Question

The \(K_{\mathrm{sp}}\) of \(\mathrm{Al}(\mathrm{OH})_{3}\) is \(2 \times 10^{-32} .\) At what \(\mathrm{pH}\) will a \(0.2-M\) \(\mathrm{Al}^{3+}\) solution begin to show precipitation of \(\mathrm{Al}(\mathrm{OH})_{3} ?\)

Short answer

The pH at which a 0.2-M Al³⁺ solution will begin to show precipitation of Al(OH)₃ is approximately 3.33.

Step-by-step solution

Write down the balanced chemical equation, soluble and precipitate species

First, we need to write down the balanced chemical equation for the dissolving of Al(OH)₃ in water: \[ Al(OH)_{3(s)} \rightleftharpoons Al^{3+}_{(aq)} + 3 OH^{-}_{(aq)} \]

Ksp expression and substitution

Now, we can write the equilibrium expression for this reaction, which is the Ksp expression, and substitute the given values into it: \[ K_{sp} = [Al^{3+}][OH^-]^3 \] Given that the concentration of Al³⁺ is 0.2 M: \[ 2 \times 10^{-32} = (0.2)([OH^-])^3 \]

Solve for [OH⁻]

Now, we can solve for [OH⁻] by dividing both sides by 0.2: \[ [OH^-]^3 = \frac{2 \times 10^{-32}}{0.2} \] \[ [OH^-]^3 = 10^{-32} \] Take the cube root of both sides: \[ [OH^-] = 10^{-32/3} \] \[ [OH^-] = 10^{-10.67} \]

Calculate pOH and then pH

Now that we have the [OH⁻] concentration, we can find the pOH and then the pH of the solution. First, using the relationship between pOH and [OH⁻]: \[ pOH = -\log{([OH^-])} \] Substitute the value of [OH⁻] found in step 3: \[ pOH = -\log(10^{-10.67}) \] \[ pOH = 10.67 \] Now, to find the pH, we use the relationship between pH and pOH: \[ pH + pOH = 14 \] Substitute the pOH value found in step 4: \[ pH + 10.67 = 14 \] Solve for pH: \[ pH = 14 - 10.67 \] \[ pH = 3.33 \]

Final Answer

The pH at which a 0.2-M Al³⁺ solution will begin to show precipitation of Al(OH)₃ is approximately 3.33.

Key concepts

Ksp Calculation

Understanding the solubility product constant, given the shorthand notation 'Ksp', is crucial when studying the solubility of sparingly soluble salts. The 'Ksp' reflects the maximum product of the molar concentrations of the ions that form the solid salt in a saturated solution.

Let's dive into an example. The Ksp for aluminum hydroxide, Al(OH)₃, is given as a very small number, typically in the range of 10^-32, indicating low solubility. When calculating the Ksp, we follow a systematic approach:

• Identify all the ions in the dissolving process.
• Write down the balanced chemical equation for dissolution.
• Set up the Ksp expression based on the stoichiometry of the balanced equation.
• Substitute any known concentrations and solve for the unknowns.

Through these steps, you can predict how much of a salt will dissolve in solution and at what conditions you might start to see precipitation.

pH and Solubility

The solubility of a compound is often impacted by the pH of the solution, especially when the compound is a salt containing an anion or cation that is a weak acid or base. For instance, the solubility of Al(OH)₃ increases as the solution becomes more acidic, since H+ ions from the acid react with OH- ions, shifting the equilibrium to dissolve more Al(OH)₃.

In a saturated solution, where a salt has started precipitating, the pH can provide insights into the concentration of hydrogen or hydroxide ions present. By reversing the calculation of the hydroxide ion concentration ([OH-]), as shown in the exercise, one can calculate the pOH and then the pH of the solution, aiding in predicting at what pH a precipitate will form or dissolve.

Precipitation of Salts

Precipitation occurs when a salt exceeds its solubility in a particular solvent, forming a solid. This is directly associated with the Ksp value: a low Ksp signifies a salt that is less soluble and more prone to precipitating out of solution under the right conditions. Precipitation is an important concept in various fields, from wastewater treatment to medicinal drug formulation.

The exercise shows the precipitation of Al(OH)₃ from a solution with 0.2 M Al³⁺ ions. By calculating the pH at which precipitation occurs, we can control processes in industrial and laboratory settings. Understanding precipitation helps in avoiding undesired solidification or ensuring that processes that require the removal of certain ions from a solution can occur effectively.