In an-aqueous solution of volume 500 ml , when the reaction of `2Ag^(+)+Cu hArr Cu^(2+)+2Ag` reached equilibrium , the `[Cu^(2+)]` was x M. when 500 ml of water if further a added at the equilibrium `[Cu^(2+)]` will be

Options

A. `2x M`
B. `x M`
C. between `x M` and `x/2 M`
D. less than `x/2 M`

Answer

D

Explanation

`{:(2Ag^(+) +,Cu, If V is 500 mL, `K_c = (((x xx 1000)/(500)))/((a xx 1000)/(500))^2) = x/(2a^2)`
If volume is doubled, then a becomes `a//2`, Let x be x. , then if `V = 1000 mL`
`K_c = (x.)/((a//2)^2) = (4x.)/(a^2)`
`therefore x/(2a^2) = (4x.)/(a^2) " or " x. = x/8 " or " x. lt x/2`
Therefore, when 500 mL of water is further added, at the equilibrium `[Cu^(2+)]` will be `x/8` M, which is less than `x/2` M.