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Calculate the barium ion concentration and iodide ion concentration in an aqueous solution containing 9.44g of barium iodide , BaI2 in 250 mL of solution. The molar mass of BaI2 is 391.1 g/mol

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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.

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