`0.001 mol` of solid `NaC1` was added to `1.0L` of `0.01M Hg (NO_(3))_(2)`. Calculate `[c1^(Theta)]` equilibrated with newly formed `HgC1^(o+). K_(1)` for `HgC1^(o+)` formation is `5.5 xx 10^(6)`, neglect the `K_(2)` equilibrium.

Explanation

Since the equilibrium constant is so large, very little of one or both reactants will remain at equilirbium.
`{:(,Hg^(2+)+,C1^(Theta)hArr,HgC1o+,),("Initial",rArr0.01,0.001,0,),("Used up",rArr0.001-x,0.001-x,,),("Equilibrium",rArr0.009+x,,,),(,~~0.009,x,0.001-x~~0.001,):}`
`K_(1) = ([HgC1^(o+)])/([Hg^(2+)][C1^(Theta)])`
`5.5 xx 10^(6) = (0.001)/((0.009)x) rArr x = 2.0 xx 10^(-8)M`