1.1 x 10-4 g of Cr(OH)3 and 4.0 x 10-4 g of NaOH is added to 120 L of water at 25 °C. Will a precipitate form?

To determine whether a precipitate will form when you mix chromium(III) hydroxide, Cr(OH)₃, and sodium hydroxide, NaOH, in water, we need to consider the solubility of Cr(OH)₃ and the concentrations of the reactants involved.

Understanding the Components

Chromium(III) hydroxide is known to be a sparingly soluble compound in water. Its solubility product constant (Ksp) at 25 °C is approximately 6.7 x 10⁻³³. Sodium hydroxide, on the other hand, is highly soluble in water and will dissociate completely into sodium ions (Na⁺) and hydroxide ions (OH⁻).

Calculating Concentrations

First, let's calculate the number of moles of each compound present in the solution:

• For Cr(OH)₃:
  • Molar mass of Cr(OH)₃ = 52.00 (Cr) + 3 × 16.00 (O) + 3 × 1.01 (H) = 103.00 g/mol
  • Number of moles = mass (g) / molar mass (g/mol) = (1.1 x 10⁻⁴ g) / (103.00 g/mol) ≈ 1.07 x 10⁻⁶ mol
• For NaOH:
  • Molar mass of NaOH = 22.99 (Na) + 16.00 (O) + 1.01 (H) = 40.00 g/mol
  • Number of moles = (4.0 x 10⁻⁴ g) / (40.00 g/mol) = 1.00 x 10⁻⁵ mol

Determining Hydroxide Ion Concentration

Next, we need to find the concentration of hydroxide ions in the solution. Since NaOH dissociates completely, the concentration of OH⁻ ions will be equal to the number of moles of NaOH divided by the volume of the solution in liters:

• Concentration of OH⁻ = (1.00 x 10⁻⁵ mol) / (120 L) = 8.33 x 10⁻⁸ M

Calculating the Ion Product

Now, we can calculate the ion product (Q) for Cr(OH)₃ using the concentration of hydroxide ions:

• Q = [Cr³⁺][OH⁻]³
• Since we have 1.07 x 10⁻⁶ mol of Cr(OH)₃ in 120 L, the concentration of Cr³⁺ ions will be negligible compared to the concentration of OH⁻ ions.
• Assuming complete dissociation, we can estimate Q as follows:
• Q ≈ (1.07 x 10⁻⁶ mol / 120 L) × (8.33 x 10⁻⁸ M)³

Comparing Q to Ksp

Finally, we compare the calculated ion product (Q) to the solubility product constant (Ksp) of Cr(OH)₃:

• If Q > Ksp, a precipitate will form.
• If Q < Ksp, no precipitate will form.

Given that Ksp for Cr(OH)₃ is 6.7 x 10⁻³³, and considering the very low concentrations involved, it is likely that Q will be significantly less than Ksp. Therefore, we can conclude that a precipitate of Cr(OH)₃ will not form under these conditions.

Final Thoughts

In summary, the combination of 1.1 x 10⁻⁴ g of Cr(OH)₃ and 4.0 x 10⁻⁴ g of NaOH in 120 L of water at 25 °C will not result in the formation of a precipitate, as the conditions do not exceed the solubility limits of chromium(III) hydroxide.