What is the concentration of M^(2+) ions at equilibrium?

Question

The formation constant of [M(CN)₄]²⁺ is 7.70 * 10¹⁶, where M is a generic metal. A .140 mole quantity of M(NO₃)₂ is added to a liter of .950 M NaCN solution. What is the concentration of M²⁺ ions at equilibrium?

Answer 1

The reaction is M²⁺ + 4CN^- → M(CN)₄^2- (I assume there was an error in the sign of the complex ion.

K expression is C_C/(C_MC_CN⁴) Where C_C is the complex concentration.

C_M0 = .14 M C_CN0 = .95 M

Let x be the amount of complex formed. The equilibrium concentrations can be derived in terms of x via an ICE diagram:

7.7 x 10¹⁶ = x/((.14-x)(.95-4x)⁴ which can be solved numerically.

K is huge, so we could just treat this as an LR problem with 100% conversion. in which case we need 4/1 CN to M and we have .95/.14 = 6.79 (CN is in excess, M is LR). The molarity of the complex will essentially be the molarity of the initial Metal ion.