Question

What is the iron (II) ion concentration in a solution prepared by mixing 403 mL of 0.364 M iron (II) nitrate with 417 mL of 0.241 M sodium hydroxide? The Ksp of iron (II) hydroxide is 7.9 × 10-16

Answer 1

Step 1: determine the amount of moles of each compound:

Iron (II) Nitrate:

Fe(NO₃)₂ --------> Fe⁺² + 2 NO₃^-

Moles of nitrate = moles of Fe (II)

Moles of nitrate = 0.364 M * 0.403 L = 0.147 moles

Sodium Hydroxide:

NaOH -----> Na⁺   + OH^-

moles of NaOH = moles of OH^-

moles of NaOH = 0.241 M * 0.417 L = 0.1005 moles

Step 2:

Find the molarity of OH^- in solution:

[OH^-] = 0.1005 moles / (0.403 L + 0.417 L) = 0.123 M

Step 3:

Fe(OH)₂ ----> Fe⁺²  + 2OH^-

Find the concentration of Fe (II) using Ksp:

Ksp = [Fe⁺²] [OH^-]²

[Fe⁺²] = Ksp / [OH^-]² = 7.9*10^-16 / (0.123 M)² = 5.22*10^-14 M