Calculate the concentrations of each ion remaining in solution after the precipitation is complete.

Question

The precipitation reaction involves 75.0 mL of .0750 M solution of potassium phosphate and 75.0 mL of .0750 M iron (II) acetate.

I'm very confused as to how to find the concentration of the remaining solutions, particularly because I am not sure if I am supposed to regard the coefficients that are in the balanced equation (for example, when i balanced the equation, i got 6 KC₂H₃O₂ on the side of the products, and those are technically the only ions remaining in solution but i don't know whether i should say that there are 6 K⁺ ions or what....i am extremely confused.) I would appreciate a clear, helpful answer!

Answer 1

Whereas Jim L gave an answer, it appears that you may still be confused, so I thought I'd chime in and attempt to help you better understand. Also, I'm not sure Jim L considered limiting reactants in his answer, and the solution to the problem is quite a bit more complex than the one offered previously.

First, let us write the correctly balanced equation:

2K₃PO₄(aq) + 3Fe(Ac)₂(aq) ==> 6KAc(aq) + Fe₃(PO₄)₂(s) ... Ac is the abbreviation for C₂H₃O₂^- , i.e. acetate

moles K₃PO₄ present = 0.075 L x 0.075 moles/L = 0.005625 moles

moles Fe(Ac)₂ = 0.075 L x 0.075 moles/L = 0.005625 moles

Since the mole ratio of Fe(Ac)₂ : K₃PO₄ is 3:2, the Fe(Ac)₂ will be LIMITING.

Moles KAc formed: 0.005625 moles Fe(Ac)₂ x 6 moles KAc/3 moles Fe(Ac)₂ = 0.01125 moles KAc(aq)

Moles K₃PO₄used in the rx: 0.005625 moles Fe(Ac)₂ x 2 moles K₃PO₄/3 moles Fe(Ac)₂ = 0.00375 moles

Moles K₃PO₄ left over = 0.005625 moles - 0.00375 moles = 0.001875 moles K₃PO₄(aq)

Total volume after reaction = 75 ml + 75 ml = 150 ml = 0.15 L

To find concentrations of each ion in solution, you do the following:

K⁺ ion from KAc: 0.01125 moles K^+/0.15 L = 0.075 M K⁺

K⁺ ion from K₃PO₄: 0.001875 moles K₃PO₄ x 3 moles K₊/mole = 0.005625 moles K⁺/0.15 L = 0.0375 M K⁺

Total concentration of K⁺ in solution = 0.075 M + 0.0375 M = 0.1125 M K⁺

Ac^- ion from KAc: 0.01125 moles KAc x 1 mole Ac^-/mole KAc = 0.01125 moles Ac^-

Final concentration of Ac^- in solution = 0.01125 moles/0.15 L = 0.075 M Ac^-

Final concentration of PO₄^3- ion in solution = 0.001875 moles K₃PO₄ x 1 mole PO₄^3-/mole = 0.001875 moles PO₄^3-/0.15 L = 0.0125 M PO₄^3-

SUMMARY of FINAL CONCENTRATIONS:

[K⁺] = 0.1125 M

[Ac^-] = 0.075 M

[PO₄^3-] = 0.0125 M

Answer 2

Hi Valesia

The balanced equation should have two products. One will be aqueous (aq) which means that the ions in that compound remain in solution. Rereading your question, I think you go that far.

The balanced equation also shows you how many moles of each reactant you start with and the number of moles of Potassium acetate you are left with. So once you calculate the actual number of moles of reactants you start with (using the Molarity of each times the volume of each), you use the mole ratio to find the actual number of moles of potassium acetate that's left. That number, divided by 150 mL (to account for the two solutions you started with) is the concentration of the potassium acetate ions. Both have a valence of 1, so there's no need use the 6. The 6 was taken care of by the mole ratio.