Question

In: Chemistry

You mix a 116.0 mL sample of a solution that is 0.0102 M in NiCl2 with...

You mix a 116.0 mL sample of a solution that is 0.0102 M in NiCl2 with a 165.0 mL sample of a solution that is 0.232 M in NH3. After the solution reaches equilibrium, what concentration of Ni2+(aq) remains? (Formation constant is Kf=2.0×108.)

Solutions

Expert Solution

no of moles of Ni^2+   = molarity * volume in L

                                   = 0.0102*0.116   = 0.0011832moles

no of moles of NH3     = molarity * volume in L

                                   = 0.232*0.165   = 0.03828 moles

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

I       0.0011832             0.03828                        0

C     -0.0011832            -6*0.0011832            0.0011832

E        0                           0.0311808               0.0011832

          x

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

      2*10^8 = 0.0011832/[Ni^2+](0.0311808)^6

[Ni^2+]          = 0.0011832/2*10^8(0.0311808)^6    

[NI^2+]         = 0.00643M


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