The chemical EDTA forms a strong 1:1 bond with most metals, allowing it to be ti

Question

The chemical EDTA forms a strong 1:1 bond with most metals, allowing it to be titrated similar to an acid-base or spectrophotometric titration we saw on the last test. You are asked to titrate a 25.0 mL sample that contains an unknown concentration of copper ions. You titrate this sample with 18.64 mL of 0.01258 M ETDA. Use the dilution equation (which essentially makes moles equal) to determine the concentration of copper in the 25 mL sample. When reactions take a long time, excess ETDA is added to the solution to react with the first metal. For example, you would add 25.0 mL of 0.02456 M EDTA to a 50.0 solution that contains nickel. You need to understand that the amount of EDTA is excess. After sitting overnight, the excess EDTA that has not reacted with the nickel is titrated with zinc. The zinc reaction happens fast. It take 12.25 mL of 0.001278M zinc to titrate the excess EDTA. Total moles EDTA = 25.0 mL * 0.02456 M = 0.614 mmoles Total excess EDTA (as given my Zn) = 12.25 mL * 0.01278 M = 0.1565 mmol. 0.614 - 0.1565 = 0.4575 mmol of nickel in 50 mL. Molarity of nickel is 0.00915 M This is an example of a back titration. We will see more on Monday.

Explanation / Answer

1) We use H2EDTA2- to titrate with metal ions. Write the balanced chemical equation for titration of Cu2+ ions with H2EDTA2-:

Cu2+ (aq) + H2EDTA2- (aq) --------> Cu-EDTA (aq)+ 2 H+ (aq)

As stated in the problem, we see from the stoichiometric equation that the molar ratio of reaction between Cu2+ and H2EDTA2- is 1:1.

Calculate moles of EDTA used from the volume and the molarity:

Moles of EDTA used = (volume of EDTA in L)*(concentration of EDTA in mol/L) = (18.64 mL)*(1 L/1000 mL)*(0.01258 mol/L) = 2.344912*10-4 mole.

As per the stoichiometric equation, moles of Cu2+ reacted = (2.344912*10-4 mole EDTA)*(1 mole Cu2+/1 mole EDTA) = 2.344912*10-4 mole.

Volume of Cu2+ solution taken = 25.0 mL = (25.0 mL)*(1 L/1000 mL) = 0.025 L

Molarity of Cu2+ solution = moles of Cu2+/ volume of Cu2+ solution in L = (2.344912*10-4 mole)/(0.025 L) = 9.3796*10-3 mol/L 9.38*10-3 M = 0.00938 M

Ans: The concentration of Cu2+ in 25.0 mL solution is 0.00938 mol/L.

Problem 2 is solved.

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