What is the concentration of free `Cd^(2+)` in `0.005M CdC1_(2)? K_(1)` for chloride complexation of `Cd^(2+)` is `100, K_(2)` need not be considered.

Explanation

Assume that `CdC1_(2)` is fully ionised and then undergoes the reaction:
`[[Cd^(2+)] = 0.005 M [C1^(Θ)] = 2 xx 0.005 = 0.01M]`
`{:(,,Cd^(2+)+,C1^(Θ)rarr,CdC1^(o+)),("Intial mol",rArr,0.005,0.01,0),("Moles used up",rArr,0.005-x,0.005-x,-),("Moles at equilibrium",rArr,0.005-(0.005-x),0.01-(0.005-x),),(,,~~x,=0.005+x,(0.005-x)):}`
`K_(1) = ([CdC1^(o+)])/([Cd^(2+)][C1^(Θ)]) =100`
`(0.005 -x)/(x(0.005+x)) = 100 rArr 100x^(2) + 1.5 x - 0.005 = 0`
`:. x = [Cd^(2+)] = 2.8 xx 10^(-3) M`