Question

What is the concentration of free Ni 2+ in a solution (pH 12.00) prepared by mixing 50.0 mL of 0.250 M Ga(NO3)3, 40.0 mL of 0.250 M Ni(NO3)2 and 30.0 mL of 1.00 M EDTA?

Answer 1

See EDTA is  ethylenediaminetetraacetic acid.

EDTA is a hexa-dentate ligand. It will form the complex with the metal ions depending on the charge present on them,

so here we will find out the number of moles of each of the species present in the mixture.

n (GaNO3)3 = (50.0 mL / 1000) x 0.250 M = 0.0125 moles,

n (Ni(NO3)2) = (40.0 mL /1000) x 0.250 M = 0.01 moles,

n (EDTA) = (30.0 mL / 1000) x 1.00 M = 0.03 moles ,

Now here we know that EDTA will make one complex with (GaNO3)3 and one complex with (Ni(NO3)2). As the number of moles ( quantity) of EDTA is more than both of (GaNO3)3 and (Ni(NO3)2). Therefore neither free Ni nor free Ga will left behind.

Out of 0.03 EDTA moles, 0.0125 moles will react with (GaNO3)3 to form complex and 0.01 moles will react with 0.01 moles of (Ni(NO3)2 to form complex. So only free EDTA will be left behind.

Free EDTA = 0.03 - ( 0.0125 + 0.01) moles = 0.0025 moles.

Thanks !

I hope this helps.