Question

The formation constant* of [M(CN)6]4– is 2.50 × 1017, where M is a generic metal. A 0.130-mole quantity of M(NO3)2 is added to a liter of 1.150 M NaCN solution. What is the concentration of M2 ions at equilibrium?

Answer 1

[M2+] added = number of moles / volume
                    = 0.13 / 1
                    = 0.13 M

[CN-] added = 1.15 M

M2+      +        6CN-        ---------> [M(CN)6]4–
0.13                 1.15                            0    

Since Kf is very large, reaction will go to completion
From reaction, 1 mol of M2+ reacts with 6 mol og CN-

so, 0.13 mol of M2+ requires= 0.13*6 = 0.78 mol of CN-

So M2+ is limiting reagent and whole of M2+ will react

So,
there will not be any M2+ at equilibrium