Question

The formation constant* of [M(CN)2]– is 5.30 × 1018, where M is a generic metal. A 0.150-mole quantity of M(NO3) is added to a liter of 0.610 M NaCN solution. What is the concentration of M ions at equilibrium?

Answer 1

Given,

Number of moles of M(NO₃) = 0.150 mol

Concentration of NaCN = 0.610 M

Volume of NaCN solution = 1 L

Formation constant of [M(CN)₂]^- = 5.30 x 10¹⁸

Now, We have,

Moles of M(NO₃) = Moles of M⁺ = 0.150 mol

Calculating the number of moles of CN^- from the given concentration and volume of NaCN solution,

= 1 L x 0.610 M

= 0.610 mol of NaCN x ( 1 mol CN^- / 1 mol NaCN)

= 0.610 mol of CN^-

Now, the reaction between M⁺ and CN^- is,

M⁺(aq) + 2CN^-(aq) [M(CN)₂]^-(aq)

Drawing an ICE chart,

	M+(aq)	2CN-(aq)	[M(CN)2]-(aq)
I(moles)	0.150	0.610	0
C(moles)	-0.150	-2(0.150)	+0.150
E(moles)	0	0.31	0.150

Calculating the new concentrations of [CN^-] and [M(CN)₂]^-

[CN^-] = 0.31 mol / 1L = 0.31 M

[ [M(CN)₂]^-] = 0.150 mol / 1L = 0.150 M

Now, the dissociation equilbrium reaction for  [M(CN)₂]^-,

[M(CN)₂]^-(aq) M⁺(aq) + 2CN^-(aq)

Drawing an ICE chart,

	[M(CN)2]-	M+(aq)	2CN-(aq)
I(M)	0.150	0	0.31
C(M)	-x	+x	+2x
E(M)	0.150-x	x	0.31+2x

Now, the K_d expression is,

K_d = [CN^-] [M⁺] / [ [M(CN)₂]^-]

Calculating the value of K_d from the given value of K_f,

K_d = 1/K_f

K_d = 1/5.30 x 10¹⁸

K_d = 1.8868 x 10^-19

Now,

1.8868 x 10^-19 = [0.31+2x] [x] / [0.150-x]

1.8868 x 10^-19 = [0.31] [x] / [0.150] --------Here, [0.31+2x] 0.31 and [0.150-x] 0.150, since, x<<0.31 and 0.150

x = 9.13 x 10^-20

Thus, from the ICE chart,

[M⁺] ion at equilibrium = 9.13 x 10^-20 M