \(\mathrm{PbI}_{2}(s)\) in \(0.048 M\) NaI.

A solution is prepared by mixing \(100.0 \mathrm{~mL}\) of \(1.0 \times 10^{-2} M\) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) and \(100.0 \mathrm{~mL}\) of \(1.0 \times 10^{-3} M \mathrm{NaF}\). Will \(\mathrm{PbF}_{2}(s)\) \(\left(K_{\text {sp }}=4 \times 10^{-8}\right)\) precipitate?

Calcium oxalate \(\left(\mathrm{CaC}_{2} \mathrm{O}_{4}\right)\) is relatively insoluble in water \(\left(K_{\mathrm{sp}}=2 \times 10^{-9}\right) .\) However, calcium oxalate is more soluble in acidic solution. How much more soluble is calcium oxalate in \(0.10 M \mathrm{H}^{+}\) than in pure water? In pure water, ignore the basic properties of \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\).

Calculate the concentration of \(\mathrm{Pb}^{2+}\) in each of the following. a. a saturated solution of \(\mathrm{Pb}(\mathrm{OH})_{2}, K_{\mathrm{sp}}=1.2 \times 10^{-15}\) b. a saturated solution of \(\mathrm{Pb}(\mathrm{OH})_{2}\) buffered at \(\mathrm{pH}=13.00\) c. Ethylenediaminetetraacetate \(\left(\mathrm{EDTA}^{4-}\right)\) is used as a complexing agent in chemical analysis and has the following structure: Solutions of EDTA \(^{4-}\) are used to treat heavy metal poisoning by removing the heavy metal in the form of a soluble complex ion. The reaction of EDTA \(^{4-}\) with \(\mathrm{Pb}^{2+}\) is \(\begin{aligned} \mathrm{Pb}^{2+}(a q)+\mathrm{EDTA}^{4-}(a q) \rightleftharpoons \mathrm{PbEDTA}^{2-}(a q) & \\ K &=1.1 \times 10^{18} \end{aligned}\) Consider a solution with \(0.010\) mole of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) added to 1.0 L of an aqueous solution buffered at \(\mathrm{pH}=13.00\) and containing \(0.050 M\) NasEDTA. Does \(\mathrm{Pb}(\mathrm{OH})_{2}\) precipitate from this solution?

You are browsing through the Handbook of Hypothetical Chemistry when you come across a solid that is reported to have a \(K_{\text {se }}\) value of zero in water at \(25^{\circ} \mathrm{C}\). What does this mean?
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