The formation constant of $\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}$ is $1.25 \times 10^{12}$. What will be the equilibrium concentration of $\mathrm{Cu}^{2+}$ if $0.0125 \mathrm{moles}$ of $\mathrm{Cu}$ is oxidized and put into $1.0 \mathrm{~L}$ of $0.25 \mathrm{M}-\mathrm{NH}_{3}$ solution?
(a) $2.5 \times 10^{-11} \mathrm{M}$
(b) $2.5 \times 10^{-13} \mathrm{M}$
(c) $4 \times 10^{-12} \mathrm{M}$
(d) 0