What is the concentration of M+ ions at equilibrium?

Question

The formation constant* of [M(CN) 2 ]− is 5.30×1018 , where M is a generic metal. A 0.130 mole quantity of M(NO3) is added to a liter of 0.520 M NaCN solution.

Answer 1

M⁺ + 2CN^- ==> [M(CN)₂]^-

K_f = 5.30x1018 which is very large indicating that most of the 0.130 moles of M⁺ will react.

K_f = M(CN)₂^- / [M⁺][CN^-]²

M⁺ + 2CN^- ==> [M(CN)₂]^-

0.13....0.52.........0..........Initial...

-0.13....-0.26......+0.13....Change....

~0.......0.26........0.13......Equilibrium

[CN^-] = 0.26

[M(CN)₂^-] = 0.13

5.30x10¹⁸ = 0.130 / [M⁺][0.26]²

5.30x10¹⁸ = 0.130/0.0676[M⁺]

[M⁺] = 0.130/(5.30x10¹⁸)(0.0676)

[M⁺] = 3.63x10^-19 M