Question

The formation constant* of [M(CN)2]– is 5.30 × 1018, where M is a generic metal. A 0.170-mole quantity of M(NO3) is added to a liter of 0.540 M NaCN solution. What is the concentration of M ions at equilibrium?

Answer 1

MNO₃(aq) + 2NaCN <------> [M(CN)₂]^-(aq) ; K_f = 5.3*10¹⁸

K_f value being too large indicates that all the M⁺ ions are used up

Thus, CN^- left = 0.54 - 2*0.17 = 0.2 moles

Now, 5.3*10¹⁸ = [M(CN)₂^-]/{[MNO₃]*[NaCN]²}

or, 5.3*10¹⁸ = 0.17/(0.2²*[MNO₃]}

or, [MNO₃] = 8.02*10^-19 M = [M⁺]