Question

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?

Answer 1

Answer:

0 M.

Explanation:

Hello,

In this case, the undergoing reaction is:

$M(NO_3)_2+NaCN\leftrightarrow [M(CN)_4]^{-2}+NaNO_3$

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

$M^{+2}+4CN^-\leftrightarrow [M(CN)_4]^{-2}$

In such a way, the formation constant turns out:

$K_F=\frac{[[M(CN)_4]^{-2}]_{eq}}{[M^{+2}]_{eq}[CN^{-}]_{eq}^4}$

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

$[M^{+2}]_0=0.150M;[CN^-]_0=0.820M$

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

$K_F=\frac{x}{(0.150-x)(0.820-x)^4}=7.70x10^{16}$

Now, solving for $x$:

$x_1=0.15M\\x_2=0.82M$

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

$[M^{+2}]_{eq}=0.15M-0.15M=0M$

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

Best regards.