What is the concentration of M2+ ions at equilibrium?

Question

The formation constant* of [M(CN)_4]²⁻ _{is 7.70×1016, where M is a generic metal.A 0.160 mole quantity of M(NO3)2 is added to a liter of 0.910 M NaCN solution.} 

_{What is the concentration of M}²⁺ _{ions at equilibrium?}

Answer 1

M²⁺ + 4CN^- ==> [M(CN)₄]^2-

K_f = 7.70x10¹⁶ This is a very large value indicating that most of the 0.160 moles of M²⁺ will react. So the [M²⁺] ions at equilibrium will approximate 0.

K_f = [M(CN)₄]^2- / [M²⁺][CN^-]⁴ and setting up an ICE table, we have ...

M²⁺ + 4CN^- ==> [M(CN)₄]^2-

0.160.....0.910..............0............Initial

-0.160....-0.64.........+.0.160........Change

~0........0.270............0.160..........Equilibrium

K_f = [M(CN)₄]^2- / [M²⁺][CN^-]⁴

7.70x10¹⁶ = 0.160 / [M²⁺] [0.270]⁴

7.70x10¹⁶ = 0.160 / [M²⁺] [5.314x10^-3]

[M²⁺] = 0.160 / (7.70x10¹⁶)(5.314x10^-3)

[M²⁺] = 3.91x10^-16 M

(be sure to check the math)