Question

The formation constant* of [M(CN)6]4– is 2.50 × 10^17, where M is a generic metal. A 0.160-mole quantity of M(NO3)2 is added to a liter of 1.230 M NaCN solution. What is the concentration of M2 ions at equilibrium?

Answer 1

The formation constant* of [M(CN)6]4– is 2.50 × 10^17, where M is a generic metal. A 0.160-mole quantity of M(NO3)2 is added to a liter of 1.230 M NaCN solution. What is the concentration of M2 ions at equilibrium?

Solution :-

M^2+ + 6CN- ------ > [M(CN)6]^2+

Kf= 2.50*10^17

Since the Kf is very large so all the metal ions form complex

Lets find the CN- moles needed to react with M^2+ ions

0.160 mol M^2+ * 6 mol CN-/1 mol M^2+ = 0.96 mol CN-

So the moles of CN- remain = 1.230 mol – 0.96 mol = 0.27M

So moles of [M(CN)6]^2+ that can be produced = 0.160 mol

So molarities of the [M(CN)6]^2+ = 0.160 mol / 1.0 L = 0.160 M

[CN-] = 0.27 mol / L = 0.27 M

Now lets write the dissociation reaction

[M(CN)6]^2+ -------- > M^2+   + 6CN-                               K = 1/ Kf

0.160 M                              0           0.27 M

-x                                        +x            +6x

0.160-x                                x          0.27+6x

Lets write the equilibrium equation

K= [M^2+] [CN-]/[M(CN)6]^2+

(1 / 2.50*10^17) = [x][0.27+6x]^6 /[0.160-x]

(1 / 2.50*10^17) * 0.160-x = [x][0.27+6x]^6

Solving for the x we get

X=1.65*10^-5

So the concentration of the M^2+ at equilibrium = x= 1.65*10^-15 M