Suppose 38.1 g of iron(II) iodide is dissolved in 350. mL of a 0.40M aqueous sol

Question

Suppose 38.1 g of iron(II) iodide is dissolved in 350. mL of a 0.40M aqueous solution of silver nitrate. Calculate the final molarity of iodide anion in the solution. You can assume the volume of the solution doesn't change when the iron(II) iodide is dissolved in it.
Suppose 38.1 g of iron(II) iodide is dissolved in 350. mL of a 0.40M aqueous solution of silver nitrate. Calculate the final molarity of iodide anion in the solution. You can assume the volume of the solution doesn't change when the iron(II) iodide is dissolved in it.
Calculate the final molarity of iodide anion in the solution. You can assume the volume of the solution doesn't change when the iron(II) iodide is dissolved in it.

Explanation / Answer

using molar mass, find moles of iodide:
(38.1g of iron(II) iodide) / 309.65394grams iron(II)  iodide per mol ) = 0.123moles of  iron(II)  iodide
&
0.123 moles of iron(II)  iodide releases an equal 0.123 moles of iodide ions


using molarity, find moles of AgNO3:
(0.350 Litres) (0.4 mol AgNO3 / Litre) = 0.14 moles of AgNO3
&
0.14 moles of AgNO3 releases an equal 0.14 moles of Ag+ ions


1 mol Ag+1ions reacts with 1 mol iodide ions to form 1mole of AgI precipitate
so,
0.14 moles of Ag+ ions remove an equal 0.14 moles of iodide ions from the solution
(0.123 mol iodide) - (0.14 mol iodide removed) = 0.0025 moles iodide remain in solution

find molarity of iodide:
(0.0025 moles iodide) / (0.050 litres) = 0.05 Molar iodide ion

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