Question

The reaction between Al³⁺ and EDTA is too slow for direct titration. To a 25.00 mL Al³⁺ sample 50.00 mL of 0.2000 M EDTA is added and allowed to react for 30 minutes. The excess EDTA was titrated with 21.23 mL of 0.1000 M Zn²⁺. What is the concentration of Al³⁺ in the sample?

Answer 1

The general reaction between Al3+ with EDTA and Zn 2+ with EDTA are as follows:

Al3+   + EDTA ^4-   = AlEDTA-

Zn2+   + EDTA ^4-   = ZnEDTA2-

Moles of Zn^2+:

Molarity * volume in L

= 0.1000 *0.02123

= 0.002123 Moles

These moles of Zn^2+ reacted with remaining EDTA^4-.

Total moles of EDTA:

Molarity * volume in L

=0.2000 *0.050

= 0.01 Moles EDTA^4-

Moles of EDTS^4- which reacted with Al3+= Total moles of EDTA – moles of EDTA reacted with Zn2+

= 0.01- 0.002123 Moles

= 7.877*10^-3 moles EDTA

These mole reacted with Al3+ in 1:1 ratio

Therefore the moles of Al3+ = 7.877*10^-3 moles

Molarity = number of moles / volume in L

Here volume in L = 0.025 l

Molarity = number of moles / volume in L

= 7.877*10^-3 moles/0.025 l

=0.315 M