A25.0 ml solution of .0600 m edta was added to a 57.0 ml sample containing an unknown concentration of v3+. all v3+ present formed a complex, leaving excess edta in solution.

this solution was back-titrated with a 0440 m ga3+ solution until all the edta reacted requiring 14.0 ml of the ga3+ solution.

what was the original concentration of the v3+ solution?

Explanation:

Let us assume that the ratio for the given reaction is 1:1.

Therefore, we will calculate the moles of as follows.

   Moles of solution = molarity × volume (L)

                              = 0.0440 M × 0.014 L

                              = 0.000616 moles

Moles of excess EDTA = 0.000616 moles

Also, the initial moles of EDTA will be calculated as follows.

     Total initial moles of EDTA = 0.0600 M × 0.025 L

                                                 = 0.0015

Therefore, moles of EDTA reacted with will be as follows.

                = 0.0015 - 0.000616

                = 0.00088 moles

Since, we have supposed a 1 : 1 ratio between and EDTA .

So, moles of = 0.00088 moles

Now, we will calculate the molarity of as follows.

       Molarity of solution =

                             =

                             = 0.015 M

Thus, we can conclude that the original concentration of the solution is 0.015 M.