Question

a solution is prepared by dissolving 87.4 g of iron (III) iodide in 725 ml of water.

A) what is The concentration of each ion in solution?

B) what is the total concentration of ions in this solution

Answer 1

Molar mass of FeI3 = 1*MM(Fe) + 3*MM(I)

= 1*55.85 + 3*126.9

= 436.55 g/mol

mass of FeI3 = 87.4 g

we have below equation to be used:

number of mol of FeI3,

n = mass of FeI3/molar mass of FeI3

=(87.4 g)/(436.55 g/mol)

= 0.2002 mol

volume , V = 725 mL

= 0.725 L

we have below equation to be used:

Molarity,

M = number of mol / volume in L

= 0.2002/0.725

= 0.276 M

This is concentration of FeI3

A)

[Fe3+] = [FeI3] = 0.276 M

[I-] = 3*[FeI3] = 3*0.276 = 0.828 M

B)

total concentration = 0.276 M + 0.828 M

= 1.10 M