Solution:
SOLUTION
Since equilibrium constant for the formation (stability constant) of the complex is of the order of 1018, almost all Ag+ ions will enter into complex ion formation.
Hence, [Ag(CN)2]- = 0.03 M
[CN-] = 0.1 - 2 × 0.03 = 0.04 M
K = $$\cfrac{[Ag(CN)_{2}^{-}]}{[A{{g}^{+}}]{{[C{{N}^{-}}]}^{2}}}$$ = 2.5 × 1018 (given)
[Ag+] = $$\cfrac{[Ag(CN)_{2}^{-}]}{{{[C{{N}^{-}}]}^{2}}\times 2.5\times {{10}^{18}}}$$
= $$\cfrac{0.03}{{{0.04}^{2}}\times 2.5\times {{10}^{18}}}$$ = 0.75 × 10-17 M