Question

The reaction between Al3+ and EDTA is too slow for direct titration. To a 25.00 mL Al3+ sample, 50.00 mL of EDTA is added and allowed to react for 30 minutes. The excess EDTA was titrated with 21.23 mL of 0.100 M Zn2+. What is the concentration of Al3+ in the sample?

a) 0.7877 M

b) 0.3151 M

c) 0.4000 M

Answer 1

the reaction between EDTA and Al3+ and Zn2+ are as follows:

Al3+   + EDTA^-                -à AlEDTA –

Zn2+   + EDTA^-                -à ZnEDTA- -

Calculate eth moles of Zn2+

Number of moles = molarity * volume in L

= 0.100* 0.02123

= 0.002123 moles Zn2+

Or these number of EDTA (excess) which reacted with Zn2+

Total Moles of EDTA =

Number of moles = molarity * volume in L

= 0.200* 0.050

= 0.01 moles EDTA

Moles of EDTA which reacted with Al3+

= total mole s- moles of EDTA reacted with Zn2+

= 0.01 – 0.002123

= 0.007877 MOLES

These are the number of mole sof Al3+ also because reaction occurs in 1:1 ratio.

Molarity = number of moles / volume in L

= 0.007877 / 0.025

= 0.3151M