Question

Suppose 27.1g of zinc iodide is dissolved in 350.mL of a 0.70 M aqueous solution of potassium carbonate. Calculate the final molarity of iodide anion in the solution. You can assume the volume of the solution doesn't change when the zinc iodide is dissolved in it. Round your answer to 3 significant digits.

Answer 1

Final molarity of iodide ion = 0.485 M

Explanation

mass ZnI₂ = 27.1 g

moles ZnI₂ = (mass ZnI₂) / (molar mass ZnI₂)

moles ZnI₂ = (27.1 g) / (319.189 g/mol)

moles ZnI₂ = 0.08490 mol

moles I^- = 2 * (moles ZnI₂)

moles I^- = 2 * (0.08490 mol)

moles I^- = 0.1698 mol

[I^-] = (moles I^-) / (volume of solution in Liter)

[I^-] = (0.1698 mol) / (0.350 L)

[I^-] = 0.485 M