Question

What is $[Cd^(2+)]$ in $1.0L$ of solution prepared by dissolving $0.001mol Cd(NO(3))(2)$ and $1.5 mmol. NH(3) ? K(d)$ for the dissociation of $Cd(NH(3))(4)^(2+)$ into $Cd^(2+)$ and $4NH_(3)$ is $1.8 xx 10^(-7)$. Neglect the amount of $Cd$ in complexes containing fewer than $4$ ammonia molecules.

Answer 1

${:(,Cd^(2+)+,4NH(3)hArr,Cd(NH(3))(4)^(2+),),("Initial",rArr0.001,1.5,0,),("Used up",rArr0.001-x,4(0.001-x),0.001-x,),("Eq.",rArr~~x,~~1.5,~~0.001,):}$
$Cd(NH(3))(4)^(2+) hArr Cd^(2+) + 4NH(3)$.
$K(d) = ([Cd^(2+)][NH(3)]^(4))/([Cd(NH_(3))_(4)^(2+)])$
$1.8 xx 10^(-7) = ((x)(1.5)^(4))/(0.001)$
$:.x = 3.6 xx 10^(-16) = [Cd^(2+)]$