Final answer:
The barium ion (Ba²+) concentration in an aqueous solution of 9.44 g of barium iodide (BaI₂) in 250 mL is 0.0964 M, and the iodide ion (I-) concentration is 0.1928 M.
Step-by-step explanation:
To calculate the barium ion (Ba²+) concentration and iodide ion (I-) concentration in an aqueous solution containing 9.44 g of barium iodide (BaI₂), we first need to find the number of moles of BaI₂. Using the molar mass of BaI₂, which is 391.1 g/mol, we divide the mass of the solute by the molar mass:
moles of BaI₂ = 9.44 g ÷ 391.1 g/mol = 0.0241 mol
Since the formula of barium iodide is BaI₂, dissolving it in water will release one mole of Ba²+ ions and two moles of I- ions per mole of BaI₂. Therefore, the number of moles of Ba²+ ions is also 0.0241 mol, and for I- ions, it is double that, or 0.0482 mol.
The concentration is calculated by dividing the number of moles by the volume of the solution in liters:
[Ba²+] = 0.0241 mol ÷ 0.250 L = 0.0964 M
[I-] = 0.0482 mol ÷ 0.250 L = 0.1928 M
Thus, the concentration of Ba²+ ions is 0.0964 M, and the concentration of I- ions is 0.1928 M in the solution.