The formation constant* of [M(CN) 2 ]− is 5.30×1018 , where M is a generic metal. A 0.150 mole quantity of M(NO3) is added to a liter of 0.680 M NaCN solution.

Question

What is the concentration of M⁺ ions at equilibrium?

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, leaving very little.

K_f = M(CN)₂^- / [M⁺][CN^-]²

Setting up an ICE table, we have ...

M⁺ + 2CN^- ==> [M(CN)₂]^-

0.15....0.68.........0..........Initial...

-0.15....-0.30......+0.15....Change....

~0.......0.38........0.15......Equilibrium

[CN^-] = 0.38

[M(CN)₂^-] = 0.15

K_f = M(CN)₂^- / [M⁺][CN^-]²

5.30x10¹⁸ = 0.150 / [M⁺][0.38]²

5.30x10¹⁸ = 0.150 / 0.144[M⁺]

[M⁺] = 0.150 / (5.30x10¹⁸)(0.144)

[M⁺] = 1.97x10^-19 M (pretty close to 0, as stated above)

(be sure to check the math)