Question

The formation constant* of [M(CN)6]4– is 2.50 × 1017, where M is a generic metal. A 0.160-mole quantity of M(NO3)2 is added to a liter of 1.240 M NaCN solution. What is the concentration of M2 ions at equilibrium?

Answer 1

1.33×10^-15M

Explanation

Consider the complete formation of [M(CN)6]^4-

  M²⁺ (aq) + 6CN^-(aq) ------> [M(CN)6]^4- (aq)

stoichiometrically, 1 mole of M²⁺ react with 6moles of CN^- to form 1mole of [M(CN)6]^4-

0.160M of M²⁺ reacts with 6× 0.160M = 0.960M of CN^- to give 0.160M of [M(CN)6]^4-

Remaining concentration of CN^- = 1.240M - 0.960M = 0.280M

Now, consider the dissociation equillibrium of [M(CN)6]^4-

[M(CN)6]^4-(aq) <-------> M²⁺(aq) + 6CN^-(aq)

K_d = [M²⁺] [CN^-]⁶/[M(CN)6]^4-]

K_d = 1/K_f = 1/ 2.50 ×10¹⁷ = 4.0 × 10^-18

Initial concentration

[[M(CN)6]^{​​​​​​4-}] = 0.160M

[M²⁺] = 0

[CN^-] = 0.280

change in concentration

[[M(CN)6]^4-] = -x

[M²⁺] = + x

[CN^-] = +6x

equillibrium concentration

[[M(CN)6]^4-] = 0.160 - x

[M²⁺] = x

[CN^-] = 0.280 + 6x

so,

x(0.280 + 6x)⁶ / (0.160 - x) = 4.0 × 10^-18

we can assume , 0.280 + 6x = 0.280 and 0.160 -x = 0.160 because x is small value

x(0.280)⁶/0.160 = 4.0 ×10^-18

x 0.003012= 4.0 × 10^-18

x = 1.33× 10^-15

Therefore, at equillibrium

[M²⁺] = 1.33×10^-15M