Molarity of Fe2+

Question

Determine the molarity of Fe²⁺ ions in a solution prepared by mixing 56.1 mL of a 0.51 M FeCl₂ solution and 40 mL of a 0.659 M Fe(NO₃)₃.

Answer 1

Hi Daisy,

Before we can start calculating we need to identify which of our solutions contains Fe2+ ions, since the question specified that ionization. That would be the FeCl₂ since the other one has 3 NO₃ ions, which is a 1- charge, whereas the former solution has only 2 Cl ions.

Now that we know what the focus is, lets start calculating, First we need to find the moles of Fe²⁺:

56.1 mL/1000 = 0.0561 L

Molarity = moles/L

0.51 = n/(0.0561)

n = 0.028611 moles of Fe²⁺

Because there is only one Fe²⁺ ion, we don't have to do any multiplication.

Now we need to find the total new volume when combining the 2 solutions together:

V = 0.0561 + 0.040 = 0.0961 L

We will use this now to find the new molarity concentration:

Molarity = moles/L

Molarity = 0.028611/0.0961

Molarity = 0.29 M

Please let me know if that made sense and if you have any questions!