Question

A 25.0 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?

Answer 1

Explanation:

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

Therefore, we will calculate the moles of $Ga^{3+}$ as follows.

   Moles of $Ga^{3+}$ 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 $V^{3+}$ will be as follows.

                = 0.0015 - 0.000616

                = 0.00088 moles

Since, we have supposed a 1 : 1 ratio between $V^{3+}$ and EDTA .

So, moles of $V^{3+}$ = 0.00088 moles

Now, we will calculate the molarity of $V^{3+}$ as follows.

       Molarity of $V^{3+}$ solution = $\frac{\text{moles}}{\text{volume (L)}}$

                             = $\frac{0.00088 moles}{0.057 L}$

                             = 0.015 M

Thus, we can conclude that the original concentration of the $V^{3+}$ solution is 0.015 M.