The formation constant* of [M(CN) 4 ]2− is 7.70 × 10 16 , where M is a generic metal. A 0.150 mole quantity of M(NO3)2 is added to a liter of 0.820 M NaCN solution. What is the concentration of M2+ ions at equilibrium?

0 M.

Explanation:

Hello,

In this case, the undergoing reaction is:

Nonetheless, it only matters the reaction forming the given complex:

In such a way, the formation constant turns out:

Now, one could assume that the initial concentrations of the ions equals the original compounds concentrations:

In such a way, we modify the formation constant in terms of the change due to the reaction progress:

Now, solving for :

The feasible solution is 0.15M which will lead to an equilibrium concentration of M⁺² of 0M

This fact has sense since the formation constant is very large.

Best regards.

the answer is 1 i believe

a