Question

You mix a 124.0 mL sample of a solution that is 0.0121 M in NiCl2 with a 183.5 mL sample of a solution that is 0.254 M in NH3.

Part A After the solution reaches equilibrium, what concentration of Ni2+(aq) remains? (Formation constant is Kf=2.0×108.) Express your answer with the appropriate units.

Answer 1

no of moles of NiCl2   = molarity * volume in L

                                    = 0.0121*0.124   = 0.0015moles

no of moles of NH3   = molarity * volume in L

                                   = 0.254*0.1835   = 0.0466moles

                 Ni^2+ +   4NH3 ------------>[ Ni(NH3)4]^+

   I            0.0015      0.0466                  0

   C           -0.0015   -4*0.0015             0.0015

   E            0             0.0406                  0.0015

           Kf   = [Ni(NH3)4]^+/[Ni^2+][NH3]^4

           2*10^8 = 0.0015/x*(0.0406)^4

           x           = 0.0015/2*10^8*(0.0406)^4   = 2.76*10^-6

         [Ni^2+]   =   x = 2.76*10^-6 M >>>>>answer