Question

The formation constant* of [M(CN)4]2– is 7.70 × 1016, where M is a generic metal. A 0.140-mole quantity of M(NO3)2 is added to a liter of 0.890 M NaCN solution. What is the concentration of M +2 ions at equilibrium?

Answer 1

1.53×10^-16 M

Explanation

i) Given moles of M²⁺ = 0.140

given moles of CN^- = (0.890mol/1000ml) ×1000ml = 0.890mol

ii) Consider the formation of M(CN)4^2- is complete

The reaction between M²⁺ and CN^- is as follows

M²⁺ (aq) + 4CN^-(aq) -------> M(CN)4^2- (aq)

stoixhiometricaly, 4moles of CN^- are react with 1moles of M²⁺

0.140moles of react with 4×0.140mol = 0.56mol of CN^-

remaining moles of CN^- = 0.890mol - 0.56mol = 0.33mol

moles of M(CN)4^2- formed = 0.140mol

[CN^-] = 0.330M

[M(CN)4^2-] = 0.140M

iii) consider the dissociation equillibrium of M(CN)4^2-

M(CN)4^2-(aq) <------> M²⁺(aq) + 4CN^-(aq)

K_d = [M²⁺][CN^-]⁴/[M(CN)4^2-]

K_d = 1/K_f = 1/7.70×10¹⁶

K_d= 1.30×10^-17

iv) Make ICE table for the dissociation equillibrium

concentration	[M(CN)42-]	[M2+]	[CN-]
Initial concentration	0.140	0	0.330
change in concentration	-x	+x	+4x
Equillibrium concentration	0.140 -x	x	0.330 + 4x

iv) x(0.330 + 4x)⁴/(0.140 - x) = 1.30×10^-17

we can assume 0.330 + 4x ~ 0.330 and 0.140 - x ~ 0.140

x(0.330)⁴ /0.140 = 1.30×10^-17

x = 1.53 ×10^-16

Therefore,

Concentration of M²⁺ at equillibrium = 1.53×10^-16M