Question

You mix a 150.0 −mL sample of a solution that is 0.0140 M in NiCl2 with a 175.0 −mL sample of a solution that is 0.500 M in NH3. After the solution reaches equilibrium, what concentration of Ni2+(aq) remains? The value of Kf for Ni(NH3)62+ is 2.0×108.

Answer 1

The value of Kf for Ni(NH3)62+ is 2.0×10^8

Kf = 2*10^8 .it is very large value

no of moles of NiCl2 = molarity *volume in L

                                   = 0.014*0.15 = 0.0021 moles

no of moles of NH3   = molarity *volume in L

                                   = 0.5*0.175   = 0.0875 moles

                Ni^2+ (aq) + 6NH3(aq) ------------>[ Ni(NH3)6]^2+ (aq)

at equilibrium Ni^2+ = x

at equilibrium NH3 = 0.0875-6*0.0021   = 0.0749 moles

at equilibrium [ Ni(NH3)6]^2+   = 0.0021moles

          Kf        =   [Ni(NH3)6]^2+/[Ni^2+][NH3]^6

          2*10^8 = 0.0021/x*(0.0749)^6

           x            = 0.0021/2*10^8*(0.0749)^6

                            = 5.95*10^-5

             [Ni^2+]   = x   = 5.95*10^-5M

plz nay doubt comment on the BOX