In the laboratory you dissolve 16.4 g of aluminum iodide in a volumetric flask and add water to a total volume of 500 mL What is the concentration of the iodide anion?
The correct answer and explanation is:
To find the concentration of the iodide anion (I⁻), we follow these steps:
Step 1: Write the formula of aluminum iodide
The chemical formula is AlI₃ (aluminum has a 3+ charge and iodide has a 1− charge, so they combine in a 1:3 ratio).
Step 2: Molar mass of AlI₃
- Al: 26.98 g/mol
- I: 126.90 g/mol × 3 = 380.70 g/mol
Molar mass of AlI₃ = 26.98 + 380.70 = 407.68 g/mol
Step 3: Calculate moles of AlI₃
Moles of AlI₃=16.4 g407.68 g/mol≈0.04023 mol\text{Moles of AlI₃} = \frac{16.4 \, \text{g}}{407.68 \, \text{g/mol}} \approx 0.04023 \, \text{mol}
Step 4: Dissociation of AlI₃ in water
AlI₃ (s)→Al3+(aq)+3 I−(aq)\text{AlI₃ (s)} \rightarrow \text{Al}^{3+} (aq) + 3 \, \text{I}^- (aq)
So, each mole of AlI₃ produces 3 moles of iodide ions (I⁻). Moles of I⁻=0.04023 mol×3=0.1207 mol\text{Moles of I⁻} = 0.04023 \, \text{mol} \times 3 = 0.1207 \, \text{mol}
Step 5: Convert 500 mL to liters
500 mL=0.500 L500 \, \text{mL} = 0.500 \, \text{L}
Step 6: Calculate concentration of I⁻
[I⁻]=0.1207 mol0.500 L=0.2414 mol/L\text{[I⁻]} = \frac{0.1207 \, \text{mol}}{0.500 \, \text{L}} = 0.2414 \, \text{mol/L}
✅ Final Answer:
[I⁻]=0.241 M\boxed{\text{[I⁻]} = 0.241 \, \text{M}}
Explanation (300 words):
To determine the concentration of the iodide ion (I⁻) in solution, we first look at the solute, aluminum iodide (AlI₃). This compound dissociates completely in water, producing 1 Al³⁺ ion and 3 I⁻ ions per formula unit. This 1:3 dissociation ratio is key to the calculation.
Next, we find how many moles of AlI₃ are present in the solution. Using the formula mass (molar mass) of AlI₃ — which is the sum of the atomic masses of aluminum (26.98 g/mol) and three iodines (3 × 126.90 g/mol) — we get 407.68 g/mol. By dividing the given mass (16.4 g) by this molar mass, we calculate the number of moles of AlI₃ to be about 0.0402 mol.
Each mole of AlI₃ provides 3 moles of iodide ions, so multiplying the moles of AlI₃ by 3 gives us 0.1207 mol of I⁻.
To find the concentration, we divide the number of moles of I⁻ by the volume of the solution in liters. Since 500 mL is equal to 0.500 L, we perform the division: 0.1207 mol0.500 L=0.2414 mol/L\frac{0.1207 \, \text{mol}}{0.500 \, \text{L}} = 0.2414 \, \text{mol/L}
Thus, the concentration of iodide ion in the solution is 0.241 M, which expresses how many moles of I⁻ are present per liter of solution.