Question

Aluminum ion may be precipitated from aqueous solution by addition of hydroxide ion, forming \(\mathrm{Al}(\mathrm{OH})_{3} .\) A large excess of hydroxide ion must not be added, however, because the precipitate of \(\mathrm{Al}(\mathrm{OH})_{3} \quad\) will redissolve to form a soluble aluminum/hydroxide complex. How many grams of solid NaOH should be added to \(10.0 \mathrm{mL}\) of \(0.250 \mathrm{M}\) \(\mathrm{AlCl}_{3}\) to just precipitate all the aluminum?

Short answer

To just precipitate all the aluminum ions from 10.0 mL of 0.250 M AlCl3 solution, 0.2998 g of solid NaOH should be added.

Step-by-step solution

Write the balanced chemical equation for the precipitation reaction

The balanced chemical equation for the precipitation of aluminum hydroxide is: \(Al^{3+} + 3OH^{-} \rightarrow Al(OH)_{3(s)}\)

Calculate the moles of Al3+ ions in the given solution

We have 10.0 mL of 0.250 M AlCl3 solution. To calculate the moles of Al3+ ions, we will use the formula: moles of Al3+ = Volume of AlCl3 solution × Molarity of AlCl3 solution moles of Al3+ = 0.010 L × 0.250 mol/L = 0.00250 mol

Calculate the moles of OH- ions required

From the balanced chemical equation, 1 mole of Al3+ reacts with 3 moles of OH-. So, moles of OH- = moles of Al3+ × 3 moles of OH- = 0.00250 mol × 3 = 0.00750 mol

Calculate the moles of NaOH required

Each mole of NaOH provides 1 mole of OH- ions. So, moles of NaOH = moles of OH- moles of NaOH = 0.00750 mol

Calculate the grams of NaOH required

To find the grams of solid NaOH, we will use the formula: grams of NaOH = moles of NaOH × molar mass of NaOH The molar mass of NaOH = 22.99 g/mol (for Na) + 15.999 g/mol (for O) + 1.008 g/mol (for H) = 39.997 g/mol grams of NaOH = 0.00750 mol × 39.997 g/mol = 0.2998 g Hence, 0.2998 g of solid NaOH should be added to 10.0 mL of 0.250 M AlCl3 solution to just precipitate all the aluminum ions.

Key concepts

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They describe the reactants and products involved in a chemical process using chemical formulas. In our exercise, the equation for the precipitation reaction is:

• \( \text{Al}^{3+} + 3\text{OH}^{-} \rightarrow \text{Al(OH)}_{3(s)} \)

This equation shows that one aluminum ion (\( \text{Al}^{3+} \)) reacts with three hydroxide ions (\( \text{OH}^{-} \)) to form solid aluminum hydroxide (\( \text{Al(OH)}_3 \)).
The equation is considered balanced when the number of atoms for each element is equal on both sides.
Balanced equations are crucial because they reflect the law of conservation of mass.
In chemical equations, the states of matter for each compound are often indicated, as in our exercise where \( \text{(s)} \) indicates a solid precipitation.

Molarity Calculations

Molarity is a measure of the concentration of a solute in a solution, expressed as moles of solute per liter of solution. To find molarity, use the formula:

• \[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \]

In our exercise, the aluminum chloride (\( \text{AlCl}_3 \)) solution has a known molarity of 0.250 M and volume of 10.0 mL (converted to 0.010 L for calculations).
Molarity calculations help us determine how many moles of \( \text{Al}^{3+} \) ions are present.
This is important because it tells us the exact quantity of ions available to react with \( \text{OH}^{-} \) to form the precipitate.

Ion Precipitation

Ion precipitation involves forming a solid from a solution during a chemical reaction.
When ions in solution combine to form an insoluble compound, a precipitate is sometimes formed. For aluminum hydroxide:

• Aluminum ions (\( \text{Al}^{3+} \)) combine with hydroxide ions (\( \text{OH}^{-} \))
• The product is aluminum hydroxide (\( \text{Al(OH)}_3 \)) which is insoluble in water

This process often relies on the solubility product constant, which dictates conditions under which a precipitate will form.
Only a specific ratio of ions will result in precipitation, as shown in our balanced equation where three \( \text{OH}^{-} \) ions are required for every \( \text{Al}^{3+} \) ion.
Understanding precipitation allows chemists to separate and purify substances through techniques like filtration.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions.
It involves using balanced chemical equations to determine the relationships between the amounts of reactants and products.
For aluminum hydroxide precipitation from \( \text{AlCl}_3 \), stoichiometry tells us:

• Each mole of \( \text{Al}^{3+} \) reacts with 3 moles of \( \text{OH}^{-} \)
• Using stoichiometry, we calculated the moles of \( \text{OH}^{-} \) required as moles of \( \text{Al}^{3+} \) (0.00250 mol) multiplied by 3, resulting in 0.00750 mol of \( \text{OH}^{-} \)

This relationship helped us determine how much \( \text{NaOH} \) is needed, as each mole of \( \text{NaOH} \) provides one mole of \( \text{OH}^{-} \).
The stoichiometric calculations lead us to conclude that 0.2998 g of \( \text{NaOH} \) are necessary to completely precipitate all aluminum ions in the solution.