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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

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ajeigbeibraheem

Answer:

[tex]\mathbf{0.02 M}[/tex]

Explanation:

[tex]\text{So, from the given question:}[/tex]

[tex]\text{EDTA will make complex with}[/tex] [tex]V^{+3}[/tex] [tex]\text{and the remaining EDTA will react with }[/tex][tex]Ga^{+3}[/tex]

[tex]\text{Hence, the total concentration of}[/tex] [tex]V^{+3}[/tex] & [tex]Ga^{+3}[/tex] [tex]\text{will be equivalent to EDTA concentration.}[/tex]

[tex]V_{EDTA} = 25 \ mL[/tex]

[tex]V_{V^{+3}} = 59.0 \ mL[/tex]

[tex]V_{Ga^{+3}} = 13.0 \ mL[/tex]

[tex]M_{EDTA} = 0.0680 \ M[/tex]

[tex]M_{V^{+3}} = ???(unknown)[/tex]

[tex]M_{Ga^{+3}} = 0.0400 \ M[/tex]

[tex]V^{+3} + EDTA \to V[EDTA] + EDTA(Excess) \to^{CoA} \ Ga[EDTA] _{complex}[/tex]

[tex]M_{EDTA} \times V_{EDTA} = ( V_{V^+3}\times M_{V^{+3}}+ V_{Ga^{+3} }\times M_{Ga^{+3}}})[/tex]

[tex]0.0680 \times 25 = (59\times x + 13 \times 0.040) \\ \\ 1.7 = 59x + 0.52\\ \\ 1.7 - 0.52 = 59x \\ \\ 59x = 1.18[/tex]

[tex]x = \dfrac{1.18}{59}[/tex]

[tex]\mathbf{x =0.02 \ M }[/tex]

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