Question

A 25.0 mL aliquot of 0.0680 M EDTA was added to a 59.0 mL solution containing an unknown concentration of V3 . All of the V3 present in the solution formed a complex with EDTA , leaving an excess of EDTA in solution. This solution was back-titrated with a 0.0400 M Ga3 solution until all of the EDTA reacted, requiring 13.0 mL of the Ga3 solution. What was the original concentration of the V3 solution

Answer 1

`\mathbf{0.02 M}`

Step-by-step explanation:

`\text{So, from the given question:}`

`\text{EDTA will make complex with}`
`V^(+3)`
`\text{and the remaining EDTA will react with }`
`Ga^(+3)`

`\text{Hence, the total concentration of}`
`V^(+3)` &
`Ga^(+3)`
`\text{will be equivalent to EDTA concentration.}`

`V_(EDTA) = 25 \ mL`

`V_(V^(+3)) = 59.0 \ mL`

`V_(Ga^(+3)) = 13.0 \ mL`

`M_(EDTA) = 0.0680 \ M`

`M_(V^(+3)) = ???(unknown)`

`M_(Ga^(+3)) = 0.0400 \ M`

`V^(+3) + EDTA \to V[EDTA] + EDTA(Excess) \to^(CoA) \ Ga[EDTA] _(complex)`

`M_(EDTA) * V_(EDTA) = ( V_(V^+3)* M_(V^(+3))+ V_(Ga^(+3) )* M_(Ga^(+3))})`

`0.0680 * 25 = (59* x + 13 * 0.040) \\ \\ 1.7 = 59x + 0.52\\ \\ 1.7 - 0.52 = 59x \\ \\ 59x = 1.18`

`x = (1.18)/(59)`

`\mathbf{x =0.02 \ M }`