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 1.460 M NaCN solution.

What is the concentration of M2 ions at equilibrium?

Please post all of your work with it! I have tried 3 seperate times and have gotten the wrong answer each time. The first and third time I got 1.00 x 10^-20 and the second I got 4.00 x 10^-20.

Answer 1

The reaction taking place is

M²⁺ + 6 CN^- <====> [M(CN)₆]^4-

K_f = [M(CN)₆^4-]/[M²⁺][CN^-]⁶ = 2.50*10¹⁷

Since the formation constant is high, we can assume that practically all the M²⁺ ions added as M(NO₃)₂ will be complexed to form [M(CN)₆]^4-; therefore, moles of [M(CN)₆]^4- at equilibrium = 0.160 mole.

Since volume of the solution is 1 L, hence, [M(CN)₆^4-] = (0.160 mole)/(1 L) = 0.160 M

As per the balanced stoichiometric reaction above, 1 mole M²⁺ = 6 moles CN^-; therefore, 0.160 mole M²⁺ = (6*0.160) mole = 0.960 mole CN^-.

Since volume of the solution is 1 L, we must have moles of CN^- initially present = (1.460 mole/L)*(1 L) = 1.460 mole.

Moles of CN^- retained at equilibrium = (1.460 – 0.960) mole = 0.500 mole.

Since volume of the solution is still 1 L, the molar concentration of unreacted CN^- at equilibrium = (0.500 mole)/(1 L) = 0.500 M

Therefore, we must have

2.50*10¹⁷ = [M(CN)₆^4-]/[M²⁺][CN^-]⁶ = (0.160)/[M²⁺](0.500)⁶ = (0.160)/[M²⁺](0.015625)

====> [M²⁺] = (0.160)/(2.50*10¹⁷)(0.015625) = 4.096*10^-17

The molar concentration of M²⁺ at equilibrium = 4.096*10^-17 M ≈ 4.1*10^-17 M (ans).