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 0.570 M NaCN solution. What is the concentration of M ions at equilibrium?

Answer 1

M(NO3) --------------> M^+ (aq) + NO3^-(aq)

0.17 Mole                  0.17 mole

no of moles of NaCN = molarity * volume in L

                                    = 0.57*1 = 0.57 moles

NaCN -----------> Na^+ (aq)+ CN^-

0.57 mole                               0.57mole

         M^+ (aq) + 2CN^- -----------> [M(CN)2]^-

I         0.17          0.57                      0

C       -0.17       -2*0.17                   0.17

E        0             0.23                      0.17

          x             0.23                        0.17

         Kf = [M(CN)2]^-/[M^+][CN^-]^2

      5.3*10^18   = 0.17/x*(0.23)^2

        x                = 0.17/5.3*10^18*(0.23)^2   = 6.06*10^-19

    [M^+]   = x   = 6.06*10^-19M