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 \(\mathrm{Al}(\mathrm{OH})_{3}\) will redissolve as a soluble compound containing aluminum ions and hydroxide ions begins to form. How many grams of solid \(\mathrm{NaOH}\) 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 in a 10.0 mL solution of 0.250 M AlCl₃, we need to add 0.300 grams of solid NaOH.

Step-by-step solution

Write the balanced chemical equation

For the precipitation reaction of aluminum ions (Al³⁺) with hydroxide ions (OH⁻), we get the following balanced chemical equation: \[Al^{3+}(aq) + 3OH^-(aq) \rightarrow Al(OH)_3(s)\]

Calculate the moles of Al3+ ions

Given the volume of the 0.250 M AlCl3 solution is 10.0 mL or 0.010 L. We can find the moles of Al3+ ions in the solution as follows: Moles of Al3+ ions = Molarity × Volume Moles of Al3+ ions = 0.250 mol/L × 0.010 L = 0.0025 mol

Calculate the moles of OH- ions needed

From the balanced chemical equation we know that 3 moles of OH⁻ ions are needed to precipitate 1 mole of Al³⁺: Moles of OH⁻ ions = 3 × (moles of Al³⁺ ions) Moles of OH⁻ ions = 3 × 0.0025 mol = 0.0075 mol

Calculate the grams of NaOH needed

Given that 1 mole of NaOH contains 1 mole of OH⁻ ions, we can calculate the grams of NaOH, required to provide the necessary OH⁻ ions as follows: Grams of NaOH = (moles of OH⁻ ions) × (molar mass of NaOH) Grams of NaOH = 0.0075 mol × 40.00 g/mol = 0.300 g Thus, we should add 0.300 grams of solid NaOH to 10.0 mL of 0.250 M AlCl3 solution to just precipitate all the aluminum ions.

Key concepts

Chemical Equations

Chemical equations are vital tools in chemistry that represent chemical reactions symbolically. Each side of the equation shows different compounds before and after the reaction.
For instance, in the precipitation reaction given, the equation represents aluminum ions reacting with hydroxide ions to form a solid precipitate, aluminum hydroxide:

• The left side: shows reactants - aluminum ions \(Al^{3+}\) and hydroxide ions \(OH^-\).


• The right side: demonstrates the product - aluminum hydroxide, \Al(OH)_3\.

It’s essential to balance these equations, meaning the number of atoms and charge must remain constant on both sides. This balance signifies the conservation of mass and charge.
In our case:

• We need 3 hydroxide ions per aluminum ion to form \(Al(OH)_3\).
• This balanced equation allows us to understand the proportions needed to achieve the complete reaction.

Mastering chemical equations helps in predicting product formation and necessary reactants.

Molarity Calculations

Molarity, defined as the concentration of a solution, tells us how many moles of solute are present per liter of solution.
It is expressed as moles per liter (mol/L), a critical concept in preparing solutions of known concentrations. For example:

• In our exercise, the molarity of the \(AlCl_3\) solution is 0.250 M, indicating 0.250 moles of \(Al^{3+}\) ions in 1 liter of solution.
• We then calculate the moles of \(Al^{3+}\) ions in the specific volume we have (10.0 mL or 0.010 L) using the formula \(\text{Molarity} \times \text{Volume}\).

This calculation gives us a basis for determining how much of other reactants, such as hydroxide ions, will be needed.
Hence, understanding molarity is key to accurately making up solutions for reactions, especially to avoid adding too much or too little of a reactant.

Stoichiometry

Stoichiometry revolves around quantitative aspects of chemical reactions, allowing chemists to relate quantities of reactants and products.
By relying on balanced chemical equations, stoichiometry enables precise calculations.

• In the given reaction, we employ stoichiometry to determine the exact volume ratios and weights needed of reactants to yield desired products.
• The balanced equation \(Al^{3+} + 3OH^- \rightarrow Al(OH)_3\) tells us three hydroxide ions are needed for every aluminum ion precipitated.

Using stoichiometry, we convert between moles of various substances involved in the reaction.
In our problem, stoichiometry helped us calculate the grams of \(NaOH\) needed, using the relationship:

• Moles of \(OH^-\) needed are three times the moles of \(Al^{3+}\), thus needing 0.0075 moles of \(OH^-\).
• Since \(NaOH\) provides \(OH^-\), we calculated the grams of \(NaOH\) required (0.300 g) using its molar mass.

Mastering stoichiometry ensures reactions are performed efficiently and economically, providing a baseline for experimental and industrial uses.