Question

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

Short answer

The molar solubility of \(\mathrm{Al}(\mathrm{OH})_{3}\) is approximately \(7.4 \times 10^{-9} \,\text{mol/L}\).

Step-by-step solution

Write down the dissociation equation of \(\mathrm{Al}(\mathrm{OH})_{3}\)

The dissociation equation of \(\mathrm{Al}(\mathrm{OH})_{3}\) is \begin{equation*} \mathrm{Al}\left(\mathrm{OH}\right)_{3} \rightleftharpoons \mathrm{Al}^{3+}+3 \mathrm{OH}^{-} \end{equation*}

Write down the expression for \(K_{sp}\)

The expression for the solubility product constant, \(K_{sp}\), is the product of the equilibrium concentrations of the ions, raised to their coefficients in the dissociation equation: \begin{equation*} K_{sp} = [\mathrm{Al}^{3+}][\mathrm{OH}^-]^3 \end{equation*}

Relate equilibrium concentrations of ions and molar solubility

Let the molar solubility of \(\mathrm{Al}(\mathrm{OH})_{3}\) be represented by \(x\). When it dissolves, it produces 1 mol of \(\mathrm{Al}^{3+}\) and 3 mol of \(\mathrm{OH}^-\) ions. Therefore, the equilibrium concentrations of the ions are: \begin{align*} [\mathrm{Al}^{3+}] = x \\ [\mathrm{OH}^-] = 3x \end{align*} Substitute these expressions into the \(K_{sp}\) expression: \begin{align*} K_{sp} = x(3x)^3 \end{align*}

Calculate the molar solubility using the given value of \(K_{sp}\)

We're given that \(K_{sp} = 2 \times 10^{-32}\), so we can substitute this value into the \(K_{sp}\) expression, and solve for \(x\): \begin{align*} 2 \times 10^{-32} &= x(3x)^3 \\ 2 \times 10^{-32} &= 27x^4 \\ x^4 &= \frac{2 \times 10^{-32}}{27} \\ x &= \sqrt[4]{\frac{2 \times 10^{-32}}{27}} \end{align*} We can now calculate the molar solubility, \(x\): \begin{align*} x &\approx 7.4 \times 10^{-9} \end{align*}

Conclusion

The molar solubility of \(\mathrm{Al}(\mathrm{OH})_{3}\) is approximately \(7.4 \times 10^{-9} \,\text{mol/L}\).

Key concepts

Dissociation Equation

Understanding the dissociation equation is a fundamental part of grasping how solutes dissolve in solvents, which is crucial in the study of chemistry. In simple terms, a dissociation equation represents how a compound breaks down into its ions when it dissolves in water.

For the compound aluminum hydroxide, \text{Al}(\text{OH})_3, the dissociation equation can be expressed as follows:
\[ \text{Al}(\text{OH})_3 \rightleftharpoons \text{Al}^{3+} + 3\text{OH}^{-} \]
This equation shows that one molecule of aluminum hydroxide separates into one aluminum ion (\text{Al}^{3+}) and three hydroxide ions (\text{OH}^-) when dissolved. It's important to note the coefficient of 3 in front of the hydroxide ions, which indicates that three hydroxide ions are produced for each aluminum ion. This ratio will be used to calculate solubility and will impact the solubility product constant calculation as well.

Solubility Product Constant

The solubility product constant (\text{K}_{sp}) is a crucial concept in predicting the solubility of a precipitate in a solution. It is used to express the equilibrium between a solid and its respective ions in a saturated solution. The \text{K}_{sp} is determined by the concentrations of the ions at equilibrium, each raised to the power of their stoichiometric coefficients from the dissociation equation.

Formally, the expression for aluminum hydroxide's \text{K}_{sp} is written as:
\[ \text{K}_{sp} = [\text{Al}^{3+}][\text{OH}^-]^3 \]
This form illustrates that the \text{K}_{sp} involves the product of the concentration of aluminum ions times the concentration of hydroxide ions cubed, in alignment with their appearance in the dissociation equation. It's a fixed value at a given temperature and is pivotal in the calculation of molar solubility, which determines how much of the compound can be dissolved in a particular solvent.

Equilibrium Concentrations

Equilibrium concentrations refer to the stable concentrations of products and reactants in a chemical reaction at equilibrium. When a substance is at equilibrium, the rate of its dissolution process equals the rate of its precipitation process. To find the equilibrium concentrations of the ions resulting from a dissolution, one can use the molar solubility of the compound, which is the number of moles that dissolve per liter of solution.

For example, if the molar solubility of aluminum hydroxide, \text{Al}(\text{OH})_3, is represented as 'x,' the equilibrium concentrations of the ions produced would be as follows:
\[ [\text{Al}^{3+}] = x \] \[ [\text{OH}^-] = 3x \]
These values can be substituted back into the \text{K}_{sp} expression to solve for 'x,' which will yield the molar solubility. Thus, understanding equilibrium concentrations is key to relating molar solubility and the solubility product constant in quantitative chemical analysis.