posted by Ashley .
A 25.0-mL solution of 0.0660 M EDTA was added to a 33.0-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 0.0450 M Ga3 solution until all the EDTA reacted, requiring 13.0 mL of the Ga3 solution. What was the original concentration of the V3 solution?
EDTA reacts with both V3+ and Ga3+ in a 1:1 stoichiometric ratio.
Therefore, the moles of EDTA added is equal to the sum of the moles of V3+ present and Ga3+ added.
moles of EDTA= moles V3 + moles of Ga3
(0.025 L)(0.0660 M EDTA)= (x moles of V3) + (0.013 L)(0.0450 M Ga3)
moles of V3= 0.001065
M of V3 = 0.001065 moles / 0.033 L
M of V3 = 0.032 moles/L