Question

calculate the concentration of free Ni2+ in a solution prepared by mixing 0.100 mol of ethylenediamine plus 1.00 mL of 0.0100 M Ni2+ and diluting to 1.00 L with dilute base

Answer 1

Answer – Given, mol of ethylenediamine = 0.100 mole

1. mL of 0.0100 M Ni²⁺, final volume = 1.00 L = 1000 mL

we know the reaction

Ni²⁺ + 3en -----> [Ni(en)₃]²⁺

Moles of Ni²⁺ = 0.0100 M * 0.001 L

                        = 1.0*10^-5 moles

So, for 3 moles of en = 1 moles of Ni²⁺

So, 0.100 moles of en = ?

= 0.0333 moles

So all mole of Ni²⁺ are free, since there is very less number of moles of Ni ²⁺ ion

So according to dilution law

M1V1 = M2V2

M2= 0.0100 M * 1.00 mL / 1000 mL

     = 1.0*10^-5 M

So, the concentration of free Ni²⁺ in a solution is 1.0*10^-5 M