Question

The formation constant* of [M(CN)4]2– is 7.70 × 1016, where M is a generic metal. A 0.170-mole quantity of M(NO3)2 is added to a liter of 0.930 M NaCN solution. What is the concentration of M2 ions at equilibrium?

Answer 1

[M (CN)_4]^2 - K_f = 7.70 times 10^16 formation constant is huge, therefore, its to assume that the reaction goes to completion. Write the formation equation. M^2 + + 4CN^- rightwardsharpoonoverleftwardsharpoon [M (CN)_4]^2- K_f = [[M(CN)_4]^2-]/[M^2 +][CN^-]^4 From 0.170 mole of M(NO_3)_2, we will get 0.170 mole of M^+2 From 0.930 mole of NaCN we will get 0.930 moles of CN^-. Since reaction went to completion. For every mole of M^+2 we need 4 CN^- Therefore, for 0.170 mole M^+2 rightarrow 4 times 0.17 CN^- = 0.68 mole CN^- Moles of CN^- left = 0.930 - 0.68 = 0.25 mole. Moles of [M (CN)_4]^2 formed = 0.170 mole. \r\n K_f = 0.170/[M^2 +] times (0.25)^4 7.7 times 10^16 = 0.170/[M^2 +](0.25)^4 [M^2 +] = 5.65 times 10^-16 which is almost negligible.