In an aqueous solution of volume 500 mL, when the reaction of $2{\mathrm{Ag}}^{+}+\mathrm{Cu}\rightleftharpoons {\mathrm{Cu}}^{2+}+2\mathrm{Ag}$ reached equilibrium the [Cu²⁺] was x M. when 500 mL of water is further added, at the equilibrium [Cu²⁺]will be:

$\underset{\mathrm{a}}{2{\mathrm{Ag}}^{+}}+\mathrm{Cu}\rightleftharpoons \underset{\mathrm{x}}{{\mathrm{Cu}}^{2+}}+2\mathrm{Ag}$
If V is 500 mL
${\mathrm{K}}_{\mathrm{c}}=\frac{\left(\frac{\mathrm{x}\times 1000}{500}\right)}{{\left(\frac{\mathrm{a}\times 1000}{500}\right)}^2}=\frac{\mathrm{x}}{2{\mathrm{a}}^2}$
If volume is doubled, then a becomes a/2; Let x be x', then if V - 1000 mL
$\begin{array}{l}{\mathrm{K}}_{\mathrm{c}}=\frac{{\mathrm{x}}^{\mathrm{\prime }}}{(\mathrm{a}/2{)}^2}=\frac{4{\mathrm{x}}^{\mathrm{\prime }}}{{\mathrm{a}}^2}\\ \therefore \frac{\mathrm{x}}{2{\mathrm{a}}^2}=\frac{4{\mathrm{x}}^{\mathrm{\prime }}}{{\mathrm{a}}^2}\\ {\mathrm{x}}^{\mathrm{\prime }}=\frac{\mathrm{x}}{8} \mathrm{or} {\mathrm{x}}^{\mathrm{\prime }}<\frac{\mathrm{x}}{2}\end{array}$
${\mathrm{K}}_{\mathrm{c}}=\frac{\left[{\mathrm{Cu}}^{2+}\right]}{{\left[{\mathrm{Ag}}^{+}\right]}^2}=\frac{\mathrm{x}}{(\mathrm{a}{)}^2}$
to have K_c constant [Cu²⁺]will be less than $\frac{\mathrm{x}}{2}$.