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

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 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 A) what is The concentration of each ion in solution? B) what is the total concentration of ions in this solution

Explanation / Answer

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

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.